# Uniform boundedness for finite Morse index solutions
to supercritical semilinear elliptic equations

Alessio Figalli, Yi Ru-Ya Zhang  ETH Zürich, Department of Mathematics,
Rämistrasse 101, 8092, Zürich, Switzerland<EMAIL_ADDRESS>Academy of Mathematics and Systems Science, the Chinese Academy of Sciences,
Beijing 100190, China<EMAIL_ADDRESS>
###### Abstract.

We consider finite Morse index solutions to semilinear elliptic questions, and
we investigate their smoothness. It is well-known that:
\- For $n=2$, there exist Morse index $1$ solutions whose $L^{\infty}$ norm
goes to infinity.
\- For $n\geq 3$, uniform boundedness holds in the subcritical case for power-
type nonlinearities, while for critical nonlinearities the boundedness of the
Morse index does not prevent blow-up in $L^{\infty}$.
In this paper, we investigate the case of general supercritical nonlinearities
inside convex domains, and we prove an interior a priori $L^{\infty}$ bound
for finite Morse index solution in the sharp dimensional range $3\leq n\leq
9$. As a corollary, we obtain uniform bounds for finite Morse index solutions
to the Gelfand problem constructed via the continuity method.

###### Key words and phrases:

Supercritical semilinear elliptic equations, finite Morse index, boundedness
of solutions, Gelfand problem.

###### 2000 Mathematics Subject Classification:

35J61, 35B65, 35B32, 35B35

Both authors have received funding from the European Research Council under
the Grant Agreement No. 721675 “Regularity and Stability in Partial
Differential Equations (RSPDE)”. The second author is also partially funded by
the Chinese Academy of Science and NSFC grant No. 11688101

##  1\. Introduction

Given $\Omega\subset\mathbb{R}^{n}$ a bounded domain, and
$f:\mathbb{R}\to\mathbb{R}$ a nonnegative $C^{1}$ function, we consider a
solution $u:\Omega\to\mathbb{R}$ to the following semilinear equation

$\left\\{\begin{array}[]{ll}-\Delta u=f(u)&\text{ in }\Omega,\\\ u=0&\text{ on
}\partial\Omega.\end{array}\right.$ (1.1)

Note that, by the nonnegativity of $f$ and the maximum principle, $u>0$ inside
$\Omega$ (unless $u\equiv 0$).

Set $F(t):=\int_{0}^{t}f(s)\,ds$. Then (1.1) corresponds to the Euler-Lagrange
equation for the energy functional

$\mathcal{E}[u]:=\int_{\Omega}\Bigl{(}\frac{|\nabla
u|^{2}}{2}-F(u)\Bigr{)}\,dx.$

Consider the second variation of $\mathcal{E}$, that is,

$\frac{d^{2}}{d\epsilon^{2}}\Big{|}_{\epsilon=0}\mathcal{E}[u+\epsilon\xi]=\int_{\Omega}\Bigl{(}|\nabla\xi|^{2}-f^{\prime}(u)\xi^{2}\Bigr{)}\,dx.$

Given a subdomain $\Omega^{\prime}\subseteq\Omega$ and $k\in\mathbb{N}$, $u$
is said to have finite Morse index $k$ in $\Omega^{\prime}$, and we write
${\rm ind}(u,\Omega^{\prime})=k$, if $k$ is the maximal dimension of a
subspace $X_{k}\subset C^{1}_{c}(\Omega^{\prime})$ such that, for any $\xi\in
X_{k}\setminus\\{0\\}$,

$Q_{u}(\xi):=\int_{\Omega^{\prime}}\Big{(}|\nabla\xi|^{2}-f^{\prime}(u)\xi^{2}\Big{)}\,dx<0.$

Also, $u$ is said to be stable in $\Omega^{\prime}$ if ${\rm
ind}(u,\Omega^{\prime})=0$ (that is, $Q_{u}(\xi)\geq 0$ for all $\xi\in
C^{1}_{c}(\Omega^{\prime})$).

###  1.1. Finite Morse index vs uniform boundedness

The idea of using a bound on the Morse index to characterize the uniform
boundedness of a solution to a semilinear elliptic equation was first
introduced in the seminal paper [1]. In this work, as well in several other
subsequent papers (see for instance [27, 21]), the authors considered
(variants of) the subcritical case, namely

$f(t)\simeq(\alpha+t)^{\frac{n+2}{n-2}-\epsilon},\qquad\alpha,\epsilon>0,$

and they proved that the boundedness of solutions is equivalent to the
boundedness of the Morse index.111Note however that this result does not cover
the full subcritical case: as shown in [22, Lemma 5 and Theorem 1(iii))], for
$f_{\lambda}(t)=\lambda(1+t)^{p}$, $\lambda>0$ and $p<\frac{n+2}{n-2}$, there
exists a family $u_{\lambda}$ of solutions with Morse index $1$ with
$\|u_{\lambda}\|_{L^{\infty}(B_{1})}\to\infty$ as $\lambda\to 0$. In other
words, in the subcritical case, both upper and lower bounds are needed on $f$
in order to show the equivalence between boundedness in $L^{\infty}$ and
boundedness of the Morse index.

In the critical case, namely

$f(u)=(\alpha+u)^{\frac{n+2}{n-2}},\qquad\alpha>0,$

the finiteness of the Morse index does not imply the boundedness of the
solutions. Indeed it is not difficult to check that the functions

$u_{\alpha,\mu}(x)=\biggl{(}\bigg{(}\frac{\mu\sqrt{n(n-2)}}{\mu^{2}+|x|^{2}}\bigg{)}^{\frac{n-2}{2}}-\alpha\biggr{)}_{+},\
\mu>0$ (1.2)

are solutions with Morse index $1$. In particular, choosing
$\alpha_{\mu}:=\left(\frac{\mu\sqrt{n(n-2)}}{1+\mu^{2}}\right)^{\frac{n-2}{2}}$
so that $u_{\alpha_{\mu},\mu}=0$ on $\partial B_{1}$, and letting $\mu\to 0$,
one can construct a family Morse index $1$ solutions in $B_{1}$ whose
$L^{\infty}$ norm goes to infinity (see, e.g., [8]).

The supercritical case, instead, is much less understood. The special case
where $f$ is a polynomial or an exponential function has been studied in [16]
and [11], respectively. There, the uniform boundedness of solutions is
obtained by proving suitable Liouville-type results. Unfortunately, this
approach does not seem suitable for general nonlinearities.

We finally mention a recent result [15], where the authors investigate the
regularity and symmetry properties of finite Morse index solutions.

###  1.2. Main result: finite Morse index solutions with supercritical
nonlinearities are uniformly bounded

Very recently, in [9] the authors investigated the properties of stable
solutions for all nonlinearities, and they proved uniform boundedness when
$3\leq n\leq 9$,222 In the stable case, the case $n=2$ is a consequence of the
case $n=3$ by noticing that a stable solution in two dimensions is also stable
in three dimensions (by looking at it as a function constant in the third
variable). This is why the results in [9] hold for $n\leq 9$. This is not the
case anymore when ${\rm ind}(u)>0$, and indeed a change of behavior of finite
Morse index solutions between dimension $n=2$ and dimension $3\leq n\leq 9$
was already observed in [22]. In particular, as [22, Fig. 1 b, pag. 245]
shows, there exists a curve of two-dimensional solutions with Morse index $1$
that blows-up in $L^{\infty}$. and interior $W^{1,2}$ estimates in all
dimensions.

In this paper, we exploit these result to develop a series of new tools for
finite Morse index solutions (cf. Section 1.3.3 below) that allow us to prove
a universal $L^{\infty}$ bound for solutions to (1.1) when $f$ grows
superlinearly in a suitably quantified way.333Our quantitative superlinarity
assumption (1.4) already appeared in the paper [26] (see also [13]), where the
author proved the uniqueness of solutions to (1.3) for small values of
$\lambda$. As common in these problems, we assume that $f(0)>0$ (actually, we
quantify this assumption by asking that $f(0)\geq c_{0}>0$, so to better
emphasize the dependences in our $L^{\infty}$ bound). This assumption is
particularly natural in the superlinear case, since the Derrick-Pohozaev
identity prevents the existence of nontrivial solutions (see [17, Theorem 1,
Page 515]).

Actually, because of applications to the Gelfand problem described in Section
1.4 below, it will be convenient to prove a more robust result that
establishes a uniform bound whenever the nonlinearity is of the form $\lambda
f$, where $\lambda\in[0,\hat{\lambda}]$ for some fixed $\hat{\lambda}$. Also,
for the sake of generality, it is interesting to observe how the bound depends
on $f$. So, instead of considering a fixed nonlinearity $f$, we assume that
$f$ belongs to a locally compact $C^{1}$ family. As shown in Section 1.3.2
below, this assumption can be considerably weakened if $f$ is assumed to be
convex.

######  Theorem 1.1.

Let $3\leq n\leq 9$, $\Omega\subset\mathbb{R}^{n}$ a bounded convex domain,
and $c_{0}>0$. Consider

$\mathcal{K}\subset\\{h\in C^{1}(\mathbb{R}):\,h\geq 0,\,h^{\prime}\geq
0,h(0)\geq c_{0}\\},$

and assume that $\mathcal{K}$ is compact for the $C^{1}_{\rm loc}(\mathbb{R})$
topology. Let $\hat{\lambda}>0$, and let $u\in C^{2}(\Omega)$ solve

$\left\\{\begin{array}[]{ll}-\Delta u=\lambda f(u)&\text{ in }\Omega,\\\
u=0&\text{ on }\partial\Omega,\end{array}\right.$ (1.3)

for some $f\in\mathcal{K}$ and $\lambda\in[0,\hat{\lambda}]$. Finally, assume
that ${\rm ind}(u,\Omega)\leq k$ and that there exist $\epsilon,t_{0}>0$ such
that

$f(t)t\geq\left(\frac{2n}{n-2}+\epsilon\right)F(t)\quad\text{ for all }\ t\geq
t_{0},$ (1.4)

where $F(t):=\int_{0}^{t}f(s)\,ds$. Then

$\|u\|_{L^{\infty}(\Omega)}\leq
C(n,k,\mathcal{K},\hat{\lambda},\epsilon,t_{0},\Omega).$

######  Remark 1.2.

We observe that, as a consequence of (1.4), it holds

$f(t)\geq c_{1}t^{\frac{n+2}{n-2}+\epsilon}\qquad\forall\,t\geq 0,$ (1.5)

with $c_{1}:=f(0)t_{0}^{-\frac{n+2}{n-2}-\epsilon}$.

Indeed, (1.4) can be rewritten as

$F^{\prime}(t)\geq\frac{\left(\frac{2n}{n-2}+\epsilon\right)}{t}F(t)\quad\text{
for all }\ t\geq t_{0},$

so it follows from Grönwall inequality that

$F(t)\geq F(t_{0})\left(\frac{t}{t_{0}}\right)^{\frac{2n}{n-2}+\epsilon}.$

Inserting this information in (1.4), we get

$f(t)\geq\left(\frac{2n}{n-2}+\epsilon\right)F(t_{0})t_{0}^{-\frac{2n}{n-2}-\epsilon}t^{\frac{n+2}{n-2}+\epsilon}\quad\text{
for all }\ t\geq t_{0}.$

Also, since $f$ is increasing we have $F(t_{0})\geq f(0)t_{0},$ and therefore

$f(t)\geq\left\\{\begin{array}[]{cc}f(0)\left(\frac{2n}{n-2}+\epsilon\right)t_{0}^{-\frac{n+2}{n-2}-\epsilon}t^{\frac{n+2}{n-2}+\epsilon}&\text{
for}\ t\geq t_{0}\\\ f(0)&\text{ for }\ 0\leq t<t_{0},\end{array}\right.$

which implies (1.5).

######  Remark 1.3.

As mentioned before, the dimensional range $3\leq n\leq 9$ follows from [9,
Theorem 1.2], since boundedness of stable solutions for all nonlinearities is
true only under this assumption. However, for some particular choices of
nonlinearities (e.g., $f(u)=(1+u)^{p}$ for suitable values of $p$), we believe
that our ideas and techniques could be applied also in higher dimension (cf.
[12]).

###  1.3. About Theorem 1.1: extensions and tools used in the proof

We first discuss some possible extensions and generalizations of Theorem 1.1,
and then we briefly present the three key ingredients behind its proof.

####  1.3.1. On the convexity of $\Omega$

The convexity assumption on $\Omega$ in Theorem 1.1 allows us:
\- to focus only on interior regularity, since the regularity near the
boundary is handled via the moving plane method, see Lemma 2.8 below;
\- to apply the classical Derrick-Pohozaev argument on convex domains, see the
argument after (3.9).

We believe that a nontrivial modification of our techniques could be used to
analyze the boundary behavior inside general smooth domains. However our proof
strongly relies on the Derrick-Pohozaev argument, and this requires $\Omega$
to be at least star-shaped (see, e.g., [17, Theorem 1, Page 515]). Hence, it
looks likely to us that by combining the ideas developed in this paper with
the boundary regularity from [9], one should be able to extend Theorem 1.1 to
(sufficiently smooth) star-shaped domains.

####  1.3.2. A result for convex nonlinearities

The assumption that the nonlinearity $f$ belongs to a family $\mathcal{K}$
that is compact for the $C^{1}_{\rm loc}(\mathbb{R})$ topology can be removed,
if one assumes the nonlinearities to be convex and to be dominated by a fixed
continuous nonnegative function $g:\mathbb{R}\to\mathbb{R}$. More precisely,
if $f$ is a convex function such that $0\leq f\leq g$ and (1.4) holds, then
$\|u\|_{L^{\infty}(\Omega)}\leq C(n,k,g,\hat{\lambda},\epsilon,t_{0},\Omega).$

Indeed, the compactness assumption in $C^{1}_{\rm loc}$ is used only to apply
Proposition 2.3. If all the nonlinearities are convex, then the bound $0\leq
f\leq g$ guarantees compactness in $C^{0}_{\rm loc}$. Therefore, one only
needs to check that Proposition 2.3 holds if $f_{j}$ are convex functions
satisfying $f_{j}\to f_{\infty}$ in $C^{0}_{\rm loc}(\mathbb{R})$. This can be
done by suitable adapting the notion of stability for convex functions,
defining

$Q_{u}(\xi):=\int_{\Omega^{\prime}}\Big{(}|\nabla\xi|^{2}-f_{-}^{\prime}(u)\xi^{2}\Big{)}\,dx\qquad\text{with
$f^{\prime}_{-}(t):=\lim_{\tau\to
0^{+}}\frac{f(t)-f(t-\tau)}{\tau}=\sup_{\tau>0}\frac{f(t)-f(t-\tau)}{\tau}$}.$

Indeed, with this definition, the results from [9] still apply. In addition,
the following lower semicontinuity property holds:

$t_{j}\to t,\quad f_{j}\to f\text{ in }C^{0}_{\rm
loc}(\mathbb{R})\qquad\Rightarrow\qquad
f^{\prime}_{-}(t)\leq\liminf_{j\to\infty}(f_{j})_{-}^{\prime}(t_{j}),$

and this allows one to show that upper bounds on the Morse index are
preserved. We leave the details to the interested reader.

