$\widetilde{\varphi}^{\prime}_{\lambda}(u)=R_{\lambda^{\ast}}\circ\widehat{\varphi}^{\prime}_{\lambda}(u)\circ
R_{\lambda^{\ast}}^{-1}$. It follows that
$\widetilde{\varphi}^{\prime}_{\lambda}(0)=id_{X}$ and

$\|\widetilde{\varphi}^{\prime}_{\lambda}(u)-\widetilde{\varphi}^{\prime}_{\lambda}(0)\|\leq\|R_{\lambda^{\ast}}\|\|\widehat{\varphi}^{\prime}_{\lambda}(u)-\widehat{\varphi}^{\prime}_{\lambda}(0)\|\|(R_{\lambda^{\ast}})^{-1}\|.$

Since $(\lambda,u)\mapsto\widehat{\varphi}^{\prime}_{\lambda}(u)$ is $C^{0}$,
$\|\widehat{\varphi}^{\prime}_{\lambda}(u)-\widehat{\varphi}^{\prime}_{\lambda}(0)\|\to
0$ as $(\lambda,u)\to(\lambda^{\ast},0)$. Further shrinking $\Lambda_{0}$ and
$\epsilon_{0}>0$ we may get that
$\|[\widetilde{\varphi}_{\lambda}^{\prime}(0)]^{-1}\|_{X}\leq 2$ and
$\|\widetilde{\varphi}_{\lambda}^{\prime}(u)-\widetilde{\varphi}_{\lambda}^{\prime}(0)\|_{X}\leq\frac{1}{4}$
for all $(\lambda,u)\in\Lambda_{0}\times B_{X}(0,\epsilon_{0})$. By Lemma B.4,
$\widetilde{\varphi}_{\lambda}(B_{X}(0,\epsilon_{0}))$ contains
$B_{X}(0,\frac{\epsilon_{0}}{4})$ for each $\lambda\in\Lambda_{0}$.

Summarizing Steps 1, 2 we get the conclusions in the first part.

Step 3. For the “Moreover” part, by the final part in Lemma B.14 the map
$\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X)$
is $C^{1}$, so is
$(\lambda,u)\mapsto\widehat{\mathcal{B}}_{\lambda}(u)=R_{\lambda^{\ast}}{\mathcal{B}}_{\lambda}(R_{\lambda^{\ast}}u)R_{\lambda^{\ast}}$.
Then $\Phi$ and so $\psi$ in Step 1 is $C^{1}$. The desired claim is easily
seen from the above proof. ∎

######  Remark B.5.

The assumptions (f1)-(f2) and (f3) in Theorem B.3 are used to guarantee that
the map $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X)$
defined by (B.4) is $C^{1}$. When $\Lambda=\mathbb{R}$ and
$A_{\lambda}=A-\lambda\widehat{A}$, where $A,\widehat{A}:B_{X}(0,\delta)\to X$
are $C^{2}$ maps corresponding with $C^{3}$ functionals
$\mathscr{L},\widehat{\mathscr{L}}:B_{X}(0,\delta)\to\mathbb{R}$ satisfying
(a) and (b) above (B.1), the assumptions (f1)-(f3) can be replaced by
condition that each of $A^{\prime\prime}$ and $\widehat{A}^{\prime\prime}$ is
either $C^{1}$ or uniformly continuous in some neighborhood of the line
segment $[0,1]u:=\\{tu\,|\,0\leq t\leq 1\\}$ for each $u\in B_{X}(0,\delta)$.
See arguments in Remark B.1 and Lemma B.14.

By slightly strengthening assumptions of [8, Theorem 1.2] we can prove:

######  Theorem B.6.

Let $\mathscr{L}:B_{X}(0,\delta)\to\mathbb{R}$ be a $C^{2}$-functional
satisfying the conditions (a)-(b) above (B.1) and $d\mathscr{L}(0)=0$, and let
$B:B_{X}(0,\delta)\to\mathscr{L}_{s}(H)$ be defined by (B.1). Suppose that the
following two conditions are satisfied:

  1. (i)

There is a $C^{2}$ map $A:B_{X}(0,\delta)\to X$ such that
$d\mathscr{L}(x)[u]=(A(x),u)_{H}$ for all $x\in B_{X}(0,\delta)$ and $u\in X$,
and that $A^{\prime\prime}$ is either $C^{1}$ or uniformly continuous in some
neighborhood of any line segment in $B_{X}(0,\delta)$.

  2. (ii)

$0\in\sigma(B(0)|_{X})$ and $\sigma(B(0)|_{X})\setminus\\{0\\}$ is bounded
away from the imaginary axis.

Let $X=X_{0}\oplus X_{+}\oplus X_{-}$ be a direct sum decomposition of Banach
spaces, which corresponds to the spectral sets $\\{0\\}$,
$\sigma_{+}(B(0)|_{X})$ and $\sigma_{-}(B(0)|_{X})$ as above. Denote by
$P_{0}$, $P_{+}$ and $P_{-}$ the corresponding projections as in Lemma B.2.
Then there exists $\epsilon\in(0,\delta)$, a $C^{2}$ map
$h:B_{X}(0,\epsilon)\cap X_{0}\to X_{+}\oplus X_{-}$ with $h(0)=0$ (which is
also $C^{3}$ if $A$ is $C^{3}$), and a $C^{1}$ origin-preserving
diffeomorphism $\varphi$ from $B_{X}(0,\epsilon)$ onto an open neighborhood of
$0$ in $X$ such that

$\displaystyle\mathscr{L}\circ\varphi(x)=\frac{1}{2}(B(0)x,x)_{H}+\mathscr{L}^{\circ}(P_{0}x),\quad\forall
x\in B_{X}(0,\epsilon).$ (B.7)

where the functional $\mathscr{L}^{\circ}:B_{X_{0}}(0,\epsilon)\to\mathbb{R}$,
defined by $\mathscr{L}^{\circ}(z)=\mathscr{L}(z+h(z))$, is of class $C^{2}$,
and has the first-order derivative at $z_{0}\in B_{X_{0}}(0,\epsilon)$ given
by

$\displaystyle
d\mathscr{L}^{\circ}(z_{0})[z]=\bigl{(}A(z_{0}+h(z_{0})),z\bigr{)}_{H},\quad\forall
z\in X_{0},$ (B.8)

and the second-order derivative at $0\in B_{X_{0}}(0,\epsilon)$,
$d^{2}\mathscr{L}^{\circ}(0)=0$. Moreover, after suitably shrinking
$\epsilon>0$ the diffeomorphism $\varphi$ can be chosen to satisfy

$\mathscr{L}\circ\varphi(x)=\|P_{+}x\|_{H}^{2}-\|P_{-}x\|_{H}^{2}+\mathscr{L}(h(P_{0}x)+P_{0}x),\quad\forall
x\in B_{X}(0,\epsilon).$

(B.7) and (B.8) imply that $0\in X_{0}$ is an isolated critical point of
$\mathscr{L}^{\circ}$ if and only if $0\in X$ is such a critical point of of
$\mathscr{L}$.

###### Proof of Theorem B.6.

For conveniences let $P_{\pm}:=P_{+}+P_{-}$, $X_{\pm}:=X_{+}+X_{-}$ and
$H_{\pm}$ be the closure of $X_{\pm}$ in $H$. A standard implicit function
theorem argument yields a $\epsilon\in(0,\delta)$ and a $C^{2}$ map
$h:B_{X}(0,\epsilon)\cap X_{0}\to X_{\pm}$ with $h(0)=0$ (which is also
$C^{3}$ if $A$ is $C^{3}$), such that

$\displaystyle(id_{X}-P_{0})A(z+{h}(z))=0,\quad\forall z\in
B_{X_{0}}(0,\epsilon).$

For each $z\in\Lambda_{0}\times B_{X_{0}}(0,\epsilon)$, define

$\displaystyle{\bf
L}_{z}:B_{X_{\pm}}(0,\delta/2)\to\mathbb{R},\;u\mapsto\mathscr{L}(z+{h}(z)+u)-\mathscr{L}(z+{h}(z)),$
$\displaystyle{\bf A}_{z}:B_{X_{\pm}}(0,\delta/2)\to X_{\pm},\;u\mapsto
P_{\pm}A(z+{h}(z)+u),$ $\displaystyle{\bf
B}_{z}:B_{X_{\pm}}(0,\delta/2)\to\mathscr{L}_{s}(H_{\pm}),\;u\mapsto
P_{\pm}\circ(B(z+{h}(z)+u)|_{H_{\pm}}).$

Then $d{\bf L}_{z}(0)=0$, $d{\bf L}_{z}(x)[u]=({\bf A}_{z}(x),u)_{H}$ and
$d^{2}{\bf L}_{z}(x)[u,v]=({\bf B}_{z}(x)u,v)_{H}$ for all $x\in
B_{X_{\pm}}(0,\delta/2)$ and $u,v\in X_{\pm}$. By (ii), ${\bf
B}_{0}(0)|_{X_{\pm}}=P_{\pm}\circ(B(0)|_{X_{\pm}})\in\mathscr{L}(X_{\pm})$ is
hyperbolic and its spectrum is bounded away from the imaginary axis. Since
${\bf A}_{z}$ is $C^{2}$, we may define
$\mathcal{B}:B_{X_{0}}(0,\epsilon)\times B_{X_{0}}(0,\epsilon)\to X_{\pm}$ by

$\mathcal{B}(z,u):=\int^{1}_{0}(1-t){\bf
A}^{\prime}_{z}(tu)dt=\int^{1}_{0}(1-t)P_{\pm}A^{\prime}(z+{h}(z)+tu)|_{X_{\pm}}dt.$

Then ${\bf L}_{z}(u)=(\mathcal{B}(z,u)u,u)_{H}$. For any
$(z,u),(z_{0},u_{0})\in B_{X_{0}}(0,\epsilon)\times B_{X_{\pm}}(0,\delta/2)$
we have

$\mathcal{B}(z,u)-\mathcal{B}(z_{0},u_{0})=\int^{1}_{0}(1-t)P_{\pm}[A^{\prime}(z+{h}(z)+tu)|_{X_{\pm}}-A^{\prime}(z_{0}+{h}(z_{0})+tu_{0})|_{X_{\pm}}]dt.$

By (i), $A$ is $C^{2}$, $A^{\prime\prime}$ is bounded in a convex neighborhood
of the line segment $z_{0}+h(z_{0})+[0,1]u_{0}$. It follows from the mean
value theorem that $\mathcal{B}(z,u)\to\mathcal{B}(z_{0},u_{0})$ as
$(z,u)\to(z_{0},u_{0})$.

As before, since $A$ is $C^{2}$, by the differential method with parameter
integral we can get Gâteaux derivatives $D_{u}\mathcal{B}(z,u)$ and
$D_{z}\mathcal{B}(z,u)$,

$\displaystyle
D_{u}\mathcal{B}(z,u)[u^{\prime}]=\int^{1}_{0}(1-t)tP_{\pm}A^{\prime\prime}(z+{h}(z)+tu)[u^{\prime}]dt,\quad
u^{\prime}\in{X_{\pm}},$ $\displaystyle
D_{z}\mathcal{B}(z,u)[z^{\prime}]=\int^{1}_{0}(1-t)P_{\pm}A^{\prime\prime}(z+{h}(z)+tu)[z^{\prime}+h^{\prime}(z)[z^{\prime}]]dt,\quad
z^{\prime}\in{X_{0}}.$

Let $C_{0}:=\|P_{\pm}\|_{\mathscr{L}(X,X_{\pm})}$. For any
$(z,u),(z_{0},u_{0})\in B_{X_{0}}(0,\epsilon)\times B_{X_{\pm}}(0,\delta/2)$
it easily follows that

$\displaystyle\|D_{u}\mathcal{B}(z,u)-D_{u}\mathcal{B}(z_{0},u_{0})\|_{\mathscr{L}(X_{\pm},\mathscr{L}(X_{\pm}))}$
$\displaystyle\leq$ $\displaystyle
C_{0}\int^{1}_{0}\|A^{\prime\prime}(z+{h}(z)+tu)-A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}dt$

and

$\displaystyle\|D_{z}\mathcal{B}(z,u)-D_{z}\mathcal{B}(z_{0},u_{0})\|_{\mathscr{L}(X_{0},\mathscr{L}(X_{\pm}))}$
$\displaystyle\leq$ $\displaystyle
C_{0}\int^{1}_{0}\|A^{\prime\prime}(z+{h}(z)+tu)-A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}dt$
$\displaystyle+$ $\displaystyle
C_{0}\int^{1}_{0}\|A^{\prime\prime}(z+{h}(z)+tu)\circ h^{\prime}(z)$
$\displaystyle-A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\circ
h^{\prime}(z_{0})\|_{\mathscr{L}(X_{0},\mathscr{L}(X))}dt.$

Observe that

$\displaystyle\|A^{\prime\prime}(z+{h}(z)+tu)\circ
h^{\prime}(z)-A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\circ
h^{\prime}(z_{0})\|_{\mathscr{L}(X_{0},\mathscr{L}(X))}$ $\displaystyle\leq$
$\displaystyle\|A^{\prime\prime}(z+{h}(z)+tu)\circ
h^{\prime}(z)-A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\circ
h^{\prime}(z)\|_{\mathscr{L}(X_{0},\mathscr{L}(X))}$ $\displaystyle+$
$\displaystyle\|A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\circ
h^{\prime}(z)-A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\circ
h^{\prime}(z_{0})\|_{\mathscr{L}(X_{0},\mathscr{L}(X))}$ $\displaystyle\leq$
$\displaystyle\|A^{\prime\prime}(z+{h}(z)+tu)-A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}\|h^{\prime}(z)\|_{\mathscr{L}(X_{0},X)}$
$\displaystyle+$
$\displaystyle\|A^{\prime\prime}(z_{0}+{h}(z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}\|h^{\prime}(z)-h^{\prime}(z_{0})\|_{\mathscr{L}(X_{0},X)}.$

As above, by (i) we can derive from these inequalities that both
$D_{u}\mathcal{B}$ and $D_{z}\mathcal{B}$ are continuous. The other arguments
are standard, see the following proof of Theorem B.8. ∎

######  Remark B.7.

(I). (B.7) was proved in [23, page 436, Theorem 1] when $\mathscr{L}$ is of
class $C^{r}$, $A:B_{X}(0,\delta)\to X$ is $C^{r-1}$, and the operator
$B(0)|_{X}=dA(0)\in\mathscr{L}(X)$ restricts an isomorphism on $X_{+}+X_{-}$.
Correspondingly, (B.2) with $\Lambda=\\{\lambda^{\ast}\\}$ was proved in [23,
page 425, Theorem 2] ([71, Theorem 1.4] with $r=3$) under similar conditions.
As above we cannot prove that the $C^{r-2}$-smoothness of the map $A$ in the
proof of [23, page 425, Theorem 2] if $r=3$. There exists the same problem in
the proof [23, page 436, Theorem 1] as well.

(II). When $\varphi$ is only required to be a homeomorphism, using a different
method (thus not involving the above problem) Ming Jiang [33, Theorem 2.5]
proved (B.7) if the conditions (i)–(ii) are replaced by the following:

  1. 1)

$A:B_{X}(0,\delta)\to X$ is $C^{1}$,

  2. 2)

$B:B_{X}(0,\delta)\to\mathscr{L}_{s}(H)$ is continuous,

  3. 3)

C2) in Hypothesis 1.3, and either $0\notin\sigma(B(0))$ or $0$ is an isolated
point of $\sigma(B(0))$.

Here is a parametric version of Theorem B.6.

######  Theorem B.8.

Suppose that the conditions (d’) and (h) in Theorem B.3 are respectively
replaced by the following:

  1. (sd’)

For each point $\lambda_{0}\in\Lambda$ and any line segment $\ell\subset
B_{X}(0,\delta)$ there exist neighborhoods $\mathscr{N}(\lambda_{0})$ of
$\lambda_{0}$ in $\Lambda$, and $\mathscr{N}(\ell)$ of $\ell$ in
$B_{X}(0,\delta)$ such that

$\mathscr{N}(\lambda_{0})\times\mathscr{N}(\ell)\ni(\lambda,u)\mapsto
A_{\lambda}(u)\in X$

is continuous and that

$\displaystyle\mathscr{N}(\lambda_{0})\times\mathscr{N}(\ell)\ni(\lambda,u)\mapsto
dA_{\lambda}(u)\in\mathscr{L}(X)\quad\hbox{and}$
$\displaystyle\mathscr{N}(\lambda_{0})\times\mathscr{N}(\ell)\ni(\lambda,u)\mapsto
d^{2}A_{\lambda}(u)\in\mathscr{L}(X,\mathscr{L}(X))$

are uniform continuous.

  2. (wh)

For some $\lambda^{\ast}\in\Lambda$,
$\sigma(B_{\lambda^{\ast}}(0)|_{X})\setminus\\{0\\}$ is bounded away from the
imaginary axis.

Let $X=X_{0}^{\lambda^{\ast}}\oplus X_{+}^{\lambda^{\ast}}\oplus
X_{-}^{\lambda^{\ast}}$ be a direct sum decomposition of Banach spaces, which
corresponds to the spectral sets $\\{0\\}$,
$\sigma_{+}(B_{\lambda^{\ast}}(0)|_{X})$ and
$\sigma_{-}(B_{\lambda^{\ast}}(0)|_{X})$, and let $P_{0}^{\lambda^{\ast}}$,
$P_{+}^{\lambda^{\ast}}$ and $P_{-}^{\lambda^{\ast}}$ denote the corresponding
projections. Then there exists a neighborhood $\Lambda_{0}$ of
$\lambda^{\ast}$ in $\Lambda$, $\epsilon>0$, and

  1. (i)

a (unique) $C^{0}$ map

$\displaystyle\mathfrak{h}:\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\to X_{+}^{\lambda^{\ast}}\oplus
X_{-}^{\lambda^{\ast}}$ (B.9)

which is $C^{2}$ in the second variable and satisfies
$\mathfrak{h}(\lambda,0)=0\;\forall\lambda\in\Lambda_{0}$ and

$\displaystyle(id_{X}-P_{0}^{\lambda^{\ast}})A_{\lambda}(z+\mathfrak{h}(\lambda,z))=0,\quad\forall(\lambda,z)\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon),$ (B.10)

  2. (ii)

a $C^{0}$ map

$\displaystyle\Lambda_{0}\times B_{X}(0,\epsilon)\to
X,\quad({\lambda},x)\mapsto\Phi_{{\lambda}}(x)$ (B.11)

such that for each $\lambda\in\Lambda_{0}$ the map $\Phi_{\lambda}$ is an
origin-preserving $C^{1}$-differeomor-
phism from $B_{X}(0,\epsilon)$ onto an open neighborhood of zero of $X$
containing $B_{X}\\\ (0,\epsilon/4)$ and satisfies

$\displaystyle\mathscr{L}_{\lambda}(\Phi_{\lambda}(x))=\frac{1}{2}(B_{\lambda^{\ast}}(0)P_{\pm}^{\lambda^{\ast}}x,P_{\pm}^{\lambda^{\ast}}x)_{H}+\mathscr{L}_{\lambda}(P_{0}^{\lambda^{\ast}}x+\mathfrak{h}(\lambda,P_{0}^{\lambda^{\ast}}x))$
(B.12)

for any $x\in B_{X}(0,\epsilon)$; moreover
$({\lambda},x)\mapsto\Phi^{\prime}_{{\lambda}}(x)$ is continuous.

In addition, with notations
$X_{\pm}^{\lambda^{\ast}}:=X_{+}^{\lambda^{\ast}}\oplus
X_{-}^{\lambda^{\ast}}$ and
$P_{\pm}^{\lambda^{\ast}}:=P_{+}^{\lambda^{\ast}}+P_{-}^{\lambda^{\ast}}$, we
also have:

  1. (iii)

$(\lambda,z)\mapsto d_{z}\mathfrak{h}(\lambda,z)$ is continuous and

$d_{z}\mathfrak{h}(\lambda,z)=-[P_{\pm}^{\lambda^{\ast}}\circ({B}_{\lambda}(z+\mathfrak{h}(\lambda,z))|_{X_{\pm}^{\lambda^{\ast}}})]^{-1}\circ(P_{\pm}^{\lambda^{\ast}}\circ({B}_{\lambda}(z+\mathfrak{h}(\lambda,z))|_{X_{0}^{\lambda^{\ast}}})).$
(B.13)

  2. (iv)

The functional

$\mathscr{L}_{\lambda}^{\circ}:B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\to\mathbb{R},\;z\mapsto\mathscr{L}_{\lambda}(z+\mathfrak{h}({\lambda},z))$
(B.14)

is of class $C^{2}$, and has the first-order derivative at $z_{0}\in
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)$ given by

$\displaystyle
d\mathscr{L}^{\circ}_{\lambda}(z_{0})[z]=\bigl{(}A_{\lambda}(z_{0}+\mathfrak{h}(\lambda,z_{0})),z\bigr{)}_{H},\quad\forall
z\in X_{0}^{\lambda^{\ast}},$ (B.15)

and the second-order derivative at $0\in
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)$

$\displaystyle
d^{2}\mathscr{L}^{\circ}_{\lambda}(0)[z,z^{\prime}]=\left(P_{0}^{\lambda^{\ast}}\bigr{[}{B}_{\lambda}(0)-{B}_{\lambda}(0)(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0)|_{X_{\pm}^{\lambda^{\ast}}})^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))\bigr{]}z,z^{\prime}\right)_{H},$
(B.16) $\displaystyle\hskip 113.81102pt\forall z,z^{\prime}\in
X_{0}^{\lambda^{\ast}}.$

  3. (v)

The map $z\mapsto z+\mathfrak{h}({\lambda},z))$ induces an one-to-one
correspondence between the critical points of $\mathscr{L}_{\lambda}^{\circ}$
near $0\in X_{0}^{\lambda^{\ast}}$ and those of $\mathscr{L}_{\lambda}$ near
$0\in X$.

Let $G$ be a compact Lie group acting on $H$ orthogonally, which induces a
$C^{1}$ isometric action on $X$. Suppose that each $\mathscr{L}_{\lambda}$ is
$G$-invariant and that $A_{\lambda},B_{\lambda}$ are equivariant. Then for
each $\lambda\in\Lambda_{0}$, both $\mathfrak{h}(\lambda,\cdot)$ and
$\Phi_{\lambda}$ are equivariant, and $\mathscr{L}_{\lambda}^{\circ}$ is
$G$-invariant.