####  1.3.3. Main tools

As mentioned before, the proof of Theorem 1.1 is based on a series of new
important results for finite Morse index solutions. These are:
(i) A general stability result for bounded Morse index solutions stating that,
for $3\leq n\leq 9$, these families are weakly compact in $W^{1,2}$ and they
converge in $C^{2}_{\rm loc}$ outside finitely many points (see Proposition
2.3). This result relies on the smoothness of stable solutions for $n\leq 9$
obtained in [9], and on a slight improvement of it proved in Appendix A.
(ii) A uniform $W^{1,2}$ integrability estimate for finite Morse index
solutions (see Proposition 2.6). This result depends both on the
supercriticality assumption (1.4) and on the interior $W^{1,2}$ estimates for
stable solutions, cf. [9].
(iii) A $\varepsilon$-regularity theorem for finite Morse index solutions
stating that, if the $W^{1,2}$ norm of a solution inside a ball $B_{r}$ decays
sufficiently fast for $r\in[\varepsilon,1]$ with $\varepsilon\ll 1$, then it
decays all the way to the origin (see Proposition 2.7).
It is worth observing that while (i) needs the dimensional restriction $n\leq
9$, both (ii) and (iii) hold in every dimension. Besides playing a crucial
role in proving Theorem 1.1, we believe that these results have their own
interest.

###  1.4. An application to the Gelfand problem associated to analytic
supercritical nonlinearities

Given $f:\mathbb{R}\to\mathbb{R}$ nonnegative and increasing, and $\lambda\geq
0,$ the so-called Gelfand problem for $f$ consists in studying the nonlinear
elliptic problem

$\left\\{\begin{array}[]{ll}-\Delta u=\lambda f(u)&\text{ in }\Omega,\\\
u=0&\text{ on }\partial\Omega.\end{array}\right.$

This problem has a long history: it was first presented by Barenblatt in a
volume edited by Gelfand [18], and a series of authors studied it later, in
particular in the range where $u$ is stable; we refer the interested reader to
[2, 3, 4, 14, 6] for a complete account on this topic.

In this paper we want to study the solution curve associated to the Gelfand
problem: we look for a continuous curve $\lambda\colon[0,\infty)\to[0,\infty)$
with $\lambda(0)=0,$ and for a one-parameter family of solutions
$\\{u_{s}\\}_{s\geq 0}$, such that

$\left\\{\begin{array}[]{ll}-\Delta u_{s}=\lambda(s)f(u_{s})&\text{ in
}\Omega,\\\ u_{s}=0&\text{ on }\partial\Omega.\end{array}\right.$

When $\Omega=B_{1}$, the cases $f(t)=(1+\alpha t)^{\beta},\,\alpha,\beta>0,$
and $f(t)=e^{t}$, have been fully understood in [22] via ODE methods. In
particular, when $3\leq n\leq 9$, the authors proved that there are infinitely
many turning points in the solution curve
$s\mapsto(\lambda(s),\|u_{s}\|_{L^{\infty}(\Omega)},)$ for suitable values of
$\beta$. Later, similar phenomena were observed for special functions $f$ or
in low dimensional domains with suitable symmetries (see, e.g., [24, 10, 20,
11, 23] and the reference therein).

Assume now that $\Omega$ is a convex set of class $C^{3}$, let
$C^{1}_{0}(\overline{\Omega})$ denote the Banach space of $C^{1}$ functions on
$\overline{\Omega}$ that vanish on $\partial\Omega$, and consider the
following open subset of $C^{1}_{0}(\overline{\Omega})$ endowed with the
$C^{1}$ topology:

$\mathcal{O}:=\\{u\in C^{1}_{0}(\overline{\Omega})\,:\,u>0\text{ in
}\Omega,\,\partial_{\nu}u|_{\partial\Omega}<0\\},$ (1.6)

where $\nu$ denotes the outer unit normal to $\partial\Omega$. Following [10],
assume the map

$\mathcal{O}\ni u\mapsto f(u)\in C^{0}(\overline{\Omega})$

to be real analytic (as noted in [10], this is the case for instance if $f$ is
analytic). Then, thanks to our Theorem 1.1, one can apply the global analytic
bifurcation theory developed in [5, Section 2.1] to show the existence of a
piecewise analytic continuous curve $[0,\infty)\ni
s\mapsto(\lambda(s),u_{s})$, with $(\lambda(0),u_{0})=(0,0)$, such that both
$\|u_{s}\|_{L^{\infty}(\Omega)}$ and ${\rm ind}(u_{s},\Omega)$ tend to
infinity as $s\to\infty$. Moreover there exists a sequence
$(\lambda(s_{i}),u_{s_{i}})$ such that
$\|u_{s_{i}}\|_{L^{\infty}(\Omega)}\to\infty$ and each point of this sequence
is either a bifurcation or a turning point. This is a complete statement:

######  Theorem 1.4.

Let $3\leq n\leq 9$, $\Omega\subset\mathbb{R}^{n}$ a bounded convex domain of
class $C^{3}$, and $f>0$ an increasing analytic function satisfying (1.4). Let

$\mathcal{S}:=\\{(\lambda,u)\in\mathbb{R}_{+}\times
C_{0}^{1}(\overline{\Omega})\colon-\Delta u-\lambda f(u)=0\,\text{ and
}\,-\Delta-\lambda f^{\prime}(u)\text{ is invertible with bounded
inverse}\\}.$

Then there exist two continuous mappings

$h_{1}\colon\mathbb{R}_{+}\to\mathbb{R}_{+},\qquad
h_{2}\colon\mathbb{R}_{+}\to C_{0}^{1}(\overline{\Omega}),$

so that, denoting $h=(h_{1},\,h_{2})$, we have:

  1. (i)

$h\colon\mathbb{R}_{+}\to\overline{\mathcal{S}}$ and $\lim_{s\to\infty}{\rm
ind}(h_{2}(s),\Omega)=\infty.$

  2. (ii)

$h$ is injective on $h^{-1}(\mathcal{S})$ with $h_{1}^{\prime}(s)\neq 0$, and
real analytic at all points $s\in h^{-1}(\mathcal{S})$.

  3. (iii)

The set $h^{-1}(\overline{\mathcal{S}}\setminus\mathcal{S})$ consists of
isolated values.

  4. (iv)

For every point $s_{0}\in h^{-1}(\overline{\mathcal{S}}\setminus\mathcal{S})$
there exists an injective and continuous reparameterization
$s=\gamma(\sigma)$, $\sigma\in[-1,1]$, such that $s_{0}=\gamma(0)$ and
$h\circ\gamma$ is a real analytic function whose derivatives might only vanish
at $0$.

  5. (v)

There are infinitely many values of $s>0$ where
$h(s)\in\overline{\mathcal{S}}\setminus\mathcal{S}$ is either a bifurcation or
a turning point. Namely, either in every neighborhood of $h(s)$ there exists a
solution of (1.1) which is not in the image of $h$ (and then $h(s)$ is a
bifurcation point), or the previous case do not happen but $h_{1}$ is not
locally injective (and then $h(s)$ is a turning point).

###### Proof.

Let $\mathcal{O}$ be as in (1.6), and define the analytic map

$\mathcal{F}\colon\mathbb{R}_{+}\times\mathcal{O}\to
C^{1}_{0}(\overline{\Omega}),\qquad\mathcal{F}(\lambda,\,u):=-u+\lambda\mathcal{A}(u),$

where $\mathcal{A}(u):=(-\Delta)^{-1}[f(u)]$, and $(-\Delta)^{-1}$ denotes the
inverse of the Dirichlet Laplacian in $\Omega$. Arguing exactly as in the
proof of [10, Theorem 1] (see also the remark after the statement of the
theorem), the result follows from [5, Section 2.1]. ∎

######  Remark 1.5.

It was pointed out in [10, Remark 4] that, when $\Omega$ is a $C^{3}$ strongly
convex domain with certain symmetries, a careful modification of [25] gives
that, the image of $h$ is a smooth curve with only infinitely many turning
points but not bifurcation points. Also, this property in generic in a
neighborhood of such domains. We expect a similar result to hold also in our
setting.

###  1.5. Structure of the paper

The paper is organized as follows. In Section 2 we present a series of results
on finite Morse index solutions, which will be crucial for proving Theorem
1.1. Then, in Section 3 we prove Theorem 1.1. Finally, in a first appendix, we
show that [9, Theorem 1.2] holds also for $W^{1,2}$ stable solution that are
$C^{2}$ outside one point. This result is used in the proof of Proposition
2.3. Then, in a second appendix, we describe how the method in [9] implies
uniform boundedness of solutions whenever the spectrum of
$-\Delta-f^{\prime}(u)$ is bounded from below.

Acknowledgments. The authors are grateful to Xavier Cabré and Alberto Farina
for useful comments on a preliminary version of this manuscript.

##  2\. Technical tools on finite Morse index solutions

Let us fix some notation. For $x\in\mathbb{R}^{n}$ and $r>0$, we denote by
$B_{r}(x)$ the Euclidean ball centered at $x$ with radius $r$. The center is
usually omitted when $x$ is the origin. By $\alpha B$ we mean the ball with
the same center as $B$ but $\alpha$ times its radius. We write constants as
positive real numbers $C(\cdot)$, with the parentheses including all the
parameters on which the constants depend. We note that $C(\cdot)$ may vary
between appearances, even within a chain of inequalities. Sometimes we use
$C_{n},c_{n}$ to emphasize that a constant depends only on the dimension.

The goal of this section is to prove several new important results on finite
Morse index solutions that will be used in the next section to prove Theorem
1.1. First, we need to introduce a notion of weak solution with bounded Morse
index.

######  Definition 2.1.

Let ${\mathcal{U}}\subset\mathbb{R}^{n}$ be an open set, and let
$f:\mathbb{R}\to\mathbb{R}$ be nonnegative. We say that $u\in W^{1,2}_{\rm
loc}({\mathcal{U}})$ is a weak solution of $-\Delta u=f(u)$ in ${\mathcal{U}}$
if $f(u)\in L^{1}_{\rm loc}({\mathcal{U}})$ and

$\int_{{\mathcal{U}}}\nabla
u\cdot\nabla\varphi\,dx=\int_{{\mathcal{U}}}f(u)\,\varphi\,dx\qquad\forall\,\varphi\in
C^{1}_{c}({\mathcal{U}}).$

Assume in addition that $f$ is of class $C^{1}$. Then we say that $u$ has
finite Morse index $k\in\mathbb{N}$ in ${\mathcal{U}}$, and we write ${\rm
ind}(u,{\mathcal{U}})=k$, if $f^{\prime}(u)\in L^{1}_{\rm loc}({\mathcal{U}})$
and $k$ is the maximal dimension of a subspace $X_{k}\subset
C^{1}_{c}({\mathcal{U}})$ such that

$Q_{u}(\xi):=\int_{{\mathcal{U}}}\Big{(}|\nabla\xi|^{2}-f^{\prime}(u)\xi^{2}\Big{)}\,dx<0\qquad\forall\,\xi\in
X_{k}\setminus\\{0\\}.$

As we shall see below, whenever $f^{\prime}\geq 0$ it is possible to prove an
a priori bound on the $L^{1}_{\rm loc}$ norm of $f^{\prime}(u)$ in terms of
the Morse index. Then, by Fatou’s Lemma, this a priori bound holds for all
weak solutions that are limits of smooth solutions (see Proposition 2.3
below).

######  Lemma 2.2.

Let ${\mathcal{U}}\subset\mathbb{R}^{n}$ be an open set, let
$f:\mathbb{R}\to\mathbb{R}$ be nonnegative, increasing, and of class $C^{1}$,
and let $u\in W^{1,2}_{\rm loc}({\mathcal{U}})$ be a weak solution of $-\Delta
u=f(u)$ in ${\mathcal{U}}$ with ${\rm ind}(u,{\mathcal{U}})\leq k$. Then:

  * (i)

If $\\{{\mathcal{U}}_{i}\\}_{i=1}^{k+1}$ is a disjoint family of open subsets
of ${\mathcal{U}}$, then $u$ is stable in at least one set
${\mathcal{U}}_{i}$.

  * (ii)

The following uniform bound holds:

$\int_{B_{r}(\bar{x})}f^{\prime}(u)\,dx\leq
C_{n}(1+k)^{\frac{2}{n}}r^{n-2}\qquad\forall\,B_{2r(\bar{x})}\subset{\mathcal{U}}.$
(2.1)

###### Proof.

To prove (i) we note that, if by contradiction $u$ was unstable inside each
set ${\mathcal{U}}_{i}$, then there would exist functions $\xi_{i}\in
C^{1}_{c}({\mathcal{U}}_{i})$ such that

$\int_{{\mathcal{U}}}\left(|\nabla\xi_{i}|^{2}-f^{\prime}(u)\xi_{i}^{2}\right)\,dx<0.$

Since the functions $\\{\xi_{i}\\}_{i=1}^{k+1}$ have disjoint support, this
implies that

$\int_{{\mathcal{U}}}\left(|\nabla\xi|^{2}-f^{\prime}(u)\xi^{2}\right)\,dx<0\qquad\forall\,\xi\in{\rm
Span}(\xi_{1},\ldots,\xi_{k+1})\setminus\\{0\\},$

therefore ${\rm ind}(u,{\mathcal{U}})\geq k+1$, a contradiction.

We now prove (ii), following the ideas in [19, Theorem 5.9]. Given an open set
$\mathcal{O}$ and a pair of sets $E,\,F\subset\mathcal{O}$, the $p$-capacity
between $E$ and $F$ inside $\mathcal{O}$ for $p>1$ is defined as

${\rm
Cap}_{p}(E,\,F,\,\mathcal{O})=\inf\\{\|v\|^{p}_{W^{1,\,p}(\mathcal{O})}:\
v\in\Delta(E,\,F)\\},$

where $\Delta(E,\,F)$ denotes the class of all functions $v\in
W^{1,\,p}(\mathcal{O})$ that are continuous in $\mathcal{O}$ and satisfy $v=1$
on $E$, and $v=0$ on $F$. In particular, if $A=B_{\rho}(z)\setminus
B_{\rho/2}(z)$ is an annulus such that $2A:=B_{2\rho}(z)\setminus
B_{\rho/4}(z)$ is contained inside $B_{2r}(\bar{x})$, to control ${\rm
Cap}_{2}(A,\partial(2A),B_{2r}(\bar{x}))$ we can choose the function
$v_{z,\rho}(x):=\min\big{\\{}1,(4\rho^{-1}|x-z|-1)_{+},(2-\rho^{-1}|x-z|)_{+}\big{\\}}$
to obtain

${\rm
Cap}_{2}(A,\,\partial(2A),\,B_{2r}(\bar{x}))\leq\|v_{z,\rho}\|_{W^{1,2}}=C(n)|A|^{1-\frac{2}{n}}.$
(2.2)

Let us now consider the metric space $X:=\overline{B_{2r}(\bar{x})}$ endowed
with the Euclidean metric. In this space, we call “$X$-annuli” sets of the
form $\left(B_{\rho}(z)\setminus
B_{\rho/2}(z)\right)\cap\overline{B_{2r}(\bar{x})}$ for some
$z\in\overline{B_{2r}(\bar{x})}$.

Define the measure on $X$ given by
$\sigma:=\chi_{B_{r}(\bar{x})}f^{\prime}(u)dx$, and let $\kappa\in\mathbb{N}$
be a large constant to be fixed later. Since $\sigma$ has no atoms, we can
apply [19, Theorem 1.1] to deduce the existence of a family of Euclidean
annuli $\\{A_{i}:=B_{\rho_{i}}(z_{i})\setminus
B_{\rho_{i}/2}(z_{i})\\}_{i=1}^{\kappa}$, with $z_{i}\in B_{2r}(\bar{x})$,
such that

$\int_{B_{r}(\bar{x})}f^{\prime}(u)\,dx\leq
C_{0}(n)\,\kappa\int_{B_{r}(\bar{x})\cap
A_{i}}f^{\prime}(u)\,dx\qquad\forall\,i=1,\ldots,k+1$ (2.3)

and $\\{(2A_{i})\cap B_{2r}(\bar{x})\\}_{i=1}^{\kappa}$ are pairwise disjoint.

With no loss of generality we can assume that $B_{r}(\bar{x})\cap
A_{i}\neq\emptyset$ (otherwise (2.3) would imply that
$\int_{B_{r}(\bar{x})}f^{\prime}(u)\,dx=0$ and the result would be trivially
true). Let us split these annuli into two families: if $\rho_{i}<r/4$ then we
say that $i\in\mathcal{I}_{1}$, otherwise we say that $i\in\mathcal{I}_{2}$.

Note that, since $B_{r}(\bar{x})\cap A_{i}\neq\emptyset$, for
$i\in\mathcal{I}_{2}$ it holds $(2A_{i})\cap B_{2r}(\bar{x})\geq c_{n}r^{n}$
for some dimensional constant $c_{n}>0$. Hence, since the sets
$\\{(2A_{i})\cap B_{2r}(\bar{x})\\}_{i\in\mathcal{I}_{2}}$ are disjoint, we
deduce that $\\#\mathcal{I}_{2}\leq N_{n}$ for some dimensional constant
$N_{n}\geq 1$.

On the other hand, when $i\in\mathcal{I}_{1}$, since $B_{r}(\bar{x})\cap
A_{i}\neq\emptyset$ and $\rho_{i}<r/4$ it follows that $(2A_{i})\cap
B_{2r}(\bar{x})=2A_{i}$, hence the sets $\\{2A_{i}\\}_{i\in\mathcal{I}_{1}}$
are pairwise disjoint. Also, it follows from (2.2) that

${\rm Cap}_{2}(A_{i},\,\partial(2A_{i}),\,B_{2r}(\bar{x}))\leq
C(n)|A_{i}|^{1-\frac{2}{n}}.$ (2.4)

Now, fix $\kappa:=N_{n}+2(k+1)$ so that $\\#\mathcal{I}_{1}\geq 2(k+1)$. Since
the sets $\\{2A_{i}\\}_{i\in\mathcal{I}_{1}}$ are pairwise disjoint and
contained inside $B_{2r}(\bar{x})$, there exists a subset of indices
$\mathcal{I}_{1}^{\prime}\subset\mathcal{I}_{1}$ such that
$\\#\mathcal{I}_{1}^{\prime}\geq k+1$ and

$|2A_{i}|\leq\frac{1}{k+1}|B_{2r}|\qquad\forall\,i\in\mathcal{I}_{1}^{\prime},$

that combined with (2.4) gives

${\rm Cap}_{2}(A_{i},\,\partial(2A_{i}),\,B_{2r}(\bar{x}))\leq
C_{1}(n)(1+k)^{\frac{2}{n}-1}r^{n-2}\qquad\forall\,i\in\mathcal{I}_{1}^{\prime}.$
(2.5)

Now, assume by contradiction that (2.1) does not hold with
$C_{n}=4C_{0}C_{1}(N_{n}+1)$, namely

$\int_{B_{r}(\bar{x})}f^{\prime}(u)>C_{n}(1+k)^{\frac{2}{n}}r^{n-2},$ (2.6)

where $C_{0}$ and $C_{1}$ are as in (2.3) and (2.5). Then, since
$2(N_{n}+1)(k+1)\geq\kappa$, combining (2.6), (2.5), and (2.3), we get

${\rm
Cap}_{2}(A_{i},\,\partial(2A_{i}),\,B_{2r}(\bar{x}))<(2C_{0}\kappa)^{-1}\int_{B_{r}(\bar{x})}f^{\prime}(u)\,dx\leq\frac{1}{2}\int_{A_{i}}f^{\prime}(u)\,dx\qquad\forall\,i\in\mathcal{I}_{1}^{\prime}.$

Choose functions $\xi_{i}\in C^{1}_{c}(2A_{i})$ that almost minimize the
capacity ${\rm Cap}_{2}(A_{i},\,\partial(2A_{i}),\,B_{2r}(\bar{x}))$, so that

$\int_{B_{2r}(\bar{x})}|\nabla\xi_{i}|^{2}\,dx\leq\frac{2}{3}\int_{A_{i}}f^{\prime}(u)\,dx\leq\frac{2}{3}\int_{B_{2r}(\bar{x})}f^{\prime}(u)\xi_{i}^{2}\,dx\qquad\forall\,i\in\mathcal{I}_{1}^{\prime}.$

Since the sets $\\{2A_{i}\\}_{i\in\mathcal{I}_{1}^{\prime}}$ are pairwise
disjoint and $\\#\mathcal{I}_{1}^{\prime}\geq k+1$, we conclude that
$\\{\xi_{i}\\}_{i\in\mathcal{I}_{1}^{\prime}}$ spans a $(k+1)$-dimensional
subspace of $C^{1}_{c}(B_{2r}(\bar{x}))$ where the stability inequality fails.
This contradicts ${\rm ind}(u,B_{2r}(\bar{x}))\leq k$ and concludes the proof.
∎

We now prove a crucial convergence result for weak $W^{1,2}$ limits of smooth
solutions with bounded Morse index. Note that, a consequence of Proposition
2.3 below, limit of smooth solutions with bounded Morse index are still
smooth. However the result does not provide any uniform bound on the sequence
$u_{j}$. In particular, it could be that $\|u_{j}\|_{L^{\infty}}\to\infty$, as
the example provided by (1.2) shows.

######  Proposition 2.3.

Let $n\leq 9$, ${\mathcal{U}}\subset\mathbb{R}^{n}$ an open set, and $u_{j}\in
C^{2}({\mathcal{U}})$ a sequence of functions satisfying

$-\Delta u_{j}=f_{j}(u_{j})\qquad\text{ in }{\mathcal{U}}$

with $f_{j}:\mathbb{R}\to\mathbb{R}$ nonnegative, increasing, and of class
$C^{1}$. Assume that

${\rm ind}(u_{j},{\mathcal{U}})\leq k$ for some $k\in\mathbb{N}$,
$\sup_{j}\|u_{j}\|_{W^{1,2}({\mathcal{U}})}<+\infty$, $f_{j}\to f_{\infty}$ in
$C^{1}_{\rm loc}(\mathbb{R})$.

Then there exist a subsequence $u_{j(m)}$ and a discrete set
$\Sigma_{\infty}\subset{\mathcal{U}}$, with $\\#\Sigma_{\infty}\leq k$, such
that

$\text{$u_{j(m)}\rightharpoonup u_{\infty}$ in
$W^{1,2}({\mathcal{U}})$},\qquad\text{$u_{j(m)}\to u_{\infty}$ in $C^{2}_{\rm
loc}({\mathcal{U}}\setminus\Sigma_{\infty})$,}$

and $u_{\infty}$ satisfies

$-\Delta u_{\infty}=f_{\infty}(u_{\infty})\text{ in }{\mathcal{U}},\qquad
f_{\infty}^{\prime}(u_{\infty})\in L^{1}_{\rm loc}({\mathcal{U}}),\qquad{\rm
ind}(u_{\infty},\Omega)\leq k,\qquad u_{\infty}\in C^{2}({\mathcal{U}}).$

###### Proof.

Given $x\in{\mathcal{U}}$, for any $j$ we denote by $r_{j,x}$ the largest
radius where $u_{j}$ is stable around $x$:

$r_{j,x}:=\sup\\{r\in[0,{\mathop{\mathrm{missing}}{\,dist\,}}(x,\partial{\mathcal{U}}))\,:\,{\rm
ind}(u_{j},B_{r}(x))=0\\},$

Then, we define

$r_{\infty,x}:=\limsup_{j\to\infty}r_{j,x},\qquad\Sigma_{\infty}:=\\{x\in{\mathcal{U}}\,:\,r_{\infty,x}=0\\}.$

We claim that $\Sigma_{\infty}$ is a discrete set of cardinality at most $k$.

Indeed, suppose by contradiction that $\Sigma_{\infty}$ contains $k+1$ points
$x_{1},\ldots,x_{k+1}$, and fix

$0<r<\min\left\\{\min_{1\leq i\leq
k+1}{\mathop{\mathrm{missing}}{\,dist\,}}(x_{i},\partial{\mathcal{U}})),\frac{1}{2}\min_{1\leq
i,l\leq k+1}|x_{i}-x_{l}|\right\\}.$

Since $r_{j,x_{i}}\to 0$ as $j\to\infty$ (because $x_{i}\in\Sigma_{\infty}$),
for $j$ large enough $u_{j}$ is unstable inside each of the balls
$\\{B_{r}(x_{i})\\}_{i=1}^{k+1}$. However, since these balls are disjoint (by
the choice of $r$), Lemma 2.2(i) provides the desired contradiction.

Consider now a family of compact sets $\\{K_{\ell}\\}_{\ell\in\mathbb{N}}$
such that ${\mathcal{U}}\setminus\Sigma_{\infty}=\cup_{\ell}K_{\ell},$ and for
any $\ell$ consider the covering of $K_{\ell}$ given by
$\\{B_{r_{\infty/2,x}}(x)\\}_{x\in K_{\ell}}$. By compactness, there exists a
finite set of points $\\{x_{i}\\}_{i\in\mathcal{I}_{\ell}}\subset K_{\ell}$
such that
$K_{\ell}\subset\cup_{i\in\mathcal{I}_{\ell}}B_{r_{\infty}/2,x_{i}}(x_{i})$.
Note that, since each set $\mathcal{I}_{\ell}$ is finite, for each
$\ell\in\mathbb{N}$ we can choose a subsequence $j_{\ell}(m)$ such that

$r_{\infty,x_{i}}=\lim_{m\to\infty}r_{j_{\ell}(m),x_{i}}\qquad\forall\,i\in\mathcal{I}_{1}\cup\ldots\cup\mathcal{I}_{\ell}.$

Then, by a diagonal argument we can find a subsequence $j(m)$, independent of
$\ell$, such that

$r_{\infty,x_{i}}=\lim_{m\to\infty}r_{j(m),x_{i}}\qquad\forall\,i\in\cup_{\ell\in\mathbb{N}}\mathcal{I}_{\ell}.$

Since the functions $u_{j(m)}$ are uniformly bounded in
$W^{1,2}({\mathcal{U}})$, up to extracting a further subsequence, there exists
a weak limit in $W^{1,2}({\mathcal{U}})$ that we denote by $u_{\infty}$. We
now want to show that $u_{\infty}$ satisfies all the desired properties.

First of all, for each $\ell\in\mathbb{N}$ we define the open set

$\mathcal{O}_{\ell}:=\cup_{i\in\mathcal{I}_{\ell}}B_{r_{\infty}/2,x_{i}}(x_{i})\supset
K_{\ell}.$

Since $u_{j(m)}$ is stable on $B_{r_{j(m)},x_{i}}(x_{i})$ and
$r_{j(m),x_{i}}\to r_{\infty,x_{i}}$ as $m\to\infty$, it follows by [9,
Theorem 1.2] and elliptic regularity444Recall that $n\leq 9$, and note that
$f_{j}$ are uniformly $C^{1}$ on compact set since they converge to
$f_{\infty}$. that

$\|u_{j(m)}\|_{C^{2,\alpha}(\mathcal{O}_{\ell})}\leq
C_{\ell,\alpha}\qquad\forall\,m\gg 1,\,\forall\,\alpha\in(0,1),$

which implies that $u_{j(m)}\to u_{\infty}$ in $C^{2}(\mathcal{O}_{\ell})$.
Since $\cup_{\ell}\mathcal{O}_{\ell}={\mathcal{U}}\setminus\Sigma_{\infty}$,
this proves the convergence in $C^{2}_{\rm
loc}({\mathcal{U}}\setminus\Sigma_{\infty})$.

To show that $u_{\infty}$ solves the desired equation, by the $C^{2}_{\rm
loc}({\mathcal{U}}\setminus\Sigma_{\infty})$ convergence it follows
immediately that

$-\Delta u_{\infty}=f_{\infty}(u_{\infty})\qquad\text{in
}{\mathcal{U}}\setminus\Sigma_{\infty}.$

Then, since $u_{\infty}\in W^{1,2}(\mathcal{U})$ and $\Sigma_{\infty}$
consists of finitely many points (hence it has zero $W^{1,2}$-capacity), the
equation $-\Delta u_{\infty}=f_{\infty}(u_{\infty})$ must hold inside the
whole domain ${\mathcal{U}}$.

We now note that, thanks to Lemma 2.2(ii),

$\int_{B_{r}(\bar{x})}f_{j(m)}^{\prime}(u_{j(m)})\leq
C_{n}(1+k)^{\frac{2}{n}}r^{n-2}\qquad\forall\,B_{2r(\bar{x})}\subset{\mathcal{U}}.$

Since $f_{j(m)}^{\prime}(u_{j(m)})$ are nonnegative and converge pointwise to
$f_{\infty}^{\prime}(u_{\infty})$ inside
${\mathcal{U}}\setminus\Sigma_{\infty}$ (and so a.e.), Fatou’s Lemma implies
that

$\int_{B_{r}(\bar{x})}f_{\infty}^{\prime}(u_{\infty})\leq
C_{n}(1+k)^{\frac{2}{n}}r^{n-2}\qquad\forall\,B_{2r(\bar{x})}\subset{\mathcal{U}},$

thus $f^{\prime}_{\infty}(u_{\infty})\in L^{1}_{\rm loc}({\mathcal{U}})$.

Now, to prove the bound on the index, assume by contradiction that there
exists a $k+1$ dimensional subspace $X^{\prime}\subset
C^{1}_{c}({\mathcal{U}})$ where

$Q_{\infty}(\xi):=\int_{{\mathcal{U}}}\left(|\nabla\xi|^{2}-f_{\infty}^{\prime}(u_{\infty})\xi^{2}\right)\,dx<0\qquad\forall\,\xi\in
X^{\prime}\setminus\\{0\\}.$

We claim that also $Q_{j(m)}$ is strictly negative on
$X^{\prime}\setminus\\{0\\}$ for $m$ sufficiently large. Indeed, if not, by
homogeneity there exists a sequence $\xi_{m}\in X^{\prime}\setminus\\{0\\}$,
with $\|\xi_{m}\|_{C^{1}}=1$, such that $Q_{j(m)}(\xi_{m})\geq 0$. Since
$X^{\prime}\subset C^{1}_{c}({\mathcal{U}})$ is finite dimensional, all
functions $\xi_{m}$ live in a fixed compact set and, up to a subsequence, they
converge in $C^{1}_{c}({\mathcal{U}})$ to a limiting function $\xi_{\infty}\in
X^{\prime}$ with $\|\xi_{\infty}\|_{C^{1}}=1$. In particular,

$\int_{{\mathcal{U}}}|\nabla\xi_{m}|^{2}\,dx\to\int_{{\mathcal{U}}}|\nabla\xi_{\infty}|^{2}\,dx\qquad\text{as
}m\to\infty.$

Also, since $f_{j(m)}^{\prime}(u_{j(m)})\xi_{m}^{2}$ are nonnegative and
converge pointwise to $f_{\infty}^{\prime}(u_{\infty})\xi_{\infty}^{2}$ inside
${\mathcal{U}}\setminus\Sigma_{\infty}$ (and so a.e.), Fatou’s Lemma implies
that

$\liminf_{m\to\infty}\int_{{\mathcal{U}}}f_{j(m)}^{\prime}(u_{j(m)})\xi_{m}^{2}\geq\int_{{\mathcal{U}}}f_{\infty}^{\prime}(u_{\infty})\xi_{\infty}^{2}\,dx.$

Combining these two facts, we deduce that

$0\leq\limsup_{m\to\infty}{Q_{j(m)}(\xi_{m})}\leq Q_{\infty}(\xi_{\infty}),$

a contradiction since $\xi_{\infty}\in X^{\prime}\setminus\\{0\\}$. Hence
$Q_{j(m)}$ is strictly negative on $X^{\prime}\setminus\\{0\\}$ for $m$
sufficiently large, which is impossible since ${\rm
ind}(u_{j(m)},{\mathcal{U}})\leq k$. This contradiction proves that ${\rm
ind}(u_{\infty},{\mathcal{U}})\leq k$.