Moreover, if $\Lambda$ is a nonempty open subset of a Banach space $Z$, then
the maps in (B.9) and (B.11) are of class $C^{1}$ provided that (sd’) is
replaced by the following three conditions:

  1. (f1)

$\Lambda\times B_{X}(0,\delta)\ni(\lambda,u)\mapsto
A(\lambda,u):=A_{\lambda}(u)\in X$ and

$\Lambda\times B_{X}(0,\delta)\ni(\lambda,u)\mapsto
D_{2}A(\lambda,u)\in\mathscr{L}(X)$

are $C^{1}$. Here $D_{1}$ and $D_{2}$ are differentials of $A$ with respect to
$\lambda$ and $u$, respectively.

  2. (sf2)

Either $\Lambda\times B_{X}(0,\delta)\ni(\lambda,u)\mapsto
D^{2}_{2}A(\lambda,u)\in\mathscr{L}(X,\mathscr{L}(X))$ is $C^{1}$, or for each
point $\lambda_{0}\in\Lambda$ and any line segment $\ell\subset
B_{X}(0,\delta)$ there exist neighborhoods $\mathscr{N}(\lambda_{0})$ of
$\lambda_{0}$ in $\Lambda$, and $\mathscr{N}(\ell)$ of $\ell$ in
$B_{X}(0,\delta)$ such that

$\mathscr{N}(\lambda_{0})\times\mathscr{N}(\ell)\ni(\lambda,u)\mapsto
D^{2}_{2}A(\lambda,u)\in\mathscr{L}(X,\mathscr{L}(X))$

is uniform continuous.

  3. (sf3)

Either $\Lambda\times B_{X}(0,\delta)\ni(\lambda,u)\mapsto
D_{1}D_{2}A(\lambda,u)\in\mathscr{L}(Z,\mathscr{L}(X))$ is $C^{1}$, or for
each point $\lambda_{0}\in\Lambda$ and any line segment $\ell\subset
B_{X}(0,\delta)$ there exist neighborhoods $\mathscr{N}(\lambda_{0})$ of
$\lambda_{0}$ in $\Lambda$, and $\mathscr{N}(\ell)$ of $\ell$ in
$B_{X}(0,\delta)$ such that

$\mathscr{N}(\lambda_{0})\times\mathscr{N}(\ell)\ni(\lambda,u)\mapsto
D_{1}D_{2}A(\lambda,u)\in\mathscr{L}(Z,\mathscr{L}(X))$

is uniform continuous.

Clearly, (sd’) and these three assumptions hold if $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto A(\lambda,u)\in X$ is $C^{3}$.

###### Proof.

Step 1. By the first two conditions in (sd’), and (wh), a standard implicit
function theorem argument yields (i) and (iii). As in Step 2 of the proof of
[41, Lemma 3.1] we can get (B.15). Then it and (B.13) lead to (B.16).

By shrinking $\Lambda_{0}$ and $\epsilon$ we can assume that
$z+\mathfrak{h}(\lambda,z))\in B_{X}(0,\delta/2)$ for all
$(\lambda,z)\in\Lambda_{0}\times B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)$. Let
$H_{\pm}^{\lambda^{\ast}}$ be the closure of $X_{\pm}^{\lambda^{\ast}}$ in
$H$. For each $(\lambda,z)\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)$, define maps

$\displaystyle{\bf
L}_{(\lambda,z)}:B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)\to\mathbb{R},\;u\mapsto\mathscr{L}_{\lambda}(z+\mathfrak{h}(\lambda,z)+u)-\mathscr{L}_{\lambda}(z+\mathfrak{h}(\lambda,z)),$
$\displaystyle{\bf
A}_{(\lambda,z)}:B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)\to
X_{\pm}^{\lambda^{\ast}},\;u\mapsto
P_{\pm}^{\lambda^{\ast}}A_{\lambda}(z+\mathfrak{h}(\lambda,z)+u),$
$\displaystyle{\bf
B}_{(\lambda,z)}:B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)\to\mathscr{L}_{s}(H_{\pm}^{\lambda^{\ast}}),\;u\mapsto
P_{+}^{\lambda^{\ast}}\circ(B_{\lambda}(z+\mathfrak{h}(\lambda,z)+u)|_{H_{\pm}^{\lambda^{\ast}}}).$

Then $d{\bf L}_{(\lambda,z)}(0)=0$, $d{\bf L}_{(\lambda,z)}(x)[u]=({\bf
A}_{(\lambda,z)}(x),u)_{H}$ and $d^{2}{\bf L}_{(\lambda,z)}(x)[u,v]=({\bf
B}_{(\lambda,z)}(x)u,v)_{H}$ for all $x\in
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)$ and for all $u,v\in
X_{\pm}^{\lambda^{\ast}}$. Moreover, the condition (wh) implies that

${\bf
B}_{(\lambda^{\ast},0)}(0)|_{X_{\pm}^{\lambda^{\ast}}}=P_{\pm}^{\lambda^{\ast}}\circ(B_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}})\in\mathscr{L}(X_{\pm}^{\lambda^{\ast}})$

is hyperbolic and its spectrum is bounded away from the imaginary axis. Since
${\bf A}_{(\lambda,z)}$ is $C^{2}$, we can define a map
$\mathcal{B}:\Lambda_{0}\times B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)\to\mathscr{L}(X_{\pm}^{\lambda^{\ast}})$
by

$\displaystyle\mathcal{B}(\lambda,z,u):=\int^{1}_{0}(1-t){\bf
A}^{\prime}_{(\lambda,z)}(tu)dt=\int^{1}_{0}(1-t){\bf
B}_{(\lambda,z)}(tu)|_{X_{\pm}^{\lambda^{\ast}}}dt$ (B.17)

for any $(\lambda,z,u)\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)$. Then ${\bf
L}_{(\lambda,z)}(u)=(\mathcal{B}(\lambda,z,u)u,u)_{H}$. As in Step 1 of the
proof of Lemma B.14, we can use the first condition in (sd’) to show that
$\mathcal{B}$ is continuous.

By the proof of Theorem B.6 it is easily seen that $\mathcal{B}(\lambda,z,u)$
has Gâteaux derivatives $D_{u}\mathcal{B}(\lambda,z,u)$ in $u$ and
$D_{z}\mathcal{B}(\lambda,z,u)$ in $z$,

$\displaystyle
D_{u}\mathcal{B}(\lambda,z,u)[u^{\prime}]=\int^{1}_{0}(1-t)tP_{\pm}^{\lambda^{\ast}}A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)[u^{\prime}]dt,\quad
u^{\prime}\in X_{\pm}^{\lambda^{\ast}},$ $\displaystyle
D_{z}\mathcal{B}(\lambda,z,u)[z^{\prime}]$
$\displaystyle=\int^{1}_{0}(1-t)P_{\pm}^{\lambda^{\ast}}A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)[z^{\prime}+D_{z}\mathfrak{h}(\lambda,z)[z^{\prime}]]dt,\quad
z^{\prime}\in X_{0}^{\lambda^{\ast}}.$

Let
$C:=\|P_{\pm}^{\lambda^{\ast}}\|_{\mathscr{L}(X,X_{\pm}^{\lambda^{\ast}})}$.
For any $(\lambda,z,u),(\lambda_{0},z_{0},u_{0})\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)$ it easily follows from the first
equation that

$\displaystyle\|D_{u}\mathcal{B}(\lambda,z,u)-D_{u}\mathcal{B}(\lambda_{0},z_{0},u_{0})\|_{\mathscr{L}(X_{\pm}^{\lambda^{\ast}},\mathscr{L}(X_{\pm}^{\lambda^{\ast}}))}$
$\displaystyle\leq$ $\displaystyle
C\int^{1}_{0}\|A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)-A^{\prime\prime}_{\lambda_{0}}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}dt.$

As in Step 3 of the proof of Lemma B.14, we can use the second condition in
(sd’) to show that $D_{u}\mathcal{B}$ is continuous at
$(\lambda_{0},z_{0},u_{0})$.

Similarly, as in the proof of Theorem B.6 we derive

$\displaystyle\|D_{z}\mathcal{B}(\lambda,z,u)-D_{z}\mathcal{B}(\lambda_{0},z_{0},u_{0})\|_{\mathscr{L}(X_{0}^{\lambda^{\ast}},\mathscr{L}(X_{\pm}^{\lambda^{\ast}}))}$
$\displaystyle\leq
C\int^{1}_{0}\|A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)-A^{\prime\prime}_{\lambda_{0}}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}dt$
$\displaystyle+C\int^{1}_{0}\|A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)\circ\mathfrak{h}^{\prime}_{z}(\lambda,z)$
$\displaystyle-A^{\prime\prime}_{\lambda_{0}}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\circ\mathfrak{h}^{\prime}_{z}(\lambda_{0},z_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}dt$

and

$\displaystyle\|A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)\circ\mathfrak{h}^{\prime}_{z}(\lambda,z)$
$\displaystyle-A^{\prime\prime}_{\lambda_{0}}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\circ\mathfrak{h}^{\prime}_{z}(\lambda_{0},z_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}$
$\displaystyle\leq$
$\displaystyle\|A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)$
$\displaystyle-A^{\prime\prime}_{\lambda_{0}}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}\|\mathfrak{h}^{\prime}_{z}(\lambda,z)\|_{\mathscr{L}(X_{0},X)}$
$\displaystyle+$
$\displaystyle\|A^{\prime\prime}_{\lambda}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}\|\mathfrak{h}^{\prime}_{z}(\lambda,z)-\mathfrak{h}^{\prime}_{z}(\lambda_{0},z_{0})\|_{\mathscr{L}(X_{0},X)}.$

By (B.13), $(\lambda,z)\mapsto\mathfrak{h}^{\prime}_{z}(\lambda,z)$ is
continuous. Using the second condition in (sd’) again we deduce that

$\sup_{t\in[0,1]}\|A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)-A^{\prime\prime}_{\lambda_{0}}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}\to
0$

as $\lambda\to\lambda_{0}$ and $u\to u_{0}$, and that
$\|A^{\prime\prime}_{\lambda}(z+\mathfrak{h}(\lambda,z)+tu)\|_{\mathscr{L}(X,\mathscr{L}(X))}$
uniformly converges to
$\|A^{\prime\prime}_{\lambda_{0}}(z_{0}+\mathfrak{h}(\lambda_{0},z_{0})+tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}$
with respect to $t\in[0,1]$. Hence $D_{z}\mathcal{B}$ is continuous at
$(\lambda_{0},z_{0},u_{0})$.

In summary, we have proved that $(\lambda,z,u)\mapsto\mathcal{B}(\lambda,z,u)$
is continuous, $C^{1}$ in each of $z$ and $u$, and both
$(\lambda,z,u)\mapsto\mathcal{B}^{\prime}_{z}(\lambda,z,u)$ and
$(\lambda,z,u)\mapsto\mathcal{B}^{\prime}_{u}(\lambda,z,u)$ are continuous.