Finally, to prove that $u_{\infty}\in C^{2}({\mathcal{U}})$, we recall that
finite Morse index solutions are locally stable (see for instance [14,
Proposition 1.5.1] or [12, Proposition 2.1]). Hence, we can apply Proposition
A.1 and elliptic regularity around each of the points in $\Sigma_{\infty}$ to
deduce that $u_{\infty}\in C^{2}({\mathcal{U}})$. ∎

Our next goal is to show a uniform $W^{1,2}$ integrability estimate for finite
index solutions. It is for this result that the growth assumption on $f$ plays
a crucial role. Before stating and proving it, we first recall the following
simple estimate that can be found, for instance, in [9, Lemma A.1].

######  Lemma 2.4.

Let $f:\mathbb{R}\to\mathbb{R}$ be a nonnegative function, and let $v\in
C^{2}$ solve $-\Delta v=f(v)$ inside $B_{r}(\bar{x})$. Then

$\int_{B_{r/2}(\bar{x})}f(v)\,dx\leq C_{n}r^{-2}\int_{B_{r}(\bar{x})}|v|\,dx.$

We begin by proving a uniform $W^{1,2}$ integrability estimate for stable
solutions, that will be used below to address the general case.

######  Proposition 2.5.

Let $f\in C^{1}$ be nonnegative, and let $u\in C^{2}$ be a nonnegative stable
solution to $-\Delta u=f(u)$ in $B_{r}(\bar{x})$ for some $r\in(0,1]$. Assume
that $f$ satisfies (1.5) for some $c_{1}>0$. Then there exists
$\delta=\delta(n,\epsilon)>0$ such that

$\int_{B_{\rho}(\bar{x})}|\nabla u|^{2}\,dx\leq
C(c_{1},n)\,\rho^{\delta}\qquad\text{for all }0<\rho<\frac{r}{4}.$ (2.7)

###### Proof.

With no loss of generality we can assume $\bar{x}=0$.

By Hölder inequality, (1.5), and Lemma 2.4, for any ball $B_{2\rho}(z)\subset
B_{r}$ it holds

$\bigg{(}\rho^{-n}\int_{B_{\rho}(z)}u\,dx\bigg{)}^{\frac{n+2}{n-2}+\epsilon}\leq
C(n)\rho^{-n}\int_{B_{\rho}(z)}u^{\frac{n+2}{n-2}+\epsilon}\,dx\\\ \leq
C(n,c_{1})\rho^{-n}\int_{B_{\rho}(z)}f(u)\,dx\leq
C_{0}\rho^{-2-n}\int_{B_{2\rho}(z)}u\,dx,$ (2.8)

where $C_{0}=C_{0}(n,c_{1})$. Let $\delta=\delta(n,\epsilon)>0$ be small
enough so that

$\biggl{(}\frac{n-2}{2}-\delta\biggr{)}\biggl{(}\frac{n+2}{n-2}+\epsilon\biggr{)}-2\geq\frac{n-2}{2}-\delta,$
(2.9)

and define

$G(z,\rho):=\max\left\\{1,\gamma\sup_{B_{s}(y)\subset
B_{\rho}(z)}s^{-\frac{n+2}{2}-\delta}\int_{B_{s}(y)}u\,dx\right\\},$

where $\gamma\in(0,1)$ is a small constant to be fixed later. Then, thanks to
(2.8) and (2.9), whenever $B_{2\rho}(z)\subset B_{r}$ we have

$G(z,\rho)\leq G(z,\rho)^{\frac{n+2}{n-2}+\epsilon}\leq
1+\gamma^{\frac{n+2}{n-2}+\epsilon}\sup_{B_{s}(y)\subset
B_{\rho}(z)}\bigg{(}s^{-\frac{n+2}{2}-\delta}\int_{B_{s}(y)}u\,dx\bigg{)}^{\frac{n+2}{n-2}+\epsilon}\\\
\leq 1+C_{0}\gamma^{\frac{n+2}{n-2}+\epsilon}\sup_{B_{s}(y)\subset
B_{\rho}(z)}s^{-\frac{n+2}{2}-\delta}\int_{B_{2s}(y)}u\,dx\leq
1+C_{0}\gamma^{\frac{4}{n-2}+\epsilon}G(z,2\rho).$ (2.10)

We now claim that $G(0,r/2)$ is uniformly bounded.

To show this, consider the quantity

$Q:=\sup_{z\in B_{r},\\\ \rho\leq r-|z|}G(z,\rho/2).$

Note that, since $u$ is of class $C^{2}$, $Q$ is a finite constant. Also, we
can assume that $Q>2$ (otherwise there is nothing to prove). Consider now
$z\in B_{r}$ and $\rho\leq r-|z|$ such that $G(z,\rho/2)\geq Q/2$. Since
$Q/2>1$, it follows from the definition of $G$ that there exists
$B_{s}(y)\subset B_{\rho/2}(z)$ such that $\gamma
s^{-\frac{n+2}{2}-\delta}\int_{B_{s}(y)}u\geq Q/3$. We can now cover
$B_{s}(y)$ with $N_{n}$ balls $\\{B_{s/4}(y_{k})\\}_{k=1}^{N_{n}}$ with
$y_{k}\in B_{s}(y)\subset B_{\rho/2}(z)$, where $N_{n}$ is a dimensional
constant, and observe that

$\frac{Q}{3}\leq\gamma s^{-\frac{n+2}{2}-\delta}\int_{B_{s}(y)}u\,dx\leq\gamma
s^{-\frac{n+2}{2}-\delta}\sum_{k=1}^{N_{n}}\int_{B_{s/4}(y_{k})}u\,dx\\\
\leq\gamma(s/4)^{-\frac{n+2}{2}-\delta}\sum_{k=1}^{N_{n}}\int_{B_{s/4}(y_{k})}u\,dx\leq\sum_{k=1}^{N}G(y_{k},s/4).$
(2.11)

Note now that, since $s\leq\rho/2$ and $y_{k}\in B_{\rho/2}(z)$,

$|y_{k}|+s\leq|z|+\frac{\rho}{2}+\frac{\rho}{2}\leq|z|+\rho\leq r.$

In particular $B_{s/2}(y_{k})\subset B_{r}$, and it follows by (2.10) and the
definition of $Q$ that

$G(y_{k},s/4)\leq 1+C_{0}\gamma^{\frac{4}{n-2}+\epsilon}G(y_{k},s/2)\leq
1+C_{0}\gamma^{\frac{4}{n-2}+\epsilon}Q.$

Combining this bound with (2.11), this yields

$\frac{Q}{3}\leq N_{n}\left(1+C_{0}\gamma^{\frac{4}{n-2}+\epsilon}Q\right)\leq
N_{n}\left(1+C_{0}\gamma^{\frac{4}{n-2}}Q\right),$

and by choosing $\gamma$ small enough (depending only on $C_{0}$ and the
dimension), we conclude that $Q\leq 4N_{n}$, and therefore

$\gamma\sup_{s\leq r/2}s^{-\frac{n+2}{2}-\delta}\int_{B_{s}(0)}u\,dx\leq
G(0,r/2)\leq Q\leq 4N_{n},$ (2.12)

as desired.

Recall now that, by [9, Theorem 1.2], if $u$ is stable on a ball $B$ then

$\|u\|_{W^{1,2}\left(\frac{1}{2}B\right)}\leq
C(n)\,({\mathop{\mathrm{missing}}{\,diam\,}}(B))^{-\frac{n+2}{2}}\|u\|_{L^{1}(B)}.$

Combining this estimate with (2.12), we obtain (2.7).∎

We next improve this result to solutions with finite Morse index.

######  Proposition 2.6.

Let $f\in C^{1}$ be nonnegative, and let $u\in C^{2}$ be a nonnegative
solution to $-\Delta u=f(u)$ in $B_{r}(\bar{x})$ for some $r>0$. Assume that
${\rm ind}(u,B_{r}(\bar{x}))\leq k$ for some $k\in\mathbb{N}$, and that $f$
satisfies (1.5) for some $c_{1}>0$. Then

$\int_{B_{\rho}(\bar{x})}|\nabla u|^{2}\,dx\leq
C(c_{1},n,\epsilon)\,k\,\rho^{\delta}\qquad\text{for all }\,\rho\in(0,r/4),$

where $\delta=\delta(n,\epsilon)>0$ is as in Proposition 2.5.

###### Proof.

Let $M>1$ be a fixed constant555One can choose $M$ to be any constant larger
than $1$, for instance $M=2$. However, for notational convenience we prefer to
use the notation $M$ instead of fixing its value, as we believe that the
estimates become easier to follow., define the set $Q^{0}:=B_{\rho}(\bar{x})$,
and consider the covering of $Q^{0}$ given by
$\left\\{B_{M^{-1}\rho}(z)\right\\}_{z\in Q^{0}}.$ By Besicovitch Covering
Theorem, there exist a dimensional constant $N_{n}\in\mathbb{N}$ and a
subfamily of balls
$\left\\{B^{0}_{\ell}\right\\}_{\ell\in\mathcal{I}_{0}}\subset\left\\{B_{M^{-1}\rho}(z)\right\\}_{z\in
Q^{0}}$ such that

$1\leq\sum_{\ell\in\mathcal{I}_{0}}\chi_{B^{0}_{\ell}}(y)\leq
N_{n}\qquad\text{for all }y\in Q^{0}.$ (2.13)

In particular, since these balls have radius $M^{-1}\rho$ and are contained
inside $B_{2\rho}(\bar{x})$, it follows that

$\\#\mathcal{I}_{0}\,|B_{M^{-1}\rho}|\leq
N_{n}|B_{2\rho}|\qquad\Rightarrow\qquad\\#\mathcal{I}_{0}\leq
2^{n}M^{n}N_{n}.$

Moreover, since each point $y\in B_{\rho}(\bar{x})$ is covered by at most
$N_{n}$ balls of radius $M^{-1}\rho$, then the same is true if we double the
radius of the balls: more precisely, there exists a dimensional constant
$N^{\prime}_{n}\in\mathbb{N}$ such that666A simple way to see this is to note
that, as a consequence of (2.13), we can split
$\left\\{B^{0}_{\ell}\right\\}_{\ell\in\mathcal{I}_{0}}$ into $N_{n}$
subfamilies of balls, where the balls of each subfamily are disjoint. This
implies that the centers of the balls of each subfamily are at mutual distance
at least $2M^{-1}\rho$. Then, if we double the radius, the overlapping for
each of these subfamilies is bounded by a dimensional constant $C_{n}\geq 1$.

$1\leq\sum_{\ell\in\mathcal{I}_{0}}\chi_{2B^{0}_{\ell}}(y)\leq
N^{\prime}_{n}\qquad\forall\,y\in Q^{0}.$ (2.14)

Let us split $\left\\{2B^{0}_{\ell}\right\\}_{\ell\in\mathcal{I}_{0}}$ into
$N^{\prime}_{n}$ subfamilies of balls, where the balls of each subfamily are
disjoint. As ${\rm ind}(u_{s})\leq k$, we can apply Lemma 2.2(i) to each
subfamily. Then we deduce that, except for at most $N^{\prime}_{n}k$ balls,
say $B^{0}_{1},\ldots,\,B^{0}_{k_{0}}$ with $k_{0}\leq N_{n}^{\prime}k$, the
function $u$ is stable inside each ball
$\\{2B^{0}_{\ell}\\}_{\ell\in\mathcal{I}_{0}\setminus\\{1,\ldots,k_{0}\\}}$.
Thus by Proposition 2.5, we have

$\int_{B^{0}_{\ell}}|\nabla u|^{2}\,dx\leq
C(c_{1},n)M^{-\delta}\rho^{\delta}\qquad\forall\,\mathcal{I}_{0}\setminus\\{1,\ldots,k_{0}\\}.$

Now, we consider the set $Q^{1}:=\bigcup_{1\leq\ell\leq k_{0}}B^{0}_{\ell}$
and the covering $\left\\{B_{M^{-2}\rho}(z)\right\\}_{z\in Q^{1}}.$ Again by
Besicovitch Covering Theorem, there exists a subfamily of balls
$\left\\{B^{1}_{\ell}\right\\}_{\ell\in\mathcal{I}_{1}}\subset\left\\{B_{M^{-2}\rho}(z)\right\\}_{z\in
Q^{1}}$ such that

$1\leq\sum_{\ell\in\mathcal{I}_{1}}\chi_{B^{1}_{\ell}}(y)\leq
N_{n}\qquad\text{for all }y\in Q^{1}.$ (2.15)

Also, since these balls are contained inside $\bigcup_{1\leq\ell\leq
k_{0}}2B^{0}_{\ell}$, it follows that (recall that $k_{0}\leq
N_{n}^{\prime}k$)

$\\#\mathcal{I}_{1}\,|B_{M^{-2}\rho}|\leq
k_{0}N_{n}|B_{2M^{-1}\rho}|\qquad\Rightarrow\qquad\\#\mathcal{I}_{1}\leq
2^{n}M^{n}k_{0}N_{n}\leq 2^{n}M^{n}N_{n}^{\prime}N_{n}k.$

Furthermore, as before,

$1\leq\sum_{\ell\in\mathcal{I}_{1}}\chi_{2B^{1}_{\ell}}(y)\leq
N^{\prime}_{n}\qquad\forall\,y\in Q^{1}.$

Hence (up to renaming the indices) $u$ is stable inside each ball
$\\{2B^{1}_{\ell}\\}_{\ell\in\mathcal{I}_{1}\setminus\\{1,\ldots,k_{1}\\}}$
with $k_{1}\leq N_{n}^{\prime}k$, and therefore

$\int_{B^{1}_{\ell}}|\nabla u|^{2}\,dx\leq
C(c_{1},n)M^{-2\delta}\rho^{\delta}\qquad\forall\,\ell\in\mathcal{I}_{1}\setminus\\{1,\ldots,k_{1}\\}.$

To continue this construction, define

$Q^{2}:=\bigcup_{1\leq\ell\leq k_{2}}B^{2}_{\ell}.$

Then, we can apply the very same argument used for $Q^{1}$ to find a family of
balls $\\{B^{2}_{\ell}\\}_{\ell\in\mathcal{I}_{2}}$, with
$\\#\mathcal{I}_{2}\leq 2^{n}M^{n}k_{1}N_{n}\leq
2^{n}M^{n}N_{n}^{\prime}N_{n}k$, such that

$\int_{B^{2}_{\ell}}|\nabla u|^{2}\,dx\leq
C(c_{1},n)M^{-3\delta}\rho^{\delta}\qquad\forall\,\ell\in\mathcal{I}_{2}\setminus\\{1,\ldots,k_{2}\\},\qquad\text{with
}k_{2}\leq N_{n}^{\prime}k.$

Iterating this construction, we obtain that the family of balls
$\\{B^{j}_{\ell}\\}_{\ell\in\mathcal{I}_{j}\setminus\\{1,\ldots,k_{j}\\},\,j\in\mathbb{N}}$
covers $Q^{0}\setminus K$, with $K:=\cap_{j\in\mathbb{N}}Q^{j}$,777Here one
could note that, since $u$ is smooth, every ball sufficiently small is stable
and therefore $Q^{j}$ is empty for $j$ large enough, hence $K=\emptyset$.
However this information is not needed, and this proof also applies to weak
solutions with bounded index. and

$\int_{B^{j}_{\ell}}|\nabla u|^{2}\,dx\leq
C(c_{1},n)M^{-j\delta}\rho^{\delta}\quad\forall\,\ell\in\mathcal{I}_{j}\setminus\\{1,\ldots,k_{j}\\},\qquad\\#\mathcal{I}_{j}\leq
2^{n}M^{n}N_{n}^{\prime}N_{n}k,\qquad k_{j}\leq N_{n}^{\prime}k.$

Since $K$ has measure zero (because $|Q^{j}|\leq k_{j}|B_{M^{-j}\rho}|\leq
N_{n}^{\prime}k|B_{M^{-j}\rho}|\to 0$ as $j\to\infty$), we have

$\int_{Q^{0}}|\nabla u|^{2}\,dx=\int_{Q^{0}\setminus K}|\nabla
u|^{2}\,dx\leq\sum_{j=0}^{\infty}\sum_{\ell\in\mathcal{I}_{j}\setminus\\{1,\ldots,k_{j}\\}}\int_{B^{j}_{\ell}}|\nabla
u|^{2}\,dx\\\ \leq
C(c_{1},n)\biggl{(}\sum_{j=0}^{\infty}\\#\mathcal{I}_{j}M^{-j\delta}\biggr{)}\rho^{\delta}\leq
C(c_{1},n,\delta)\,k\,\rho^{\delta}.$

Recalling that $\delta=\delta(n,\epsilon),$ this concludes the lemma. ∎

The next result is a powerful $\varepsilon$-regularity theorem which shows the
following: given $\gamma\in(0,1),$ if the $W^{1,2}$ norm of a solution in a
ball $B_{r}$ decays like $r^{n-2+2\gamma}$ for $r\in[\varepsilon,1]$ with
$\varepsilon$ small enough, then it decays all the way to the origin.

######  Proposition 2.7.

Let $g:\mathbb{R}\to\mathbb{R}$ be a continuous nonnegative function, let
$u\in C^{2}(B_{1})$ solve $-\Delta u=f(u)$ for some increasing function
$f:\mathbb{R}\to\mathbb{R}$ of class $C^{1}$ satisfying $0\leq f\leq g$ and
(1.4), and assume that ${\rm ind}(u,B_{1})\leq k$ and $\int_{B_{1}}|u|\leq M$.
Then, for any $\gamma\in(0,1)$ there exists
$m_{0}=m_{0}(n,g,\epsilon,t_{0},k,M,\gamma)\in\mathbb{N}$ such that the
following holds:
Suppose that

$\int_{B_{r}}|\nabla u|^{2}\,dx\leq
r^{n-2+2\gamma}\qquad\forall\,r\in[2^{-m_{0}},1].$

Then

$\int_{B_{r}}|\nabla u|^{2}\,dx\leq r^{n-2+2\gamma}\qquad\forall\,r\in[0,1]$
(2.16)

and $|u(0)|\leq M+C(n,\gamma)$.

###### Proof.

We begin with the proof of (2.16). For that, it suffices to prove the
following implication: if

$\int_{B_{r}}|\nabla u|^{2}\,dx\leq
r^{n-2+2\gamma}\qquad\forall\,r\in[2^{-m},1]$

for some $m\geq m_{0}$, then

$\int_{B_{r}}|\nabla u|^{2}\,dx\leq
r^{n-2+2\gamma}\qquad\forall\,r\in[2^{-(m+1)},1].$

Indeed, iterating this result with $m=m_{0},m_{0}+1,\ldots$, the result
follows.

To prove the implication above, we argue by contradiction. If it was false, we
could find a sequence of functions $u_{j}\in C^{2}(B_{1})$, and $f_{j}\in
C^{1}(\mathbb{R})$ satisfying (1.4), such that

$-\Delta u_{j}=f_{j}(u_{j}),\qquad\text{$f_{j}$ increasing},\qquad 0\leq
f_{j}\leq g,\qquad{\rm ind}(u_{j},B_{1})\leq k,\qquad\int_{B_{1}}|u_{j}|\leq
M,$

and a sequence $m_{j}\to\infty$, such that

$\int_{B_{r}}|\nabla u_{j}|^{2}\,dx\leq
r^{n-2+2\gamma}\qquad\forall\,r\in[2^{-m_{j}},1]$ (2.17)

but

$\int_{B_{r_{j}}}|\nabla u_{j}|^{2}\,dx\geq r^{n-2+2\gamma}\qquad\text{for
some }r_{j}\in[2^{-(m_{j}+1)},2^{-m_{j}}].$ (2.18)

We introduce the notation $A_{r}:=B_{r}\setminus B_{r/2}$.

We first note that, as a consequence of (2.17) and the bound
$\int_{B_{1}}|u_{j}|\leq M$, it follows that

${\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{2^{-(m_{j}+1)}}}|u_{j}|\,dx\leq
M+C(n,\gamma).$ (2.19)

Indeed, thanks to (2.17), for any $2^{-m_{j}}\leq s\leq r\leq 1$ we have

$\begin{split}\bigg{|}{\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{\partial B_{r}}|u_{j}|-{\hbox
to0.0pt{\bf\kern 4.49997pt--\hss}\int}\ignorespaces_{\partial
B_{s}}|u_{j}|\biggr{|}&\leq{\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{\partial
B_{1}}|u_{j}(ry)-u_{j}(sy)|\,dy\leq\int_{s}^{r}\biggl{(}{\hbox
to0.0pt{\bf\kern 4.49997pt--\hss}\int}\ignorespaces_{\partial B_{1}}|\nabla
u_{j}(\tau y)|\,dy\biggr{)}\,d\tau\\\ &=\int_{B_{r}\setminus
B_{s}}\frac{|\nabla
u_{j}(x)|}{|x|^{n-1}}\,dx\leq\sum_{\ell=1}^{m_{j}}\int_{A_{2^{-\ell}}}\frac{|\nabla
u_{j}(x)|}{|x|^{n-1}}\,dx\leq\sum_{\ell=1}^{m_{j}}2^{\ell(n-1)}\int_{A_{2^{-\ell}}}|\nabla
u_{j}|\,dx\\\ &=C(n)\sum_{\ell=1}^{m_{j}}2^{-\ell}{\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{2^{-\ell}}}|\nabla u_{j}|\,dx\leq
C(n)\sum_{\ell=1}^{m_{j}}2^{-\ell}\biggl{(}{\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{2^{-\ell}}}|\nabla
u_{j}|^{2}\,dx\biggr{)}^{1/2}\\\ &\leq
C(n)\sum_{\ell=1}^{m_{j}}2^{-\ell}2^{-\ell(\gamma-1)}=C(n)\sum_{\ell=1}^{m_{j}}2^{-\ell\gamma}\leq
C(n,\gamma),\end{split}$ (2.20)

therefore

$\biggl{|}{\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{1}}|u_{j}|\,dx-{\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{2^{-(m_{j}+1)}}}|u_{j}|\,dx\biggr{|}\leq
C(n,\gamma),$

and (2.19) follows.

To simplify the notation, we set $r_{j}:=2^{-m_{j}}$, and we define

$a_{j}:={\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{2r_{j}}}u_{j}\,dx,\qquad
w_{j}(x):=r_{j}^{-\gamma}[u_{j}(r_{j}x)-a_{j}],$

so that

${\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{2}}w_{j}=0,\qquad-\Delta
w_{j}=h_{j}(w_{j}),\qquad
h_{j}(t):=r_{j}^{2-\gamma}f_{j}(a_{j}+r_{j}^{\gamma}t).$ (2.21)

Note that ${\rm ind}(w_{j},B_{2^{m_{j}}})\leq k$ and $0\leq h_{j}\leq
r_{j}^{2-\gamma}g(a_{j}+r_{j}^{\gamma}t),$ so it follows from (2.19) that
$h_{j}\to 0$ in $C^{1}_{\rm loc}$. Also (2.17) and (2.18) imply that

$\int_{B_{2^{\ell}}}|\nabla w_{j}|^{2}dx\leq
2^{\ell(n-2+2\gamma)}\qquad\forall\,0\leq\ell\leq m_{j}$ (2.22)

and

$\int_{B_{1}}|\nabla w_{j}|^{2}dx\geq 2^{-(n-2+2\gamma)}.$ (2.23)

Thus, thanks to Proposition 2.3 and a diagonal argument we deduce that, up to
a subsequence,

$\text{$w_{j}\rightharpoonup w_{\infty}$ in $W^{1,2}_{\rm
loc}(\mathbb{R}^{n})$},\qquad\text{$w_{j}\to w_{\infty}$ in $C^{2}_{\rm
loc}(\mathbb{R}^{n}\setminus\Sigma_{\infty})$,}$

where $\Sigma_{\infty}$ has cardinality at most $k$, and $w_{\infty}$
satisfies (by (2.22) and (2.21))

$-\Delta w_{\infty}=0\quad\text{in }\mathbb{R}^{n},\qquad{\hbox
to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{2}}w_{\infty}=0,\qquad\int_{B_{2^{\ell}}}|\nabla
w_{\infty}|^{2}dx\leq 2^{\ell(n-2+2\gamma)}\qquad\forall\,\ell\geq 0.$

Then it follows from Liouville Theorem for harmonic functions that
$w_{\infty}\equiv 0$,888Indeed, by Liouville Theorem $w_{\infty}$ must be a
harmonic polynomial, and the bound $\int_{B_{2^{\ell}}}|\nabla
w_{\infty}|^{2}dx\leq 2^{\ell(n-2+2\gamma)}$ for $\ell\geq 0$ implies that
$w_{\infty}$ must be constant (recall that $\gamma<1$). Finally, since
$\int_{A_{2}}w_{\infty}=0$ we deduce that $w_{\infty}\equiv 0$. and therefore

$\text{$w_{j}\rightharpoonup 0$ in $W^{1,2}_{\rm
loc}(\mathbb{R}^{n})$},\qquad\text{$w_{j}\to 0$ in $C^{2}_{\rm
loc}(\mathbb{R}^{n}\setminus\Sigma_{\infty})$,}\qquad\\#\Sigma_{\infty}\leq
k.$

We now want to get a contradiction with (2.23).

Consider the annuli

$B_{2}\setminus B_{1},\qquad B_{3}\setminus B_{2},\qquad\ldots,\qquad
B_{k+2}\setminus B_{k+1}.$

Since $\\#\Sigma_{\infty}\leq k$, there exists $\hat{i}\in\\{1,\ldots,k+1\\}$
such that $(B_{\hat{i}+1}\setminus B_{\hat{i}})\cap\Sigma_{\infty}=\emptyset.$
In particular, if we fix $\varphi\in C^{\infty}_{c}(B_{\hat{i}+3/4})$
nonnegative such that $\varphi|_{B_{\hat{i}+1/4}}=1$, then $w_{j}\to 0$ in
$C^{2}$ on $\\{\nabla\varphi\neq 0\\}.$

Now we first test the equation for $w_{j}$ (see (2.21)) with $w_{j}\varphi$ to
get

$\begin{split}r_{j}^{2-\gamma}\int_{B_{\hat{i}+1}}f_{j}\bigl{(}a_{j}+r_{j}^{\gamma}w_{j}\bigr{)}w_{j}\varphi\,dx&=\int_{B_{\hat{i}+1}}-w_{j}\Delta
w_{j}\varphi\,dx\\\ &=\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\varphi+w_{j}\nabla w_{j}\cdot\nabla\varphi\,dx\\\
&=\int_{B_{\hat{i}+1}}|\nabla w_{j}|^{2}\varphi\,dx+o(1),\end{split}$ (2.24)

where $o(1)$ denotes a quantity that goes to $0$ as $j\to\infty$, and the last
equality follows from the $C^{2}$ convergence of $w_{j}$ to $0$ on the set
$\\{\nabla\varphi\neq 0\\}.$

Similarly, testing (2.21) with $(\nabla w_{j}\cdot x)\varphi$, we obtain

$\displaystyle\int_{B_{\hat{i}+1}}r_{j}^{2-\gamma}f_{j}\bigl{(}a_{j}+r_{j}^{\gamma}w_{j}\bigr{)}(\nabla
w_{j}\cdot x)\varphi\,dx$ $\displaystyle=\int_{B_{\hat{i}+1}}-\Delta
w_{j}(\nabla w_{j}\cdot x)\varphi\,dx$
$\displaystyle=\int_{B_{\hat{i}+1}}D^{2}w_{j}\nabla w_{j}\cdot
x\varphi+|\nabla w_{j}|^{2}\varphi+(x\cdot\nabla w_{j})\nabla
w_{j}\cdot\nabla\varphi\,dx$
$\displaystyle=\left(1-\frac{n}{2}\right)\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\varphi\,dx-\frac{1}{2}\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\nabla\varphi\cdot x\,dx+o(1)$
$\displaystyle=\left(1-\frac{n}{2}\right)\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\varphi\,dx+o(1).$

Also, if we define $F_{j}(t):=\int_{0}^{t}f_{j}(\tau)\,d\tau,$ then we can
rewrite the first term above as follows:

$\displaystyle\int_{B_{\hat{i}+1}}r_{j}^{2-\gamma}f_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)(\nabla
w_{j}\cdot x)\varphi\,dx$
$\displaystyle=\int_{B_{\hat{i}+1}}r_{j}^{2-2\gamma}\nabla\left[F_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)-F_{j}(a_{j})\right]\cdot
x\,\varphi\,dx$
$\displaystyle=-n\int_{B_{\hat{i}+1}}r_{j}^{2-2\gamma}\left[F_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)-F_{j}(a_{j})\right]\varphi\,dx+\int_{B_{\hat{i}+1}}r_{j}^{2-\gamma}\left[F_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)-F_{j}(a_{j})\right]x\cdot\nabla\varphi\,dx$
$\displaystyle=-n\int_{B_{\hat{i}+1}}r_{j}^{2-2\gamma}\left[F_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)-F_{j}(a_{j})\right]\varphi\,dx+o(1),$

and we eventually get

$r_{j}^{2-2\gamma}\int_{B_{\hat{i}+1}}\left[F_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)-F_{j}(a_{j})\right]\varphi\,dx=\frac{n-2}{2n}\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\varphi\,dx+o(1).$ (2.25)