Step 2. Since $\mathcal{B}(\lambda^{\ast},0,0)=\frac{1}{2}{\bf
B}_{(\lambda^{\ast},0)}(0)|_{X_{\pm}^{\lambda^{\ast}}}$ we can take $M>0$ such
that

$\displaystyle\sigma(\mathcal{B}({\lambda^{\ast}},0,0))_{+}\cap(-M^{2}\sigma(\mathcal{B}({\lambda^{\ast}},0,0))_{-})=\emptyset.$
(B.18)

Set
$R_{\lambda^{\ast}}:=P_{+}^{\lambda^{\ast}}|_{X_{\pm}^{\lambda^{\ast}}}+MP_{-}^{\lambda^{\ast}}|_{X_{\pm}^{\lambda^{\ast}}}$
and $\widehat{\bf L}_{(\lambda,z)}:={\bf L}_{(\lambda,z)}\circ
R_{\lambda^{\ast}}$. As before we have

$\displaystyle\widehat{\bf
L}_{(\lambda,z)}(u)=(\widehat{\mathcal{B}}(\lambda,z,u)u,u)_{H},\quad\forall
u\in B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{1}),$ (B.19)

where $\delta_{1}>0$ is such that
$R_{\lambda^{\ast}}\left(B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{1})\right)\subset
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)$, and

$\displaystyle\widehat{\mathcal{B}}(\lambda,z,u):=R_{\lambda^{\ast}}{\mathcal{B}}(\lambda,z,R_{\lambda^{\ast}}u)R_{\lambda^{\ast}}.$
(B.20)

The latter implies
$\sigma(\widehat{\mathcal{B}}({\lambda^{\ast}},0,0))=\sigma(\mathcal{B}({\lambda^{\ast}},0,0))_{+}\cup(M^{2}\sigma(\mathcal{B}({\lambda^{\ast}},0,0))_{-})$
and hence

$\displaystyle\sigma(\widehat{\mathcal{B}}({\lambda^{\ast}},0,0))\cap(-\sigma(\widehat{\mathcal{B}}({\lambda^{\ast}},0,0)))=\emptyset$
(B.21)

by (B.18). Consider the map

$\Psi:\Lambda_{0}\times B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{1})\times\mathscr{S}(X_{\pm}^{\lambda^{\ast}})\to\mathscr{S}(X_{\pm}^{\lambda^{\ast}})$

defined by
$\Psi(\lambda,z,u,Y)=Y\widehat{\mathcal{B}}(\lambda,z,Yu)Y-\widehat{\mathcal{B}}(\lambda^{\ast},0,0)$.
Then $\Psi$ is continuous, $C^{1}$ in $(z,u,Y)$, and
$\Psi^{\prime}_{z}(\lambda,z,u,Y)$, $\Psi^{\prime}_{u}(\lambda,z,u,Y)$ and
$\Psi^{\prime}_{Y}(\lambda,z,u,Y)$ continuously depend on $(\lambda,z,u,Y)$.
Since

$D_{Y}\Psi(\lambda^{\ast},0,0,id_{X}):\mathscr{S}(X_{\pm}^{\lambda^{\ast}})\to\mathscr{S}(X_{\pm}^{\lambda^{\ast}}),\;Z\mapsto\widehat{\mathcal{B}}({\lambda^{\ast}},0,0)Z+Z\widehat{\mathcal{B}}({\lambda^{\ast}},0,0)$

is a Banach space isomorphism by (B.21) and [8, Lemma 1.4], as in the proof of
Theorem B.3 using the implicit function theorem we can, by shrinking the
neighborhood $\Lambda_{0}$ of $\lambda^{\ast}$ and $\epsilon>0$, get
$\delta_{2}\in(0,\delta_{1})$ and a continuous map

$\psi:\Lambda_{0}\times B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{2})\to\mathscr{S}(X_{\pm}^{\lambda^{\ast}})$

such that
$\bullet$ for all $(\lambda,z,u)\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{2})$,

$\left.\begin{array}[]{ll}&\psi(\lambda^{\ast},0,0)=id_{X_{\pm}^{\lambda^{\ast}}},\quad\Psi(\lambda,z,u,\psi(\lambda,z,u))\equiv
0\quad\hbox{and}\\\
&\|\psi(\lambda,z,u)-\psi(\lambda^{\ast},0,0)\|<1/2,\end{array}\right\\}$
(B.22)

$\bullet$ $\psi(\lambda,z,u)$ is $C^{1}$ in $(z,u)$, both
$\psi^{\prime}_{z}(\lambda,z,u)$ and $\psi^{\prime}_{u}(\lambda,z,u)$
continuously depend on $(\lambda,z,u)$. Clearly, (B.22) implies that each
operator $\psi(\lambda,z,u):X_{\pm}^{\lambda^{\ast}}\to
X_{\pm}^{\lambda^{\ast}}$ has a linear bounded inverse. Now for any
$(\lambda,z,u)\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{2})$ it holds that

$\displaystyle\psi(\lambda,z,u)\widehat{\mathcal{B}}(\lambda,z,\psi(\lambda,z,u)u)\psi(\lambda,z,u)\equiv\widehat{\mathcal{B}}(\lambda^{\ast},0,0)$
$\displaystyle=\frac{1}{2}R_{\lambda^{\ast}}\left({\bf
B}_{(\lambda^{\ast},0)}(0)|_{X_{\pm}^{\lambda^{\ast}}}\right)R_{\lambda^{\ast}}.$
(B.23)

Moreover, from (B.19) we derive

$\displaystyle\mathscr{L}_{\lambda}(z+\mathfrak{h}(\lambda,z)+R_{\lambda^{\ast}}u)-\mathscr{L}_{\lambda}(z+\mathfrak{h}(\lambda,z)),$
(B.24) $\displaystyle=$ $\displaystyle{\bf
L}_{(\lambda,z)}(R_{\lambda^{\ast}}u)=(\widehat{\mathcal{B}}(\lambda,z,u)u,u)_{H},\quad\forall
u\in B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{1}).$

By further shrinking the neighborhood $\Lambda_{0}$ of $\lambda^{\ast}$ and
$\epsilon>0$, we can get $\delta_{3}\in(0,\delta_{2})$ such that

$\psi(\lambda,z,u)u\in
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{1}),\quad\forall(\lambda,z,u)\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{3})$

because of (B.22). Then (B) and (B.24) lead to

$\displaystyle\mathscr{L}_{\lambda}(z+\mathfrak{h}(\lambda,z)+R_{\lambda^{\ast}}\psi(\lambda,z,u)u)-\mathscr{L}_{\lambda}(z+\mathfrak{h}(\lambda,z)),$
(B.25) $\displaystyle=$
$\displaystyle(\widehat{\mathcal{B}}(\lambda,z,\psi(\lambda,z,u)u)\psi(\lambda,z,u)u,\psi(\lambda,z,u)u)_{H}$
$\displaystyle=$
$\displaystyle(\psi(\lambda,z,u)\widehat{\mathcal{B}}(\lambda,z,\psi(\lambda,z,u)u)\psi(\lambda,z,u)u,u)_{H}$
$\displaystyle=$ $\displaystyle\frac{1}{2}(({\bf
B}_{(\lambda^{\ast},0)}(0)|_{X_{\pm}^{\lambda^{\ast}}})R_{\lambda^{\ast}}u,R_{\lambda^{\ast}}u)_{H}.$

Take $\delta_{4}\in(0,\delta_{3})$ such that
$R_{\lambda^{\ast}}\left(B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{4})\right)\subset
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{3})$. Then (B.25) yields

$\displaystyle\mathscr{L}_{\lambda}\big{(}z+\mathfrak{h}(\lambda,z)+R_{\lambda^{\ast}}\psi(\lambda,z,(R_{\lambda^{\ast}})^{-1}u)(R_{\lambda^{\ast}})^{-1}u\big{)}-\mathscr{L}_{\lambda}(z+\mathfrak{h}(\lambda,z)),$
(B.26) $\displaystyle=$ $\displaystyle\frac{1}{2}(({\bf
B}_{(\lambda^{\ast},0)}(0)|_{X_{\pm}^{\lambda^{\ast}}})u,u)_{H}=\frac{1}{2}(B_{\lambda^{\ast}}(0)u,u)_{H}$

for all $(\lambda,z,u)\in\Lambda_{0}\times
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{4})$, where
$(R_{\lambda^{\ast}})^{-1}=P_{+}^{\lambda^{\ast}}|_{X_{\pm}^{\lambda^{\ast}}}+M^{-1}P_{-}^{\lambda^{\ast}}|_{X_{\pm}^{\lambda^{\ast}}}$
is the inverse of $R_{\lambda^{\ast}}$.

Take $\delta_{5}=\min\\{\epsilon,\delta_{4}\\}$. Then for $x\in
B_{X}(0,\delta_{5})$ we have $z=P_{0}^{\lambda^{\ast}}x\in
B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)$ and $u=P_{\pm}^{\lambda^{\ast}}x\in
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta_{4})$. For each $\lambda\in\Lambda_{0}$,
define $\Phi_{\lambda}:B_{X}(0,\delta_{5})\to X$ by

$\displaystyle\Phi_{\lambda}(x)=z+\mathfrak{h}(\lambda,z)+R_{\lambda^{\ast}}\psi(\lambda,z,(R_{\lambda^{\ast}})^{-1}u)(R_{\lambda^{\ast}})^{-1}u,$
(B.27)

where $z=P_{0}^{\lambda^{\ast}}x$ and $u=P_{\pm}^{\lambda^{\ast}}x$. Then
$\Phi_{\lambda}(0)=0$, $\Phi_{\lambda}$ is $C^{1}$, and both
$(\lambda,x)\mapsto\Phi_{\lambda}(x)$ and
$(\lambda,x)\mapsto\Phi^{\prime}_{\lambda}(x)$ are continuous. For
$z^{\prime}\in X_{0}^{\lambda^{\ast}}$ and $y^{\prime}\in
X_{\pm}^{\lambda^{\ast}}$ a direct computation leads to

$\displaystyle d\Phi_{\lambda}(0)[z^{\prime}+y^{\prime}]$ $\displaystyle=$
$\displaystyle
z^{\prime}+d_{z}\mathfrak{h}(\lambda,0)[z^{\prime}]+R_{\lambda^{\ast}}\psi(\lambda,0,0)(R_{\lambda^{\ast}})^{-1}y^{\prime}$
(B.28)

and so
$d\Phi_{\lambda^{\ast}}(0)[z^{\prime}+y^{\prime}]=z^{\prime}+y^{\prime}$
because $d_{z}\mathfrak{h}(\lambda^{\ast},0)=0$ by (B.13). From these and
(B.22) we get

$\displaystyle
d\Phi_{\lambda}(0)[z^{\prime}+y^{\prime}]-z^{\prime}-y=d_{z}\mathfrak{h}(\lambda,0)[z^{\prime}]+R_{\lambda^{\ast}}(\psi(\lambda,0,0)-\psi(\lambda^{\ast},0,0))(R_{\lambda^{\ast}})^{-1}y^{\prime}$

and therefore we may shrink $\Lambda_{0}$ so that
$\|d\Phi_{\lambda}(0)-id_{X_{\pm}^{\lambda^{\ast}}}\|<\frac{1}{2}$ for all
$\lambda\in\Lambda_{0}$. Then we have

$\|[d\Phi_{\lambda}(0)]^{-1}\|\leq 2\quad\forall\lambda\in\Lambda_{0}$

as before. Since $\psi^{\prime}_{z}(\lambda,z,u)$ and
$\psi^{\prime}_{u}(\lambda,z,u)$ continuously depend on $(\lambda,z,u)$,
shrinking $\Lambda_{0}$ and $\delta_{5}>0$ again, it follows from (B.27) and
(B.28) that

$\|d\Phi_{\lambda}(x)-d\Phi_{\lambda}(0)\|\leq\frac{1}{4}\quad\forall(\lambda,x)\in\Lambda_{0}\times
B_{X}(\theta,\delta_{5}).$

By Lemma B.4, for each $\lambda\in\Lambda_{0}$,
$\Phi_{\lambda}(B_{X}(0,\delta_{5}))\subset X$ is an open neighborhood of zero
in $X$ containing $B_{X}(0,\frac{\delta_{5}}{4})$ and $\Phi_{\lambda}$ is a
$C^{1}$ diffeomorphism from $B_{X}(0,\delta_{5})$ to
$\Phi_{\lambda}(B_{X}(0,\delta_{5}))$. Moreover, (B.26) yields (B.12) with
$\epsilon=\delta_{5}$.