Now, given a constant $N>0$, we define the set

$S_{N}:=\left\\{x\in B_{\hat{i}+1}\colon r_{j}^{\gamma}|w_{j}(x)|\leq
N\right\\}.$

Since $w_{j}$ is uniformly bounded in $W^{1,\,2}(B_{\hat{i}+1})$ and $a_{j}$
is uniformly bounded, for any $N>0$ fixed we have

$r_{j}^{2-\gamma}\int_{S_{N}}\left[F_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)-F_{j}(a_{j})\right]\varphi\,dx\to
0,\qquad
r_{j}^{2-\frac{\gamma}{2}}\int_{S_{N}}f_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)w_{j}\varphi\,dx\to
0,$

so it follows from (2.24) and (2.25) that

$r_{j}^{2-\gamma}\int_{B_{\hat{i}+1}\setminus
S_{N}}f_{j}\bigl{(}a_{j}+r_{j}^{\gamma}w_{j}\bigr{)}w_{j}\varphi\,dx=\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\varphi\,dx+o(1)$ (2.26)

and

$r_{j}^{2-2\gamma}\int_{B_{\hat{i}+1}\setminus
S_{N}}\left[F_{j}\left(a_{j}+r_{j}^{\gamma}w_{j}\right)-F_{j}(a_{j})\right]\varphi\,dx=\frac{n-2}{2n}\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\varphi\,dx+o(1).$ (2.27)

Note now that, thanks to (1.4), the fact that $a_{j}$ is uniformly bounded
(see (2.19)), and that $0\leq f_{j}\leq g$, there exists a large constant
$N=N(M,n,\gamma,t_{0},\epsilon)$ such that, for all $j$,

$\left(\frac{2n}{n-2}+\frac{\epsilon}{2}\right)\bigl{[}F_{j}(t+a_{j})-F_{j}(a_{j})\bigr{]}\leq
f_{j}(t+a_{j})t\qquad\forall\,t\geq N.$

Combining this inequality with (2.26) and (2.27), we get

$\left(\frac{2n}{n-2}+\frac{\epsilon}{2}\right)\frac{n-2}{2n}\int_{B_{\hat{i}+1}}|\nabla
w_{j}|^{2}\varphi\leq\int_{B_{\hat{i}+1}}|\nabla w_{j}|^{2}\varphi+o(1),$

or equivalently

$\frac{(n-2)\epsilon}{4n}\int_{B_{\hat{i}+1}}|\nabla w_{j}|^{2}\varphi\leq
o(1).$

This contradicts (2.23) and concludes the proof of (2.16).

Now, to prove that bound on $|u(0)|,$ we observe that (2.16) allows us to
deduce the validity of (2.20) for all $0\leq s\leq r\leq 1.$ In particular
this implies that

$\biggl{|}{\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{A_{1}}|u_{j}|\,dx-|u(0)|\biggr{|}\leq
C(n,\gamma),$

so $|u(0)|\leq M+C(n,\gamma)$ as desired. ∎

Finally, we conclude this section with a useful consequence of the moving
plane method.

######  Lemma 2.8.

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded convex domain, and let $u\in
C^{2}(\Omega)$ solve (1.1) for some increasing positive function
$f:\mathbb{R}\to\mathbb{R}$ of class $C^{1}$. Then there exists
$\rho_{0}=\rho_{0}(\Omega)\in(0,1)$ such that

$\max_{\Omega}u=\max_{\Omega_{0}}u,$

where
$\Omega_{0}:=\left\\{x\in\Omega\colon{\mathop{\mathrm{missing}}{\,dist\,}}(x,\,\partial\Omega)>\rho_{0}\right\\}.$

###### Proof.

Recall that, since $f\geq 0$, the maximum principle implies that $u>0$ (unless
$u\equiv 0$, in which case the result is trivially true). Then, since $\Omega$
is bounded and convex, the result follows by the classical moving plane method
(see also the footnote inside [9, Proof of Corollary 1.4] for more details). ∎

##  3\. Uniform finite Morse index: Proof of Theorem 1.1

Let us assume, by contradiction, that there exists a sequence of $C^{2}$
solutions $u_{j}$

$\left\\{\begin{array}[]{ll}-\Delta u_{j}=\lambda_{j}f_{j}(u_{j})&\text{ in
}\Omega,\\\ u_{j}=0&\text{ on }\partial\Omega,\end{array}\right.$

where ${\rm ind}(u_{j},\Omega)\leq k$, the functions
$f_{j}:\mathbb{R}\to\mathbb{R}$ satisfy all the assumptions in the statement
of the theorem, $0\leq\lambda_{j}\leq\hat{\lambda},$ but
$\|u_{j}\|_{L^{\infty}(\Omega)}\to\infty$ as $j\to\infty$. Since
$f_{j}\in\mathcal{K}$ which is a compact family, up to a subsequence $f_{j}\to
f_{\infty}$ in $C^{1}_{\rm loc}(\mathbb{R})$ and
$\lambda_{j}\to\lambda_{\infty}\in[0,\hat{\lambda}]$. Define

$\hat{f}(t):=\sup_{j}f_{j}(t)\qquad\forall\,t\in\mathbb{R},$ (3.1)

so that $0\leq f_{j}\leq\hat{f}$ for all $j$. Since the functions $f_{j}$ are
locally uniformly Lipschitz (by the $C^{1}_{\rm loc}$ compactness), $\hat{f}$
is a continuous function.

We distinguish two cases, depending on the value of $\lambda_{\infty}$.

###  3.1. The case $\lambda_{\infty}>0$

Since $\lambda_{j}\to\lambda_{\infty}$, it follows from Remark 1.2 that, for
$j$ large enough,

$\lambda_{j}f_{j}(t)\geq c_{1}t^{\frac{n+2}{n-2}+\epsilon}\qquad\forall\,t\geq
0,\qquad\text{for some }c_{1}>0.$ (3.2)

Thanks to this bound, it follows from [9, Proposition B.1] that

$\|u_{j}\|_{L^{1}(\Omega)}\leq C_{0}=C_{0}(c_{1},\Omega).$ (3.3)

Also, if we define
$\Omega_{\tau}:=\left\\{x\in\Omega\colon{\mathop{\mathrm{missing}}{\,dist\,}}(x,\,\partial\Omega)>\tau\right\\}$,
then (3.2) and Proposition 2.6 yield

$\|\nabla u_{j}\|_{L^{2}(\Omega_{\tau})}\leq
C_{1}=C_{1}(c_{1},\Omega,\rho)\qquad\forall\,\tau>0.$

Since $\tau>0$ is arbitrary, Proposition 2.3 and a diagonal argument imply
that, up to a subsequence,

$\text{$u_{j}\rightharpoonup u_{\infty}$ in $W_{\rm
loc}^{1,2}({\mathcal{U}})$},\qquad\text{$u_{j}\to u_{\infty}$ in $C^{2}_{\rm
loc}(\Omega\setminus\Sigma_{\infty})$,}$ (3.4)

for some discrete set $\Sigma_{\infty}\subset\Omega$ with
$\\#\Sigma_{\infty}\leq k$, and some function $u_{\infty}\in C^{2}(\Omega)$.

Let $\rho_{0}\in(0,1)$ and $\Omega_{0}$ be given by Lemma 2.8, and define

$\Sigma_{\infty}^{0}:=\Omega_{0}\cap\Sigma_{\infty}=\\{\hat{x}_{1},\ldots,\hat{x}_{\ell}\\}\quad(\ell\leq
k),\qquad r_{0}:=\frac{1}{2}\min\left\\{\rho_{0},\min_{1\leq
i,l\leq\ell}|x_{i}-x_{l}|\right\\}.$

Then it follows from (3.4) that, for any $\rho\in(0,r_{0})$,

$\max_{1\leq i\leq\ell}\|u_{j}-u_{0}\|_{C^{2}(B_{r_{0}}(\hat{x}_{i})\setminus
B_{\rho}(\hat{x}_{i}))}\to 0\qquad\text{as }j\to\infty.$

In particular, since $u_{\infty}\in C^{2}(\Omega)$, there exists a constant
$\bar{C}>0$ such that that following holds: for any $\rho\in(0,r_{0})$ there
exists $j_{\rho}\in\mathbb{N}$ such that

$\max_{1\leq i\leq\ell}\|\nabla
u_{j}\|_{L^{\infty}(B_{r_{0}}(\hat{x}_{i})\setminus
B_{\rho}(\hat{x}_{i}))}\leq\bar{C}\qquad\forall\,j\geq j_{\rho}.$ (3.5)

We now make the following:
Claim: There exist $\hat{C},\hat{r}>0$ such that $\max_{1\leq
i\leq\ell}\|u_{j}\|_{L^{\infty}(B_{\hat{r}}(\hat{x}_{i}))}\leq\hat{C}$ for all
$j$ sufficiently large.

Assuming for a moment that the claim is proved, since $u_{j}\to u_{\infty}$ in
$C^{2}_{\rm loc}(\Omega\setminus\Sigma_{\infty})$ and $u_{\infty}\in
C^{2}(\Omega)$, it follows from the claim that

$\sup_{j}\|u_{j}\|_{L^{\infty}(\Omega_{0})}<\infty,$

where $\Omega_{0}$ is given by Lemma 2.8. But then Lemma 2.8 implies that
$\sup_{j}\|u_{j}\|_{L^{\infty}(\Omega)}<\infty$, a contradiction to our
initial assumption. Hence, in the case $f_{\infty}(0)>0$, the theorem is
proved provided we can show the claim.

To prove the claim, it suffices to control
$\|u_{j}\|_{L^{\infty}(B_{r_{0}}(\hat{x}_{i}))}$ for each $i$. With no loss of
generality, we can fix $i=1$ and assume that $\hat{x}_{1}=0$. Then, thanks to
(3.2) we can apply Proposition 2.6 to get the following estimate: for any
$r\in(0,r_{0}/2)$ and any $z\in B_{r}$, given $\rho\in(0,2r)$ it follows from
(3.5) that

$\int_{B_{r}(z)}|\nabla u_{j}|^{2}\,dx\leq\int_{B_{2r}}|\nabla
u_{j}|^{2}\,dx=\int_{B_{\rho}}|\nabla u_{j}|^{2}\,dx+\int_{B_{r}\setminus
B_{\rho}}|\nabla u_{j}|^{2}\,dx\\\ \leq C\rho^{\delta}+\bar{C}|B_{r}|\leq
C^{\prime}\left(\rho^{\delta}+r^{n}\right)\qquad\forall\,j\geq j_{\rho}.$
(3.6)

Also, it follows from (3.3) that

${\hbox to0.0pt{\bf\kern
4.49997pt--\hss}\int}\ignorespaces_{B_{r_{0}}}u_{j}\leq|B_{r_{0}}|^{-1}C_{0}=:M,\qquad
M=M(n,r_{0},c_{1},\Omega).$ (3.7)

Fix $\gamma:=1/2$, and let $m_{0}\in\mathbb{N}$ be the constant provided by
Proposition 2.7 with $g=\hat{\lambda}\hat{f}$ (see (3.1)). Then, with
$C^{\prime}$ as in (3.6), we choose first $\bar{r}\in(0,r_{0})$ such that
$C^{\prime}\bar{r}\leq\frac{1}{2}$, and then we fix $\rho\in(0,2\bar{r})$ such
that $C^{\prime}\rho^{\delta}\leq 2^{-m_{0}(n-1)-1}\bar{r}^{n-1}$. With these
choices it follows from (3.6) that, for any $r\in[2^{-m_{0}}\bar{r},\bar{r}]$
and any $z\in B_{2^{-m_{0}}\bar{r}}$,

$\int_{B_{r}(z)}|\nabla u_{j}|^{2}\,dx\leq
C^{\prime}\left(\rho^{\delta}+r^{n}\right)\leq
2^{-m_{0}(n-1)-1}\bar{r}^{n-1}+C^{\prime}\bar{r}r^{n-1}\leq\frac{1}{2}r^{n-1}+\frac{1}{2}r^{n-1}=r^{n-1},$
(3.8)

provided $j$ is sufficiently large. Hence, applying Proposition 2.7 to the
functions $u_{j,z}(x):=u_{j}(z+\bar{r}x)$ with $f=\bar{r}^{2}\lambda_{j}f_{j}$
(note that
$0\leq\bar{r}^{2}\lambda_{j}f_{j}\leq\lambda_{j}f_{j}\leq\hat{\lambda}\hat{f}$),
thanks to (3.7) we conclude that $|u_{j}(z)|=|u_{j,z}(0)|\leq M+C(n)$ for all
$j$ sufficiently large, for all $z\in B_{2^{-m_{0}}\bar{r}}$. Choosing
$\hat{r}:=2^{-m_{0}}\bar{r}$, this proves the claim and concludes the proof of
this case.

###  3.2. The case $\lambda_{\infty}=0$

Let $M_{j}=\|\nabla u_{j}\|_{L^{2}(\Omega)}.$ We prove the result by
contradiction, distinguishing between two cases.

Case 1: ${M_{j}}\to 0$ as $j\to\infty$. In this case, Proposition 2.3 implies
that $u_{j}\to 0$ in $C^{2}_{\rm loc}$ outside a set $\Sigma_{\infty}$
consisting of at most $k$ points. Since $\|\nabla u_{j}\|_{L^{2}(\Omega)}\to
0$, with the same notation as in the case $\lambda_{\infty}>0$, we deduce that
(3.8) holds around each point $\hat{x}_{i}\in\Sigma_{\infty}\cap\Omega_{0}$.
Also, by Poincaré and Hölder inequalities,

$\|u_{j}\|_{L^{1}(\Omega)}\leq C(n,\Omega)\|u_{j}\|_{L^{2}(\Omega)}\leq
C(n,\Omega)\|\nabla u_{j}\|_{L^{2}(\Omega)}\to 0.$

Hence, arguing exactly as the previous case, thanks to Proposition 2.7 we
deduce that $|u_{j}(z)|\leq o(1)+C(n)$ for all $j$ sufficiently large, for all
$z\in B_{2^{-m_{0}}\bar{r}}(\hat{x}_{i})$. This implies that
$\sup_{j}\|u_{j}\|_{L^{\infty}(\Omega)}<\infty$, a contradiction.