It is clear that the positive number $\epsilon$ can be replaced by
$\delta_{5}$ in all conclusions of Theorem B.8 which have been proved.

Claims in (iv) and (v) easily follows as in [41, 45].

Under the assumptions of the “Moreover” part, by the above arguments and the
final part of proof of Lemma B.14 it is not hard to check that the maps
$\mathcal{B}:\Lambda_{0}\times B_{X_{0}^{\lambda^{\ast}}}(0,\epsilon)\times
B_{X_{\pm}^{\lambda^{\ast}}}(0,\delta/2)\to\mathscr{L}(X_{\pm}^{\lambda^{\ast}})$
and $\widehat{\mathcal{B}}$ in (B.20) are $C^{1}$. Therefore the desired
claims are easily seen from the above proof. ∎

######  Remark B.9.

The assumptions (f1) and (sf2)-(sf3) in Theorem B.8 guarantee that the map
$\mathcal{B}$ in (B.17) (and hence $\widehat{\mathcal{B}}$ in (B.20)) is
$C^{1}$. When $\Lambda=\mathbb{R}$ and $A_{\lambda}=A-\lambda\widehat{A}$,
where $A,\widehat{A}:B_{X}(0,\delta)\to X$ are $C^{2}$ maps corresponding with
$C^{3}$ functionals
$\mathscr{L},\widehat{\mathscr{L}}:B_{X}(0,\delta)\to\mathbb{R}$ satisfying
(a) and (b) above (B.1), if each of $A^{\prime\prime}$ and
$\widehat{A}^{\prime\prime}$ is either $C^{1}$ or uniformly continuous in some
neighborhood of any segment in $B_{X}(0,\delta)$ it is easily seen from the
above proof that the corresponding maps $\mathcal{B}$ and
$\widehat{\mathcal{B}}$ are also $C^{1}$. Hence Theorem B.8 is still true in
this situation.

Since $0\in X_{0}^{\lambda^{\ast}}$ is an isolated critical point of
$\mathscr{L}_{\lambda}^{\circ}$ if and only if $0\in X$ is such a critical
point of $\mathscr{L}_{\lambda}$, as usual Theorem B.8 lead to:

######  Corollary B.10 (Shifting).

Under the assumptions of Theorem B.8, let $\nu_{\lambda^{\ast}}:=\dim
X_{0}^{\lambda^{\ast}}<\infty$ and $\mu_{\lambda^{\ast}}:=\dim
X_{-}^{\lambda^{\ast}}<\infty$. For any Abel group ${\bf K}$, and for any
given $\lambda\in\Lambda_{0}$, if $0\in X$ is an isolated critical point of
$\mathscr{L}$ then $C_{q}(\mathscr{L}_{\lambda},0;{\bf K})\cong
C_{q-\mu_{\lambda^{\ast}}}(\mathscr{L}^{\circ}_{\lambda},0;{\bf K})$ for all
$q\in\mathbb{N}\cup\\{0\\}$.

######  Claim B.11.

In Theorem B.8, if for some
$\lambda\in\Lambda_{0}\setminus\\{\lambda^{\ast}\\}$, $0\in X$ is a
nondegenerate critical point of $\mathscr{L}_{\lambda}$, i.e., the operator
$B_{\lambda}(0)|_{X}\in\mathscr{L}(X)$ is hyperbolic ([72]), then $0\in
X_{0}^{\lambda^{\ast}}$ is such a critical point of
$\mathscr{L}^{\circ}_{\lambda}$ too.

###### Proof.

Let $z\in X_{0}^{\lambda^{\ast}}$ be such that
$d^{2}\mathscr{L}^{\circ}_{\lambda}(0)[z,z^{\prime}]=0\;\forall z^{\prime}\in
X_{0}^{\lambda^{\ast}}$. By (B.16) we deduce

$\displaystyle
P_{0}^{\lambda^{\ast}}\bigr{[}{B}_{\lambda}(0)-{B}_{\lambda}(0)(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}})^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))\bigr{]}z=0.$
(B.29)

Moreover, differentiating the equality in (B.10) at $z=0$ and using (B.13) we
get that

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(id_{X}-P_{0}^{\lambda^{\ast}})B_{\lambda}(0)|_{X}(z+d_{z}\mathfrak{h}(\lambda,0)z)$
$\displaystyle=$
$\displaystyle(id_{X}-P_{0}^{\lambda^{\ast}})B_{\lambda}(0)|_{X}(z-[P_{\pm}^{\lambda^{\ast}}{B}_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}}]^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))|_{X_{0}^{\lambda^{\ast}}}z)$
$\displaystyle=$
$\displaystyle(id_{X}-P_{0}^{\lambda^{\ast}})[B_{\lambda}(0)-B_{\lambda}(0)(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}})^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))|_{X_{0}^{\lambda^{\ast}}})]z$
$\displaystyle=$
$\displaystyle[B_{\lambda}(0)-B_{\lambda}(0)(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}})^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))|_{X_{0}^{\lambda^{\ast}}})]z$
$\displaystyle-$ $\displaystyle
P_{0}^{\lambda^{\ast}}[B_{\lambda}(0)-B_{\lambda}(0)(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}})^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))|_{X_{0}^{\lambda^{\ast}}})]z$
$\displaystyle=$
$\displaystyle[B_{\lambda}(0)-B_{\lambda}(0)(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}})^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))|_{X_{0}^{\lambda^{\ast}}})]z,$

where the final equality comes from (B.29). Since $B_{\lambda}(0)$ is
invertible, we deduce

$z=(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda^{\ast}}(0)|_{X_{\pm}^{\lambda^{\ast}}})^{-1}(P_{\pm}^{\lambda^{\ast}}{B}_{\lambda}(0))|_{X_{0}^{\lambda^{\ast}}})z.$

Note that $z\in X_{0}^{\lambda^{\ast}}$ and that the right side belongs to
$X_{\pm}^{\lambda^{\ast}}$. Hence $z=0$. ∎

In general, it is not easy to judge whether
$\sigma(B(0)|_{X})\setminus\\{0\\}$ is bounded away from the imaginary axis.
However, it has the following equivalent version:

  1. (*)

$0$ is at most an isolated point of $\sigma(B(0)|_{X})$ and $B(0)|_{X}$
induces a hyperbolic operator on the quotient space $X/X_{0}$, where
$X_{0}={\rm Ker}(B(0)|_{X})$.

In fact, it is clear if $0\notin\sigma(B(0)|_{X})$. Assume that
$\sigma_{0}:=\\{0\\}\subset\sigma(B(0)|_{X})$. Since
$\sigma_{1}:=\sigma(B(0)|_{X})\setminus\\{0\\}$ is bounded away from the
imaginary axis, $0$ is an isolated point of $\sigma(B(0)|_{X})$. By Lemma B.2
there exists a direct sum decomposition of the Banach space $X$,
$X=X_{0}+X_{1}$, where $X_{0}={\rm Ker}(B(0)|_{X})$ and $X_{1}$ is an
invariant subspace on which $B(0)|_{X}$ has the spectrum $\sigma_{1}$. Note
that the quotient space $X/X_{0}$ and the induced operator of $B(0)|_{X}$ on
it can be identified with $X_{1}$ and $B(0)|_{X_{1}}$, respectively. Hence (*)
is satisfied. Conversely, since $0$ is an isolated point of
$\sigma(B(0)|_{X})$ we have a decomposition $X=X_{0}+X_{1}$ as above, and
$B(0)|_{X_{1}}$ is a hyperbolic operator on $X_{1}\equiv X/X_{0}$. It follows
that $\sigma(B(0)|_{X})\setminus\\{0\\}$ is bounded away from the imaginary
axis.

For a class of operators used in this paper the following lemma determines
when $0$ is at most an isolated point of $\sigma(B(0))$.

######  Lemma B.12.

Let a self-adjoined operator $\mathfrak{B}\in\mathscr{L}_{s}(H)$ be a sum
$\mathfrak{B}=\mathfrak{P}+\mathfrak{Q}$, where
$\mathfrak{P}\in\mathscr{L}(H)$ is invertible, and
$\mathfrak{Q}\in\mathscr{L}(H)$ is compact. Let $0\in\sigma(\mathfrak{B})$.
Then $0$ is an isolated point of $\sigma(\mathfrak{B})$ and an eigenvalue of
$\mathfrak{B}$ of the finite multiplicity. ([8, Lemma 2.2])

In addition, if $\mathfrak{P}$ is also positive definite, then every
$\lambda<\inf\\{(\mathfrak{P}u,u)_{H}\,|\,\|u\|_{H}=1\\}$ is either a regular
value of $\mathfrak{B}$ or an isolated point of $\sigma(\mathfrak{B})$, which
is also an eigenvalue of finite multiplicity. ([41, Proposition B.2])

As before let $H^{0}$, $H^{+}$ and $H^{-}$ be null, positive and negative
definite spaces of $B(0)$, and let $P^{\ast}:H\to H^{\ast}$ for $\ast=0,+,-$,
be the orthogonal projections. Clearly, $X_{0}\subset H^{0}$. The following
lemma shows how the Morse index $\mu$ is computed.

######  Lemma B.13.

If $X_{0}=H^{0}$ and either $\dim H^{-}<\infty$ or $\dim H^{+}<\infty$, then
$X_{\ast}$ are dense subspaces of $H^{\ast}$ for $\ast=+,-$, moreover
$X_{-}=H^{-}$ (resp. $X_{+}=H^{+}$) as $\dim H^{-}<\infty$ (resp. $\dim
H^{+}<\infty$). (These imply $X_{\ast}=H^{\ast}\cap X$ and
$P_{\ast}=P^{\ast}|_{X}$ for $\ast=0,+,-$, and hence $X_{\pm}=X\cap
H^{\pm}=(P^{+}+P^{-})(X)$, where $H^{\pm}:=H^{+}+H^{-}$.)

###### Proof.

Since $X$ is dense in $H$, it is easily proved that $X_{\pm}:=X_{+}\oplus
X_{-}$ is dense in $H^{\pm}$. Let $X_{+}^{H}$ and $X_{-}^{H}$ be the closure
of $X_{+}$ and $X_{-}$ in $H$, respectively. They are orthogonal in $H$,
invariant with respect to $B(0)$ and $H^{\pm}=X_{+}^{H}+X_{-}^{H}$. Since
$B(0)$ is semi-positive (resp. semi-negative) on $X_{+}^{H}$ (resp.
$X_{-}^{H}$) by Lemma B.2, and $B(0)$ has no nontrivial kernel on $H^{\pm}$,
we deduce that $B(0)$ must be positive (resp. negative) on $X_{+}^{H}$ (resp.
$X_{-}^{H}$). Let $\dim H^{-}<\infty$. If $X^{H}_{-}\cap H^{-}\neq\\{0\\}$, it
is an invariant subspace of $B(0)$. Denote by $X^{H}_{--}$ and $H^{--}$ the
orthogonal complements of $X^{H}_{-}\cap H^{-}$ in $X^{H}_{-}$ and $H^{-}$,
respectively. Then $X^{H}_{--}+H^{--}+X^{H}_{-}\cap H^{-}$ is a negative
definite subspace of $B(0)$. But $\dim H^{-}$ is the maximal dimension of
negative definite subspaces of $B(0)$. Hence $X^{H}_{--}=\\{0\\}$, and thus
$X^{H}_{-}=X^{H}_{-}\cap H^{-}$. The latter means $X^{H}_{-}\subset H^{-}$ and
so $X^{H}_{+}\supseteq H^{+}$. For any $x\in H^{-}$, since
$X_{\pm}=X_{+}\oplus X_{-}$ is dense in $H^{\pm}=H^{+}+H^{-}$, we have a
sequence $(x_{n}^{+}+x_{n}^{-})\subset X_{+}\oplus X_{-}$ such that
$x_{n}^{+}+x_{n}^{-}\to x$. Note that
$\|x_{n}^{+}+x_{n}^{-}-x_{m}^{+}-x_{m}^{-}\|^{2}=\|x_{n}^{+}-x_{m}^{+}\|^{2}+\|x_{n}^{-}-x_{m}^{-}\|^{2}$.
We get $x_{n}^{+}\to x^{+}\in X^{H}_{+}$ and $x_{n}^{-}\to x^{-}\in
X^{H}_{-}$. Hence $x=x^{+}+x^{-}$. But $(x,x^{+})_{H}=0$. So $x^{+}=0$ and
$x=x^{-}$. This shows that $X_{-}$ is dense in $H^{-}$ and thus
$X_{-}=X_{-}^{H}=H^{-}$ (since $\dim H^{-}<\infty$). The latter implies
$X_{+}^{H}=H^{+}$. Similarly, we can deal with the case that $\dim
H^{+}<\infty$. ∎

######  Lemma B.14.

Let $\Lambda$ be a metric space, and
$\mathscr{L}_{\lambda}:B_{X}(0,\delta)\to\mathbb{R}$, $\lambda\in\Lambda$, be
a family of $C^{2}$-functionals satisfying the conditions (a)-(b) and
$d\mathscr{L}_{\lambda}(0)=0\;\forall\lambda$. Suppose that the corresponding
operators $A_{\lambda}$ satisfies the assumptions (c’)-(d’) in Theorem B.3.
Then $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X)$
is continuous, $C^{1}$ in $u$, and $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto\mathcal{B}^{\prime}_{\lambda}(u)\in\mathscr{L}(X,\mathscr{L}(X))$
is also continuous. Moreover, if $\Lambda$ is a nonempty open subset of a
Banach space $Z$ and the assumptions (c’) and (f1)-(f3) in Theorem B.3 are
satisfied, then $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X)$
is $C^{1}$. In particular, for $C^{2}$ maps $A,\widehat{A}:B_{X}(0,\delta)\to
X$, if each of $A^{\prime\prime}$ and $\widehat{A}^{\prime\prime}$ is either
$C^{1}$ or uniformly continuous in some neighborhood of the line segment
$[0,1]u:=\\{tu\,|\,0\leq t\leq 1\\}$ for each $u\in B_{X}(0,\delta)$, then

$\mathbb{R}\times
B_{X}(0,\delta)\to\mathscr{L}(X),\;(\lambda,u)\mapsto\mathcal{B}(\lambda,u)=\int^{1}_{0}(1-t)A^{\prime}(tu)dt-\lambda\int^{1}_{0}(1-t)\widehat{A}^{\prime}(tu)dt$

is $C^{1}$.

###### Proof.

Step 1 (Prove that $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X)$
is continuous). Fix a point $(\lambda_{0},u_{0})\in\Lambda\times
B_{X}(0,\delta)$. Then for any $(\lambda,u)\in\Lambda\times B_{X}(0,\delta)$
it holds that

$\displaystyle\|\mathcal{B}_{\lambda}(u)-\mathcal{B}_{\lambda_{0}}(u_{0})\|_{\mathscr{L}(X)}\leq\int^{1}_{0}(1-t)\|A^{\prime}_{\lambda}(tu)-A^{\prime}_{\lambda_{0}}(tu_{0})\|_{\mathscr{L}(X)}dt.$
(B.30)

By (d’) there exist neighborhoods $\mathscr{N}(\lambda_{0})$ of $\lambda_{0}$
in $\Lambda$, and $\mathscr{N}([0,1]u_{0})$ of $[0,1]u_{0}$ in
$B_{X}(0,\delta)$ such that
$\mathscr{N}(\lambda_{0})\times\mathscr{N}([0,1]u_{0})\ni(\lambda,u)\mapsto
A^{\prime}_{\lambda}(x)\in\mathscr{L}(X)$ is uniform continuous, that is, for
any $\varepsilon>0$, we have $\nu>0$ such that
$\|A^{\prime}_{\lambda_{1}}(u_{1})-A^{\prime}_{\lambda_{2}}(u_{2})\|_{\mathscr{L}(X)}<\varepsilon$
provided that
$(\lambda_{i},u_{i})\in\mathscr{N}(\lambda_{0})\times\mathscr{N}([0,1]u_{0})$,
$i=1,2$, satisfies $d(\lambda_{1},\lambda_{2})<\nu$ and
$\|u_{1}-u_{2}\|_{X}<\nu$. Thus if $(\lambda,u)\in\Lambda\times
B_{X}(\theta,\delta)$ satisfies $d(\lambda,\lambda_{0})<\nu$ and
$\|u-u_{0}\|_{X}<\nu$ then
$\|A^{\prime}_{\lambda}(tu)-A^{\prime}_{\lambda_{0}}(tu_{0})\|_{\mathscr{L}(X)}<\varepsilon$
for all $t\in[0,1]$. It follows that

$\|\mathcal{B}_{\lambda}(u)-\mathcal{B}_{\lambda_{0}}(u_{0})\|_{\mathscr{L}(X)}\leq\varepsilon\int^{1}_{0}(1-t)dt=\frac{1}{2}\varepsilon.$

Step 2 (Prove that $\mathcal{B}_{\lambda}:B_{X}(0,\delta)\to\mathscr{L}(X)$ is
Gâteaux differentiable). For fixed $u\in B_{X}(0,\delta)$ and $h\in X$ we have
$\rho>0$ such that $u+sh\in B_{X}(0,\delta)$ for all $s\in[-\rho,\rho]$. Then
the function
$[-\rho,\rho]\times[0,1]\ni(s,t)\mapsto(1-t)A^{\prime}_{\lambda}(t(u+sh))\in\mathscr{L}(X)$
is continuous and has a continuous partial derivative in $s$

$[-\rho,\rho]\times[0,1]\ni(s,t)\mapsto\frac{\partial}{\partial
s}((1-t)A^{\prime}_{\lambda}(t(u+sh)))$
$=(1-t)tA^{\prime\prime}_{\lambda}(t(u+sh))h\in\mathscr{L}(X,\mathscr{L}(X)).$

By the differential method with parameter integral we get

$\frac{\partial}{\partial
s}\mathcal{B}_{\lambda}(u+sh)=\int^{1}_{0}(1-t)tA^{\prime\prime}_{\lambda}(t(u+sh))hdt.$

In particular, $\mathcal{B}_{\lambda}$ has a Gâteaux derivative
$D\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X,\mathscr{L}(X))$ at $u$ given by

$D\mathcal{B}_{\lambda}(u)[h]=\int^{1}_{0}(1-t)tA^{\prime\prime}_{\lambda}(tu)hdt=\left(\int^{1}_{0}(1-t)tA^{\prime\prime}_{\lambda}(tu)dt\right)h,\quad\forall
h\in X.$

Step 3 (Prove that $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto\mathcal{B}^{\prime}_{\lambda}(u)\in\mathscr{L}(X,\mathscr{L}(X))$
is continuous). For a fixed point $(\lambda_{0},u_{0})\in\Lambda\times
B_{X}(0,\delta)$, by (d’) we have neighborhoods $\mathscr{N}(\lambda_{0})$ of
$\lambda_{0}$ in $\Lambda$, and $\mathscr{N}([0,1]u_{0})$ of $[0,1]u_{0}$ in
$B_{X}(0,\delta)$ such that

$\mathscr{N}(\lambda_{0})\times\mathscr{N}([0,1]u_{0})\ni(\lambda,u)\mapsto
A^{\prime\prime}_{\lambda}(x)\in\mathscr{L}(X,\mathscr{L}(X))$

is uniform continuous. Then there exists $\nu>0$ such that when
$(\lambda,u)\in\Lambda\times B_{X}(0,\delta)$ satisfies
$d(\lambda,\lambda_{0})<\nu$ and $\|u-u_{0}\|_{X}<\nu$ the following holds:

$\|A^{\prime\prime}_{\lambda}(tu)-A^{\prime\prime}_{\lambda_{0}}(tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}<\varepsilon,\quad\forall
t\in[0,1].$

Hence for such $(\lambda,u)$ we deduce

$\displaystyle\|D\mathcal{B}_{\lambda}(u)-D\mathcal{B}_{\lambda_{0}}(u_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}$
$\displaystyle\leq$
$\displaystyle\int^{1}_{0}(1-t)t\|A^{\prime\prime}_{\lambda}(tu)-A^{\prime\prime}_{\lambda_{0}}(tu_{0})\|_{\mathscr{L}(X,\mathscr{L}(X))}dt$
(B.31) $\displaystyle\leq$ $\displaystyle\frac{1}{6}\varepsilon.$

That is, $\Lambda\times B_{X}(0,\delta)\ni(\lambda,u)\mapsto
D\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X,\mathscr{L}(X))$ is continuous. In
particular, $B_{X}(0,\delta)\ni u\mapsto
D\mathcal{B}_{\lambda}(u)\in\mathscr{L}(X,\mathscr{L}(X))$ is continuous, and
so $\mathcal{B}_{\lambda}$ has a Fréchet derivative
$\mathcal{B}^{\prime}_{\lambda}(u)=D\mathcal{B}_{\lambda}(u)$ at $u$. Then
$\mathcal{B}_{\lambda}$ is $C^{1}$.