Case 2: ${M_{j}}$ are uniformly bounded away from $0$. Consider
$v_{j}:=\frac{u_{j}}{M_{j}}$, so that

$-\Delta v_{j}=\lambda_{j}h_{j}(v_{j}),\quad\quad
h_{j}(t):=M_{j}^{-1}f_{j}(M_{j}t),\qquad\|\nabla v_{j}\|_{L^{2}(\Omega)}=1.$
(3.9)

As in the proof of Proposition 2.7, we multiply the equation satisfied by
$v_{j}$ both by $v_{j}$ and by $x\cdot\nabla v_{j}$. Since $v_{j}\geq 0$,
$v_{j}|_{\partial\Omega}=0$, and $\Omega$ convex, as in the classical Derrick-
Pohozaev argument (see, e.g., [17, Proof of Theorem 1, Page 515]) the boundary
terms “have the right sign”, and we get

$\frac{\lambda_{j}}{M_{j}^{2}}\int_{\Omega}f_{j}(M_{j}v_{j})\,M_{j}v_{j}\,dx=\int_{\Omega}|\nabla
v_{j}|^{2}dx+o(1)=1+o(1),$

and

$\frac{\lambda_{j}}{M_{j}^{2}}\int_{\Omega}F_{j}(M_{j}v_{j})\,dx\geq\frac{n-2}{2n}\int_{\Omega}|\nabla
v_{j}|^{2}dx+o(1)=\frac{n-2}{2n}+o(1).$

Now, set $S_{0}:=\\{x\in\Omega\,:\,M_{j}v_{j}\leq t_{0}\\}$ and note that,
since $f_{j}$ satisfies (1.4),

$f_{j}(M_{j}v_{j})\,M_{j}v_{j}\geq\left(\frac{2n}{n-2}+\epsilon\right)F_{j}(M_{j}v_{j})\quad\text{
in }\Omega\setminus S_{0}.$

Also, since $\lambda_{j}\to 0$ and $M_{j}$ is bounded away from $0$,

$\frac{\lambda_{j}}{M_{j}^{2}}\int_{S_{0}}f_{j}(M_{j}v_{j})\,M_{j}v_{j}\,dx\leq\frac{\lambda_{j}}{M_{j}^{2}}\int_{S_{0}}f_{j}(t_{0})\,t_{0}\,dx\to
0,$
$\frac{\lambda_{j}}{M_{j}^{2}}\int_{S_{0}}F_{j}(M_{j}v_{j})\,dx\leq\frac{\lambda_{j}}{M_{j}^{2}}\int_{S_{0}}F_{j}(t_{0})\,dx\to
0.$

Therefore, combining all together,

$1+o(1)=\frac{\lambda_{j}}{M_{j}^{2}}\int_{\Omega\setminus
S_{0}}f_{j}(M_{j}v_{j})\,M_{j}v_{j}\,dx\\\
\geq\left(\frac{2n}{n-2}+\epsilon\right)\frac{\lambda_{j}}{M_{j}^{2}}\int_{\Omega\setminus
S_{0}}F_{j}(M_{j}v_{j})\,dx\geq\left(\frac{2n}{n-2}+\epsilon\right)\frac{n-2}{2n}+o(1),$

a contradiction for $j$ large enough, which concludes the proof of the
theorem.

##  Appendix A Boundedness of stable solutions in $B_{2}\setminus\\{0\\}$ for
$3\leq n\leq 9$

It was shown in [7] that, if $3\leq n\leq 9$ and $u\in W^{1,\,2}(B_{2})\cap
C^{2}(B_{2}\setminus\\{0\\})$ is a radially symmetric stable solution to (1.1)
in $\Omega=B_{2}\setminus\\{0\\}$, then

$\|u\|_{L^{\infty}(B_{1})}\leq C\|u\|_{L^{1}(B_{2})}.$

Namely removing a point does not influence the interior estimate of the
radially symmetric stable solutions.

The aim of this appendix is to show that, combining the approximation argument
in [7] with some modifications of the arguments in [9], we can prove the
following generalization of [9, Theorem 1.2] which is used in the proof of
Proposition 2.3:

######  Proposition A.1.

Let $3\leq n\leq 9$, and let $u\in W^{1,\,2}(B_{2})\cap
C^{2}(B_{2}\setminus\\{0\\})$ be a stable solution to

$-\Delta u=f(u)\qquad\text{in }B_{2}\setminus\\{0\\},$

with $f:\mathbb{R}\to\mathbb{R}$ nonnegative, increasing, and of class
$C^{1}$. Then

$\|u\|_{L^{\infty}(B_{1})}\leq C\|u\|_{L^{1}(B_{2})}$

for some universal constant $C>0.$

We begin by proving the following generalization of [9, Lemma 2.1]:

######  Lemma A.2.

Let $3\leq n\leq 9$, and let $u$ and $f$ be as in Proposition A.1. Then, for
any $y\in B_{1}$ and $0<\rho<2-|y|$ we have

$\int_{B_{2\rho/3}(y)}|(x-y)\cdot\nabla u|^{2}|x-y|^{-n}\,dx\leq
C\rho^{2-n}\int_{B_{\rho}(y)\setminus B_{2\rho/3}(y)}|\nabla u|^{2}\,dx,$
(A.1)

and

$\int_{B_{1}}|\nabla u|^{2}\,dx\leq C\int_{B_{3/2}\setminus B_{1}}|\nabla
u|^{2}\,dx.$ (A.2)

###### Proof.

For simplicity of notation, given $0<r<s<1$, we define $A(s,r):=B_{r}\setminus
B_{s}.$

We shall first prove the following improved version of (A.1): for any $y\in
B_{1}$ and $0<\rho<2-|y|$ we have

$\int_{B_{7\rho/8}(y)}|(x-y)\cdot\nabla u|^{2}|x-y|^{-n}\,dx\leq
C\rho^{2-n}\int_{B_{\rho}(y)\setminus B_{7\rho/8}(y)}|\nabla u|^{2}\,dx.$
(A.3)

We only prove (A.3) in the case $y=0$, the general case being
analogous999Actually the case $y\neq 0$ is simpler, since for $y=0$ the
function $x\mapsto|x-y|^{-n/2}(x-y)\cdot\nabla u(x)$ (that is used as a test
function is the stability inequality) is more singular at the origin..

Fix $0<\theta\ll\epsilon\ll\rho$, $\eta\in C_{c}^{1}(A(\theta,2))$, and
consider $\xi=(x\cdot\nabla u)\eta$ as test function in the stability
inequality for $u$. Then, by the very same computation as the one in [9, Proof
of Lemma 2.1, Step 1], we have

$0\leq\int_{A(\theta,\,2)}\Big{(}(x\cdot\nabla
u)^{2}|\nabla\eta|^{2}+2(x\cdot\nabla u)\nabla u\cdot\nabla(\eta^{2})-|\nabla
u|^{2}x\cdot\nabla(\eta^{2})-(n-2)|\nabla u|^{2}\eta^{2}\Big{)}\,dx.$

If we now choose
$\eta=\min\big{\\{}|x|^{1-\frac{n}{2}},\epsilon^{1-\frac{n}{2}}\big{\\}}\zeta$
with $\zeta\in C^{1}_{c}(A(\theta,2))$, then inside $A(\epsilon,2)$ the
formulas are identical to the ones in [9, Proof of Lemma 2.1, Step 2].
Therefore, in the integrals over $A(\epsilon,2)$ we have exactly all the terms
appearing in [9, Equation (2.2)], and the only difference concerns the
integrals over $A(\theta,\epsilon)$. Note that, inside $A(\theta,\epsilon)$,
it holds

$|\nabla\eta|^{2}=\epsilon^{2-n}|\nabla\zeta|^{2},\qquad\nabla(\eta^{2})=2\epsilon^{2-n}\zeta\nabla\zeta.$

Thus, we obtain

$\displaystyle 0$
$\displaystyle\leq\int_{A(\epsilon,\,2)}\Big{(}-\frac{(n-2)(10-n)}{4}|x|^{-n}(x\cdot\nabla
u)^{2}\zeta^{2}-2|\nabla u|^{2}|x|^{2-n}\zeta\
x\cdot\nabla\zeta+4|x|^{2-n}(x\cdot\nabla u)\zeta\ \nabla u\cdot\nabla\zeta$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad-(n-2)|x|^{-n}\zeta(x\cdot\nabla\zeta)(x\cdot\nabla
u)^{2}+|x|^{2-n}(x\cdot\nabla u)^{2}|\nabla\zeta|^{2}\Big{)}\,dx$
$\displaystyle+\epsilon^{2-n}\int_{A(\theta,\,\epsilon)}\Big{(}(x\cdot\nabla
u)^{2}|\nabla\zeta|^{2}+4(x\cdot\nabla u)\zeta\,\nabla
u\cdot\nabla\zeta-2|\nabla u|^{2}\zeta(x\cdot\nabla\zeta)-(n-2)|\nabla
u|^{2}\zeta^{2}\Big{)}\,dx.$

We now choose $\zeta\in C^{1}_{c}(A(\theta,\rho))$ such that $0\leq\zeta\leq
1$, $\zeta=1$ inside $A(2\theta,\rho/2)$, $|\nabla\zeta|\leq C/\theta$ in
$A(\theta,2\theta)$, and $|\nabla\zeta|\leq C/\rho$ in $A(7\rho/8,\rho)$. With
this choice, the formula above implies

$\displaystyle\frac{(n-2)(10-n)}{4}\int_{A(\epsilon,\,7\rho/8)}|x|^{-n}(x\cdot\nabla
u)^{2}\,dx$ $\displaystyle\leq C\rho^{2-n}\int_{A(7\rho/8,\,\rho)}|\nabla
u|^{2}+C\epsilon^{2-n}\int_{A(\theta,\,2\theta)}|\nabla u|^{2}\,dx,$

so (A.3) follows by letting first $\theta\to 0$ and then $\epsilon\to 0$
(recall that $3\leq n\leq 9$).

Note now that (A.3) readily implies (A.1). Also, as a consequence of (A.3)
applied with $y\in B_{1/8}$ and $\rho=\frac{11}{8}$, we have

$\int_{B_{1}}\Big{|}\frac{x-y}{|x-y|}\cdot\nabla
u\Big{|}^{2}\,dx\leq\int_{B_{\frac{77}{64}}(y)}\Big{|}\frac{x-y}{|x-y|}\cdot\nabla
u\Big{|}^{2}\,dx\leq C\int_{B_{\frac{11}{8}}(y)\setminus
B_{\frac{77}{64}}(y)}|\nabla u|^{2}\,dx\leq C\int_{B_{3/2}\setminus
B_{1}}|\nabla u|^{2}\,dx.$

Hence (A.2) follows by averaging the inequality above with respect to $y\in
B_{1/8}$. ∎

We can now prove the following analogue of [9, Lemma 3.1]:101010In Lemma A.3
we require $3\leq n\leq 9$ since we proved (A.2) as a consequence of (A.3),
and the latter bound requires this dimensional restriction. However, for
$n\geq 10$ one could combine our approximation argument with [9, Proof of
Theorem 7.1] to show that  $\int_{B_{7\rho/8}(y)}|(x-y)\cdot\nabla
u|^{2}|x-y|^{-a}\,dx\leq C\rho^{2-a}\int_{B_{\rho}(y)\setminus
B_{7\rho/8}(y)}|\nabla u|^{2}\,dx.$ for any $a<2(1+\sqrt{n-1}).$ In
particular, choosing $a=2$ and arguing as in the proof of Lemma A.2, one
proves the validity of (A.2) in every dimension. As a consequence, one can
show that Lemma A.3 holds in every dimension.

######  Lemma A.3.

Let $3\leq n\leq 9$, and let $u\in W^{1,\,2}(B_{2})\cap C^{2}(B_{2}\setminus
B_{1/2})$ be a stable solution to

$-\Delta u=f(u)\qquad\text{in }B_{2}\setminus B_{1/2},$

with $f:\mathbb{R}\to\mathbb{R}$ nonnegative, increasing, and of class
$C^{1}$. Assume that

$\int_{B_{1}}|\nabla u|^{2}\,dx\geq\delta\int_{B_{2}}|\nabla u|^{2}\,dx$

for some $\delta>0$. Then there exists a constant $C_{\delta}$ such that

$\int_{B_{3/2}\setminus B_{1}}|\nabla u|^{2}\,dx\leq
C_{\delta}\int_{B_{3/2}\setminus B_{1}}|x\cdot\nabla u|^{2}\,dx.$

###### Proof.

Assume the result to be false. Then, there exists a sequence of stable
solutions to $-\Delta u_{k}=f_{k}(u_{k})$ in $B_{2}\setminus\\{0\\}$, with
$f_{k}:\mathbb{R}\to\mathbb{R}$ nonnegative, increasing, and of class $C^{1}$,
such that

$\int_{B_{1}}|\nabla u_{k}|^{2}\,dx\geq\delta\int_{B_{2}}|\nabla
u_{k}|^{2}\,dx,\qquad\int_{B_{3/2}\setminus B_{1}}|\nabla
u_{k}|^{2}\,dx=1,\quad\text{and}\quad\int_{B_{3/2}\setminus
B_{1}}|x\cdot\nabla u_{k}|^{2}\,dx\to 0.$ (A.4)

Now, thanks to (A.4) and (A.2),

$\int_{B_{2}}|\nabla u_{k}|^{2}\,dx\leq\frac{1}{\delta}\int_{B_{1}}|\nabla
u_{k}|^{2}\,dx\leq\frac{C}{\delta}\int_{B_{3/2}\setminus B_{1}}|\nabla
u_{k}|^{2}\,dx=\frac{C}{\delta}.$ (A.5)

Therefore, since $u_{k}$ is stable in $B_{2}\setminus B_{1/2}$, it follows
from [9, Proposition 2.4] and a standard scaling and covering argument that

$\|\nabla u_{k}\|_{L^{2+\gamma}({B_{3/2}\setminus B_{1}})}\leq C\|\nabla
u_{k}\|_{L^{2}({B_{2}\setminus B_{1/2}})}\leq\frac{C}{\delta}.$

This implies that the sequence of superharmonic functions

$v_{k}:=u_{k}-{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{3/2}\setminus
B_{1}}u_{k}$

satisfies

$\|v_{k}\|_{L^{1}(B_{3/2}\setminus B_{1})}\leq
C\|v_{k}\|_{L^{2}(B_{3/2}\setminus B_{1})}\leq C$

(thanks to (A.5) and Hölder and Poincaré inequalities), as well as

$\|\nabla v_{k}\|_{L^{2}(B_{3/2}\setminus
B_{1})}=1,\quad\|v_{k}\|_{W^{1,2+\gamma}(B_{3/2}\setminus B_{1})}\leq
C,\qquad\int_{B_{3/2}\setminus B_{1}}|x\cdot\nabla v_{k}|^{2}\,dx\to 0.$

Thus, as in the proof of [9, Proposition 2.4], up to a subsequence we have
that $v_{k}\to v$ strongly in $W^{1,2}(B_{3/2}\setminus B_{1})$, where $v$ is
a superharmonic function in $B_{3/2}\setminus B_{1}$ satisfying

$\|\nabla v\|_{L^{2}(B_{3/2}\setminus B_{1})}=1\qquad\mbox{and}\qquad
x\cdot\nabla v\equiv 0\quad\mbox{a.e. in }B_{3/2}\setminus B_{1}.$

Again as in the proof of [9, Proposition 2.4], this implies that $v$ is
constant in $B_{3/2}\setminus B_{1}$, a contradiction that proves the result.
∎

We can now prove the main result of this appendix.

###### Proof of Proposition A.1.

The argument is similar to the one in [9, Proof of Theorem 1.2], with some
minor modifications.