Step 4 (Prove the conclusion in the “Moreover” part). Write

$\displaystyle\mathcal{B}(\lambda,u)=\int^{1}_{0}(1-t)D_{2}A(\lambda,tu)dt.$

Since (f1) and (f2) imply (c’) and (d’), $\mathcal{B}$ is continuous, $C^{1}$
in $u$, and

$\Lambda\times B_{X}(0,\delta)\ni(\lambda,u)\mapsto
D_{2}\mathcal{B}(\lambda,u)\in\mathscr{L}(X,\mathscr{L}(X))$

is also continuous. Similarly, from the second condition in (f1) we may derive
that $\mathcal{B}(\lambda,u)$ has a Gâteaux derivative in $\lambda$ given by

$D_{1}\mathcal{B}(\lambda,u)=\int^{1}_{0}(1-t)D_{1}D_{2}A(\lambda,tu)dt.$

As above using (f3) we get that $\Lambda\times
B_{X}(0,\delta)\ni(\lambda,u)\mapsto
D_{1}\mathcal{B}(\lambda,u)\in\mathscr{L}(Z,\mathscr{L}(X))$ is continuous.
Hence $\mathcal{B}$ is $C^{1}$.

The final claim is easily seen from the above arguments. ∎

Acknowledgments. The author is deeply grateful to Professor Xiaochun Rong for
his invitation. I also deeply thank the anonymous referees for useful remarks.

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School of Mathematical Sciences, Beijing Normal University
---
Laboratory of Mathematics and Complex Systems, Ministry of Education
Beijing 100875, The People’s Republic of China
E-mail address<EMAIL_ADDRESS>


# Schrödinger’s Ballot: Quantum Information and the Violation of Arrow’s
Impossibility Theorem

Xin Sun Department of the Foundations of Computer Science, the John Paul II
Catholic University of Lublin, Lublin, PolandQuantum Blockchains Inc., Pulawy,
Poland<EMAIL_ADDRESS>Institute of Logic and Cognition, Sun Yat-sen
University, Guangzhou, ChinaDepartment of the Foundations of Computer Science,
the John Paul II Catholic University of Lublin, Lublin, PolandQuantum
Blockchains Inc., Pulawy, PolandMakoLab SA, Lodz, PolandQuantum Blockchains
Inc., Pulawy, Poland Feifei He Corresponding authorInstitute of Logic and
Cognition, Sun Yat-sen University, Guangzhou, China<EMAIL_ADDRESS>Department of the Foundations of Computer Science, the John Paul II Catholic
University of Lublin, Lublin, PolandQuantum Blockchains Inc., Pulawy,
PolandMakoLab SA, Lodz, PolandQuantum Blockchains Inc., Pulawy, Poland Piotr
Kulicki Department of the Foundations of Computer Science, the John Paul II
Catholic University of Lublin, Lublin, PolandQuantum Blockchains Inc., Pulawy,
Poland<EMAIL_ADDRESS>MakoLab SA, Lodz, PolandQuantum Blockchains Inc.,
Pulawy, Poland Mirek Sopek MakoLab SA, Lodz, PolandQuantum Blockchains Inc.,
Pulawy, Poland<EMAIL_ADDRESS>
###### Abstract

In this paper we study Arrow’s Impossibility Theorem in the quantum setting.
Our work is based on the work of Bao and Halpern [4], in which it is proved
that the quantum analogue of Arrow’s Impossibility Theorem is not valid.
However, we feel unsatisfied about the proof presented in [4]. Moreover, the
definition of Quantum Independence of Irrelevant Alternatives (QIIA) in [4]
seems not appropriate to us. In this paper, we give a better definition of
QIIA, which properly captures the idea of the independence of irrelevant
alternatives, and a detailed proof of the violation of Arrow’s Impossibility
Theorem in the quantum setting with the modified definition.

Keywords: vote, quantum information, Arrow’s Impossibility Theorem, social
choice

##  1 Introduction

Many voting protocols based on classical cryptography have been developed and
successfully applied since Chaum et al. [6]. However, the security of
protocols based on classical cryptography is based on the unproven complexity
of some computational algorithms, such as factoring of large numbers. The
research in quantum computation shows that quantum computers are able to
factor large numbers in a short time, which means that classical protocols
based on such algorithms are already insecure. To react to the risk posed by
forthcoming quantum computers, a number of quantum voting protocols have been
developed in the last decade [9, 25, 12, 10, 13, 11, 23, 26, 22, 21].

While all these works have focused on the security problems of voting from a
cryptographic perspective, Bao and Halpern [4] studied quantum voting from a
social choice theoretic perspective by showing that the quantum analog of
Arrow’s Impossibility Theorem is violated in the quantum setting. Bao and
Halpern’s [4] idea and results are equally interesting. However, we feel
unsatisfied about the proof presented in [4]. Moreover, the definition of the
Quantum Independence of Irrelevant Alternatives (QIIA) in [4] seems not
appropriate to us. In this paper, we give a better definition of QIIA, which
properly captures the idea of independence of irrelevant alternatives, and a
detailed proof of the violation of Arrow’s Impossibility Theorem in the
quantum setting with the modified definition.

The structure of this paper is as follows. We review some background knowledge
on classical and quantum voting in Section 2. In section 3 we introduce a
voting rule called quantum Condorcet voting and proves that Arrow’s
Impossibility Theorem is violated by quantum Condorcet voting. We discuss
related work in Section 4 and conclude this paper with the plan for the future
work in Section 5. Some primitives of the quantum information theory which are
used in this paper are collected in the appendix.

##  2 Background

###  2.1 Classical voting system

Now we succinctly review the theory of classical voting. A more detailed
introduction to the classical voting and social choice theory can be found in
Zwicker [29], Pacuit [16] and Brandt et al. [5].

Let $\mathcal{V}=\\{v_{\\_}1,\ldots,v_{\\_}n\\}$ be a finite set of (at least
two) voters and $\mathcal{C}=\\{c_{\\_}1,c_{\\_}2,\ldots,c_{\\_}n\\}$ be a
non-empty set of candidates. Each voter $v_{\\_}i\in\mathcal{V}$ is endowed
with preference $\succ_{\\_}i$ over $\mathcal{C}$. The preference
$\succ_{\\_}i$ is a binary relation on $\mathcal{C}$ that is irreflexive,
transitive, and complete. In other words, $\succ_{\\_}i$ is a linear order on
$\mathcal{C}$. Let $\mathfrak{L}(\mathcal{C})$ denote the set of all linear
orders on $\mathcal{C}$. A profile
$\mathbf{R}=(R_{\\_}1,\ldots,R_{\\_}n)\in\mathfrak{L}(\mathcal{C})^{\mathcal{V}}$
is a vector of linear orders (i.e., preferences), where $R_{\\_}i$ is the
linear order supplied by voter $v_{\\_}i$. We write
$\mathcal{V}^{\mathbf{R}}_{\\_}{x\succ y}$ to denote the set of voters that
rank candidate $x$ above candidate $y$ under profile $\mathbf{R}$. A Social
Welfare Function (SWF) is a function
$F:\mathfrak{L}(\mathcal{C})^{\mathcal{V}}\mapsto\mathfrak{L}(\mathcal{C})$.
Two widely accepted properties of SWF are unanimity and independence of
irrelevant alternatives.

######  Definition 1 (Unanimity)

A SWF $F$ satisfies the unanimity condition if, whenever all voters rank $x$
above $y$, then so does society:

$\mathcal{V}^{\mathbf{R}}_{\\_}{x\succ y}=\mathcal{V}$ implies $(x,y)\in
F(\mathbf{R})$.

######  Definition 2 (Independence of Irrelevant Alternatives (IIA))

An SWF $F$ satisfies IIA if the relative social ranking of two candidates only
depends on their relative voter rankings:

$\mathcal{V}^{\mathbf{R}}_{\\_}{x\succ
y}=\mathcal{V}^{\mathbf{R}^{\prime}}_{\\_}{x\succ y}$ implies $(x,y)\in
F(\mathbf{R})\Leftrightarrow(x,y)\in F(\mathbf{R}^{\prime})$.

The intuition about IIA is that two ballot profiles that are similar according
to $(x,y)$ should produce the same ranking for $(x,y)$. This intuition will be
used later to define the quantum analogue of IIA.

The celebrated Arrow’s Impossibility Theorem states that any SWF that
satisfies both unanimity and the IIA must also satisfy a property which any
SWF should not satisfy: dictatorship.

######  Definition 3 (Dictatorship)

An SWF $F$ satisfies dictatorship if there is a voter $v_{\\_}i\in\mathcal{V}$
such that $F(\mathbf{R})=R_{\\_}i$ for every profile
$\mathbf{R}=(R_{\\_}1,\ldots,R_{\\_}n)$.

######  Theorem 1 (Arrow [3])

Any SWF for three or more candidates that satisfies unanimity and the IIA
cannot satisfy nondictatorship.

###  2.2 Quantum voting system

Now we introduce our formalism of quantum voting system, which is similar to
the formalism of Bao and Halpern [4]. In a quantum voting system with
candidates $\mathcal{C}=\\{c_{\\_}1,c_{\\_}2,\ldots,c_{\\_}n\\}$, we specify a
Hilbert space $\mathsf{H}$ of which the dimension is
$|\mathfrak{L}(\mathcal{C})|$. That is,
$\mathsf{H}=\mathbb{C}^{|\mathfrak{L}(\mathcal{C})|}$. Every voter
$v_{\\_}i\in\mathcal{V}$ is associated with a Hilbert space
$\mathsf{H}_{\\_}i$ which is isomorphic to $\mathsf{H}$. Every linear order
$R\in\mathfrak{L}(\mathcal{C})$ is naturally viewed as basis vector
$|R\rangle$ of $\mathsf{H}$. The basis
$\mathsf{B}=\\{|R^{1}\rangle_{\\_}i,\ldots,|R^{|\mathfrak{L}(\mathcal{C})|}\rangle_{\\_}i\\}$
is called the preference basis for $\mathsf{H}_{\\_}i$.

Consider a pair $(x,y)$ of candidates. $\mathsf{H}$ decomposes into subspaces
associated with the possible relationships between $x$ and $y$. By
$\mathsf{S}^{x\succ y}$, we denote the subspace spanned by the $\mathsf{B}$
elements that encode $x\succ y$ (e.g. $|x\succ y\succ z\rangle$, $|z\succ
x\succ y\rangle$). We use $\Pi^{x\succ y}$ to denote the projector onto the
subspace $\mathsf{S}^{x\succ y}$.

A quantum ballot of voter $v_{\\_}i$ is a density operator $\rho\in
D(\mathsf{H}_{\\_}i)$. A quantum ballot profile is a density operator
$\rho\in\mathsf{H}_{\\_}1\otimes\ldots\otimes\mathsf{H}_{\\_}n$. A basis
quantum ballot profile is a profile in which every component is a density
operator of a basis vector. A quantum social welfare function (QSWF) is linear
map
$\mathcal{E}:D(\mathsf{H}_{\\_}1\otimes\ldots\otimes\mathsf{H}_{\\_}n)\mapsto
D(\mathsf{H})$. The result of voting with quantum ballot profile $\rho$ is
obtained by measuring $\mathcal{E}(\rho)$ on the preference basis.

######  Definition 4 (Quantum Unanimity [4])

A QSWF $\mathcal{E}$ satisfies the sharp unanimity condition if it satisfies
the following:

  * •

For all quantum ballot profile $\rho$ and all pairs of candidates $(x,y)$, if
$Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))=1$ for each voter $v_{\\_}i$, then
$Tr(\Pi^{x\succ y}(\mathcal{E}(\rho)))=1$.

A QSWF $\mathcal{E}$ satisfies the unsharp unanimity condition if it satisfies
the following:

  * •

For all quantum ballot profile $\rho$ and all pairs of candidates $(x,y)$, if
$Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))>0$ for each voter $v_{\\_}i$, then
$Tr(\Pi^{x\succ y}(\mathcal{E}(\rho)))>0$.