Given $y\in B_{1}$ and $\rho\in(0,1)$ we define the quantities

$\mathcal{D}(\rho,y):=\rho^{2-n}\int_{B_{\rho}(y)}|\nabla
u|^{2}\,dx\qquad\textrm{and}\qquad\mathcal{R}(\rho,y):=\int_{B_{\rho}(y)}|x-y|^{-n}|(x-y)\cdot\nabla
u|^{2}\,dx.$

We claim that there exists a dimensional exponent $\alpha>0$ such that

$\mathcal{R}(\rho,y)\leq C\rho^{2\alpha}\|\nabla
u\|^{2}_{L^{2}(B_{3/2})}\qquad\forall\,\rho\in(0,1/4),\,y\in B_{1}.$ (A.6)

To prove this claim, note that (A.3) implies that

$\mathcal{R}(\rho,y)\leq C\rho^{2-n}\int_{B_{3\rho/2(y)}\setminus
B_{\rho}(y)}|\nabla u|^{2}\,dx\qquad\forall\,\rho\in(0,1/4),\,y\in B_{1}.$
(A.7)

Hence, if $\mathcal{D}(\rho,y)\geq\frac{1}{2}\mathcal{D}(2\rho,y)$ and
$0\not\in B_{2\rho(y)}\setminus B_{\rho/2(y)}$, then we can apply Lemma A.3
with $\delta=1/2$ to the function $u(y+\rho\,\cdot)$ to we deduce that

$\rho^{2-n}\int_{B_{3\rho/2}(y)\setminus B_{\rho}(y)}|\nabla u|^{2}\,dx\leq
C\rho^{-n}\int_{B_{3\rho/2}(y)\setminus B_{\rho}(y)}|(x-y)\cdot\nabla
u|^{2}\,dx\leq C\bigl{(}\mathcal{R}(3\rho/2,y)-\mathcal{R}(\rho,y)\bigr{)}$

for some universal constant $C$. Combining this bound with (A.7) and using
that $\mathcal{R}$ is nondecreasing, we deduce that

$\mathcal{R}(\rho,y)\leq
C\bigl{(}\mathcal{R}(2\rho,y)-\mathcal{R}(\rho,y)\bigr{)}\qquad\text{provided
$\mathcal{D}(\rho,y)\geq{\textstyle\frac{1}{2}}\mathcal{D}(2\rho,y)$ and
$0\not\in B_{2\rho(y)}\setminus B_{\rho/2(y)}$}.$ (A.8)

Note that $0\not\in B_{2\rho(y)}\setminus B_{\rho/2(y)}$ is equivalent to
saying that either $\rho\geq 2|y|$ or $\rho\leq|y|.$

Thus, fixed $y\in B_{1}$, if we define $a_{j}:=\mathcal{D}(2^{-j-2},y)$,
$b_{j}:=\mathcal{R}(2^{-j-2},y)$, and $N:=\lfloor-\log_{2}|y|\rfloor$ (so
$N=\infty$ if $y=0$), then there exists a universal constant $L>1$ such that:

  * (i)

$b_{j}\leq b_{j-1}$ for all $j\geq 1$ (since $\mathcal{R}$ is nondecreasing);

  * (ii)

$a_{j}+b_{j}\leq La_{j-1}$ for all $j\geq 1$ (by (A.7));

  * (iii)

if $a_{j}\geq\frac{1}{2}a_{j-1}$ then $b_{j}\leq L(b_{j-1}-b_{j})$, for all
$j\in\mathbb{N}\setminus\\{N-2,N-1\\}$111111The condition on $j$ guarantees
that either $2^{-j-2}\leq|y|$ or $2^{-j-2}\geq 2|y|$ . (by (A.8)).

Therefore, if we choose $\epsilon>0$ such that
$2^{-\epsilon}=\frac{L^{1+\epsilon}}{1+L}$, and we define
$c_{j}:=a_{j}^{\epsilon}b_{j}$ and
$\theta:=(2^{-\epsilon})^{\frac{1}{1+\epsilon}}\in(0,1)$, then the proof of
[9, Lemma 3.2] shows that

$c_{j+1}\leq\theta c_{j}\qquad\text{for all
$j\in\mathbb{N}\setminus\\{N-2,N-1\\}$},$

which implies that

$c_{j}\leq\theta^{j}c_{0}\quad\text{for }1\leq j\leq N-2,\qquad
c_{j}\leq\theta^{j-N}c_{N}\quad\text{for }j\geq N.$ (A.9)

Also, as a consequence of (i) and (ii) above, we have

$c_{N-1}=a_{N-1}^{\epsilon}b_{N-1}\leq(La_{N-2})^{\epsilon}b_{N-2}\leq
L^{\epsilon}c_{N-2},\qquad
c_{N}=a_{N}^{\epsilon}b_{N}\leq(L^{2}a_{N-2})^{\epsilon}b_{N-2}\leq
L^{2\epsilon}c_{N-2}.$ (A.10)

Hence, combining (A.9) and (A.10) we easily deduce that

$c_{j}\leq L^{2\epsilon}\theta^{j-2}c_{0}\qquad\forall\,j\geq 1.$

As in the proof of [9, Lemma 3.2], this implies that

$b_{j}\leq C(a_{0}+b_{0})\theta^{j}\leq C\|\nabla
u\|^{2}_{L^{2}(B_{3/2})}\theta^{j}\qquad\forall\,j\geq 1,$

so (A.6) follows by choosing $\alpha>0$ so that $2^{-2\alpha}=\theta$.

We now observe that, thanks [9, Proposition 2.5] and a standard scaling and
covering argument, we have $\|\nabla u\|_{L^{2}(B_{3/2}\setminus B_{1})}\leq
C\|u\|_{L^{1}(B_{2}\setminus B_{1/2})}.$ Hence, combining this bound with
(A.6) and (A.2), we obtain

$\mathcal{R}(\rho,y)\leq
C\rho^{2\alpha}\|u\|^{2}_{L^{1}(B_{2})}\qquad\forall\,\rho\in(0,1/4),\,y\in
B_{1}.$

Thanks to this estimate, the argument in [9, Proof of Theorem 1.2, Step 2]
implies that

$[u]_{C^{\alpha}(B_{1})}:=\sup_{x\neq y\in
B_{1}}\frac{|u(x)-u(y)|}{|x-y|^{\alpha}}\leq C\|u\|_{L^{1}(B_{2})},$

and the conclusion follows by the interpolation estimate

$\|u\|_{L^{\infty}(B_{1})}\leq
C\Big{(}[u]_{C^{\alpha}(B_{1})}+\|u\|_{L^{1}(B_{1})}\Big{)}.$

∎

##  Appendix B Uniform boundedness of solutions with spectrum bounded below

Although not relevant for this paper, it is interesting to observe that, by
simply adapting the arguments in [9], one can deduce an a priori bound in
$L^{\infty}$ for solutions of $-\Delta u=f(u)$ whenever the spectrum of the
linearized operator $-\Delta-f^{\prime}(u)$ is contained inside
$[-\Lambda,+\infty)$ for some finite constant $\Lambda\geq 0$. Also, for
finite Morse index solutions, the constant $\Lambda$ depends only on $n$ and
on a maximal finite dimensional subspace $X_{k}$ on which $Q_{u}$ is negative
definite. Unfortunately one cannot hope in general to control $\Lambda$ in
terms only on the Morse index, as can be seen by considering the family of
solutions (1.2) (which has index $1$).

To present this result, consider $u\in C^{2}(B_{2})$ a solution to $-\Delta
u=f(u)$ in $B_{2}$ with ${\rm ind}(u,B_{2})\leq k$, and define

$\widehat{Q}_{u}[\xi,\zeta]:=\int_{B_{2}}\Big{(}\nabla\xi\cdot\nabla\zeta-f^{\prime}(u)\,\xi\,\zeta\Big{)}\,dx.$

Since ${\rm ind}(u,B_{1})\leq k$, there exists a $k$-dimensional set
$X_{k}\subset C^{1}_{c}(B_{1})$ such that, for any $\xi\in C^{1}_{c}(B_{1})$,
we can write $\xi=\xi_{k}+\xi^{\prime}$ with $\xi_{k}\in X_{k}$,
$\widehat{Q}_{u}[\xi^{\prime},\xi^{\prime}]\geq 0$, and
$\widehat{Q}_{u}[\xi^{\prime},\xi_{k}]=0$.

Now, since $X_{k}$ is finite dimensional,

$\sup_{\xi\in
X_{k},\,\|\xi\|_{L^{2}(B_{1})}}\|\xi\|_{L^{\infty}(B_{1})}=:A_{k}<\infty,$

so it follows from Lemma 2.2(ii) (and a covering argument) that

$\inf_{\xi\in
X_{k},\,\|\xi\|_{L^{2}(B_{1})}=1}\int_{B_{1}}\Bigl{(}|\nabla\xi|^{2}-f^{\prime}(u)\xi^{2}\Bigl{)}\,dx\geq-\sup_{\xi\in
X_{k},\,\|\xi\|_{L^{2}(B_{1})}}\|\xi\|_{L^{\infty}(B_{3/4})}^{2}\int_{B_{1}}f^{\prime}(u)\,dx\geq-
C_{n}A_{k}^{2},$

which implies that

$\int_{B_{1}}|\nabla\xi|^{2}\,dx\geq\int_{B_{1}}\big{(}f^{\prime}(u)-\Lambda\big{)}\xi^{2}\,dx\qquad\forall\,\xi\in
C^{1}_{c}(B_{1}),$ (B.1)

where $\Lambda:=C_{n}A_{k}^{2}$. In other words, the spectrum of the operator
$-\Delta-f^{\prime}(u)$ on $L^{2}(B_{1})$ is bounded from below by $-\Lambda$.

In [9, Theorem 1.1], whenever $3\leq n\leq 9$, the authors proved an a priori
$L^{\infty}$ estimate121212Actually, [9, Theorem 1.1] provides a universal
$C^{\alpha}$ bound for some $\alpha>0$. Analogously, also in the general case
$\Lambda\geq 0$ one can prove an interior bound on $\|u\|_{C^{\alpha}}$. for
solutions of $-\Delta u=f(u)$ satisfying (B.1) with $\Lambda=0$. The goal of
this appendix is to show how to extend such a result to the general case
$\Lambda\geq 0.$

######  Proposition B.1.

Let $3\leq n\leq 9$, let $f:\mathbb{R}\to\mathbb{R}$ be nonnegative,
increasing, and of class $C^{1}$, and let $u\in C^{2}(B_{2})$ solve $-\Delta
u=f(u)$ and satisfy (B.1) for some $\Lambda\geq 0$. Then

$\|u\|_{L^{\infty}(B_{1/2})}\leq C(\Lambda)\|u\|_{L^{1}(B_{1})}.$

The proof of Proposition B.1 is very similar to that in [9, Sections 2 & 3],
the main differences being in two interior estimates that we present here.
Once the two lemmas below are available, the proof follows by the same
argument as in [9], and we leave the details to the interested reader.

######  Lemma B.2.

Let $u\in C^{2}(B_{1})$ be as in Proposition B.1. Then, for any $\eta\in
C^{1}_{c}(B_{1})$, we have

$\int_{B_{1}}((n-2)\eta+2x\cdot\nabla\eta)\eta|\nabla
u|^{2}-2(x\cdot\nabla\eta)\nabla u\cdot\nabla(\eta^{2})-|x\cdot\nabla
u|^{2}(|\nabla\eta|^{2}+\Lambda\eta^{2})\,dx\leq 0.$ (B.2)

Thus, for any $\varphi\in C^{1}_{c}(B_{1})$, we have

$\displaystyle\frac{(n-2)(10-n)}{4}\int_{B_{1}}|x|^{-n}|x\cdot\nabla
u|^{2}(1-\Lambda|x|^{2})\varphi^{2}\,dx$
$\displaystyle\leq\int_{B_{1}}\Big{(}-2|x|^{2-n}|\nabla
u|^{2}\varphi(x\cdot\nabla\varphi)+4|x|^{2-n}(x\cdot\nabla u)\varphi\nabla
u\cdot\nabla\varphi\,dx$ $\displaystyle\qquad\qquad+(2-n)|x|^{-n}|x\cdot\nabla
u|^{2}\varphi(x\cdot\nabla\varphi)+|x|^{2-n}|x\cdot\nabla
u|^{2}|\nabla\varphi|^{2}\Big{)}\,dx$ (B.3)

In particular, for $0<r<\frac{1}{2}\min\left\\{1,\Lambda^{-1/2}\right\\}$,

$\int_{B_{r}}|x|^{-n}|x\cdot\nabla u|^{2}\,dx\leq
C(n)r^{2-n}\int_{B_{3r/2}\setminus B_{r}}|\nabla u|^{2}\,dx.$ (B.4)

###### Proof.

We proceed as in [9, Lemma 2.1] and sketch the proof here. First we choose
$\xi=(x\cdot\nabla u)\eta$ in (B.1), with $\eta\in C^{1}_{c}(B_{1})$, to get

$\int_{B_{1}}\left(\Delta(x\cdot\nabla u)+f^{\prime}(u)(x\cdot\nabla
u)\right)(x\cdot\nabla u)\eta^{2}\,dx\leq\int_{B_{1}}(x\cdot\nabla
u)^{2}(|\nabla\eta|^{2}+\Lambda\eta^{2})\,dx.$ (B.5)

Then by noticing that

$\Delta(x\cdot\nabla u)=x\cdot\nabla\Delta u+2\Delta
u=-f^{\prime}(u)(x\cdot\nabla u)+2\Delta u,$ (B.6)

we conclude (B.2). Then (B.2) follows by taking
$\eta=|x|^{-\frac{n-2}{2}}\varphi$, and (B.4) follows by further choosing
$\varphi$ as a cut-off function supported in $B_{3r/2}$ with $\varphi=1$ on
$B_{r}$. ∎

######  Lemma B.3.

Let $u\in C^{2}(B_{1})$ be as in Proposition B.1. Write

$\mathcal{A}=\left(|D^{2}u|^{2}-\frac{|D^{2}u\cdot\nabla u|^{2}}{|\nabla
u|^{2}}\right)^{\frac{1}{2}}\,\text{ when $|\nabla u|\neq 0$,}\qquad\text{ and
}\ \mathcal{A}=0\ \text{ when $|\nabla u|=0$. }$

Then, for any $\eta\in C^{1}_{c}(B_{1})$, we have

$\int_{B_{1}}\mathcal{A}^{2}\eta^{2}\,dx\leq(1+\Lambda)\int_{B_{1}}|\nabla
u|^{2}|\nabla\eta|^{2}\,dx$

###### Proof.

We follow the argument of [9, Lemma 2.3] and again sketch the proof. Set
$u_{i}:=\partial_{i}u$. Multiplying both side of the equation $-\Delta
u_{i}=f^{\prime}(u)u_{i}$ by $u_{i}\eta^{2}$, and summing over $i=1,\ldots,n$,
we get

$\int_{B_{1}}\bigg{(}\sum_{i}|\nabla(u_{i}\eta)|^{2}-|\nabla
u|^{2}|\eta|^{2}\bigg{)}\,dx=\int_{B_{1}}f^{\prime}(u)|\nabla
u|^{2}\eta^{2}\,dx.$

On the other hand, choosing $\xi=|\nabla u|\eta$ in (B.1), we have

$\int_{B_{1}}|\nabla(|\nabla u|\eta)|^{2}+\Lambda|\nabla
u|^{2}\eta^{2}\,dx\geq\int_{B_{1}}f^{\prime}(u)|\nabla u|^{2}\eta^{2}\,dx.$

Thus we obtain

$\int_{B_{1}}|\nabla u|^{2}|\eta|^{2}+\Lambda|\nabla
u|^{2}\eta^{2}\,dx\geq\int_{B_{1}}\left(\sum_{i}|\nabla(u_{i}\eta)|^{2}-|\nabla(|\nabla
u|\eta)|^{2}\right)\,dx,$

and we aconclude the lemma by noticing that

$\sum_{i}|\nabla(u_{i}\eta)|^{2}-|\nabla(|\nabla
u|\eta)|^{2}=\mathcal{A}^{2}\eta^{2}.$

∎

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