A QSWF $\mathcal{E}$ satisfies the quantum unanimity condition if it satisfies
both sharp and unsharp unanimity conditions.

######  Definition 5 (Quantum Independence of Irrelevant Alternatives (QIIA))

A QSWF $\mathcal{E}$ satisfies the sharp IIA condition if it satisfies the
following:

  * •

For all quantum ballot profile $\rho$ and $\rho^{\prime}$ and all pairs of
candidates $(x,y)$, if $Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq
i}(\rho)))=Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho^{\prime})))$ and
$Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq i}(\rho)))=Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq
i}(\rho^{\prime})))$ for each voter $v_{\\_}i$, then

$Tr(\Pi^{x\succ y}(\mathcal{E}(\rho)))=1$ implies that $Tr(\Pi^{x\succ
y}(\mathcal{E}(\rho^{\prime})))=1$.

A QSWF $\mathcal{E}$ satisfies unsharp IIA if the following condition is
satisfied:

  * •

For all quantum ballot profile $\rho,\rho^{\prime}$ and all pairs of
candidates $(x,y)$, if $Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq
i}(\rho)))=Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho^{\prime})))$ and
$Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq i}(\rho)))=Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq
i}(\rho^{\prime})))$ for each voter $v_{\\_}i$, then

$Tr(\Pi^{x\succ y}(\mathcal{E}(\rho)))>0$ implies that $Tr(\Pi^{x\succ
y}(\mathcal{E}(\rho^{\prime})))>0$.

A QSWF $\mathcal{E}$ satisfies the QIIA condition if it satisfies both sharp
and unsharp IIA conditions.

Note that QIIA in our definition is different from the QIIA in [4]. QIIA in
[4] states that whether $\mathcal{E}(\rho)$ has support on $\mathsf{S}^{x\succ
y}$ only depends on whether each $\rho_{\\_}i$ has support on
$\mathsf{S}^{x\succ y}$ and $\mathsf{S}^{y\succ x}$. More precisely, it states
that for all quantum ballot profile $\rho,\rho^{\prime}$ and all pairs of
candidates $(x,y)$, if $Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))>0$ iff
$Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho^{\prime})))>0$ and $Tr(\Pi^{y\succ
x}(Tr_{\\_}{\neq i}(\rho)))>0$ iff $Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq
i}(\rho^{\prime})))>0$ for each voter $v_{\\_}i$, then $Tr(\Pi^{x\succ
y}(\mathcal{E}(\rho)))>0$ implies that $Tr(\Pi^{x\succ
y}(\mathcal{E}(\rho^{\prime})))>0$.

The intuition of QIIA is the same as the intuition of classical IIA: it states
that two ballot profiles that are similar according to $(x,y)$ should produce
the same ranking for $(x,y)$. It seems Bao and Halpern [4] considered two
ballot profiles $\rho$ and $\rho^{\prime}$ to be similar according to $(x,y)$
as long as $Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))>0$ iff $Tr(\Pi^{x\succ
y}(Tr_{\\_}{\neq i}(\rho^{\prime})))>0$ and $Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq
i}(\rho)))>0$ iff $Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq i}(\rho^{\prime})))>0$. To
us this requirement is too weak. For example, $\rho=0.99|x\succ y\succ
z\rangle\langle x\succ y\succ z|+0.01|y\succ x\succ z\rangle\langle y\succ
x\succ z|$ and $\rho^{\prime}=0.01|x\succ y\succ z\rangle\langle x\succ y\succ
z|+0.99|y\succ x\succ z\rangle\langle y\succ x\succ z|$ are similar according
to $(x,y)$ in Bao and Halpern’s definition, but intuitively they should not
be. On the other hand, $\rho$ and $\rho^{\prime}$ are not similar according to
$(x,y)$ in our definition. Indeed, two profiles $\rho$ and $\rho^{\prime}$ are
similar according to $(x,y)$ in our definition only if $Tr(\Pi^{x\succ
y}(Tr_{\\_}{\neq i}(\rho)))=Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq
i}(\rho^{\prime})))$ and $Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq
i}(\rho)))=Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq i}(\rho^{\prime})))$. We believe
our definition of QIIA properly capture the idea of independence of irrelevant
alternatives. That’s why we use it to replace the QIIA of Bao and Halpern [4].

######  Definition 6 (Quantum Dictatorship [4])

A QSWF $\mathcal{E}$ satisfies sharp dictatorship if there is a voter
$v_{\\_}i$ such that:

  * •

For all quantum ballot profile $\rho=(\rho_{\\_}1,\ldots,\rho_{\\_}n)$ and all
pairs of candidates $(x,y)$, $Tr(\Pi^{x\succ y}\rho_{\\_}i)=1$ iff
$Tr(\Pi^{x\succ y}(\mathcal{E}(\rho)))=1$.

A QSWF $\mathcal{E}$ satisfies unsharp dictatorship if there is a voter
$v_{\\_}i$ such that:

  * •

For all quantum ballot profile $\rho=(\rho_{\\_}1,\ldots,\rho_{\\_}n)$ and all
pairs of candidates $(x,y)$, $Tr(\Pi^{x\succ y}\rho_{\\_}i)>0$ iff
$Tr(\Pi^{x\succ y}(\mathcal{E}(\rho)))>0$.

A QSWF $\mathcal{E}$ satisfies quantum dictatorship if it satisfies both sharp
and unsharp dictatorship.

##  3 Quantum Condorcet voting and Arrow’s Impossibility Theorem

We will use a special voting rule called Quantum Condorcet Voting
$\mathcal{E}_{\\_}{qcv}$ to disprove Arrow’s Impossibility Theorem in the
quantum setting. Since $\mathcal{E}_{\\_}{qcv}$ is a linear map from
$D(\mathsf{H}_{\\_}1\otimes\ldots\otimes\mathsf{H}_{\\_}n)$ to
$D(\mathsf{H})$, we only need to specify how $\mathcal{E}_{\\_}{qcv}$ operates
on a basis quantum ballot profile.

######  Definition 7 (Quantum Condorcet Voting)

Let $\rho_{\\_}1\otimes\ldots\otimes\rho_{\\_}n$ be a basis quantum ballot
profile. The Quantum Condorcet Voting $\mathcal{E}_{\\_}{qcv}$ operates in the
following steps:

  1. 1.

Calculates the Condorcet score of each candidate according to
$\rho_{\\_}1\otimes\ldots\otimes\rho_{\\_}n$. The Condorcet score of a
candidate is the number of winning in pairwise comparison with other
candidates. That is, for a candidate $x$, his Condorcet score $S_{\\_}c(x)$ is
$|\\{y\in\mathcal{C}:|\mathcal{V}^{\mathbf{R}}_{\\_}{x\succ
y}|\geq|\mathcal{V}^{\mathbf{R}}_{\\_}{y\succ x}|\\}|$ where $R$ is the
classical ballot profile corresponding to
$\rho_{\\_}1\otimes\ldots\otimes\rho_{\\_}n$.

  2. 2.

Generate a weak order $\succeq$ over all candidates according to their
Condorcet score. That is, $x\succeq y$ iff $S_{\\_}c(x)\geq S_{\\_}c(y)$.

  3. 3.

Complete the weak order. That is, generate the set
$\\{\succ^{1},\ldots,\succ^{m}\\}$ in which each $\succ^{i}$ is a linear order
that extends $\succeq$ and $\\{\succ^{1},\ldots,\succ^{m}\\}$ contains all
extensions of $\succeq$.

  4. 4.

Transform the linear order into quantum states. That is, for
$\\{\succ^{1},\ldots,\succ^{m}\\}$ we create a quantum state
$\sigma^{1}=\frac{1}{m}\sum\limits_{\\_}{i}\sigma_{\\_}i$, where each
$\sigma_{\\_}i$ is a basis ballot corresponds to $\succ^{i}$.

  5. 5.

Give the minority a shot. For any candidate pair $(x,y)$ which is encoded by
at least one $\rho_{\\_}i$, we spread an amount $\delta\in(0,1)$ of weight
across the $x\succ y$ subspace. That is, $\sigma^{1}$ is changed to
$\sigma^{2}=(1-k\delta)\sigma^{1}+\delta\Pi^{x_{\\_}1\succ
y_{\\_}1}+\ldots+\delta\Pi^{x_{\\_}k\succ y_{\\_}k}$, where
$(x_{\\_}1,y_{\\_}1),\ldots,(x_{\\_}k,y_{\\_}k)$ ranges over all candidate
pairs that are encoded by at least one $\rho_{\\_}i$. The parameter $\delta$
is required to satisfy that $\delta<\frac{1}{|\mathcal{C}|^{2}}$.

  6. 6.

Enforce unanimity. For any candidate pair $(x,y)$ which is encoded by all
$\rho_{\\_}i$, we project $\sigma^{2}$ onto the $x\succ y$ subspace. That is,
$\sigma^{2}$ is changed to $\sigma^{3}=\Pi^{x_{\\_}k\succ
y_{\\_}k}\ldots\Pi^{x_{\\_}1\succ y_{\\_}1}\sigma^{2}\Pi^{x_{\\_}1\succ
y_{\\_}1}\ldots\Pi^{x_{\\_}k\succ y_{\\_}k}$, where
$(x_{\\_}1,y_{\\_}1),\ldots,(x_{\\_}k,y_{\\_}k)$ ranges over all candidate
pairs that are encoded by all $\rho_{\\_}i$.

Both giving the minority a shot and enforcing unanimity are first introduced
in Bao and Halpern [4]. While they may look strange at first sight, both of
them will be useful in disproving Arrow’s Impossibility Theorem.

###### Theorem 2

The Quantum Condorcet Voting $\mathcal{E}_{\\_}{qcv}$ satisfies sharp
unanimity.

Proof: Let $\rho=\rho_{\\_}1\otimes\ldots\otimes\rho_{\\_}n$ be a basis
quantum ballot profile. If $Tr(\Pi^{x\succ y}(\rho_{\\_}i))=1$ for each voter
$v_{\\_}i$, then each $\rho_{\\_}i$ encodes $x\succ y$ since $\rho_{\\_}i$ is
a basis ballot. Then the projector $\Pi^{x\succ y}$ will be applied in the
step of enforcing unanimity. Therefore, $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(\rho)))=1$.

Now let $\rho$ be a quantum ballot profile such that $Tr(\Pi^{x\succ
y}(Tr_{\\_}{\neq i}(\rho)))=1$ for each voter $v_{\\_}i$. Note that
$Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))=1$ implies that $Tr_{\\_}{\neq
i}(\rho)=x^{i}_{\\_}1|\phi^{i}_{\\_}1\rangle\langle\phi^{i}_{\\_}1|+\ldots+x^{i}_{\\_}m|\phi^{i}_{\\_}m\rangle\langle\phi^{i}_{\\_}m|$
where each $\phi^{i}_{\\_}j$ is a basis vector that encodes $x\succ y$ and
$\sum\limits_{\\_}{j}x^{i}_{\\_}j=1$. Therefore, $\rho=x^{1}_{\\_}1\ldots
x^{n}_{\\_}1|\phi^{1}_{\\_}1\otimes\ldots\otimes\phi^{n}_{\\_}1\rangle\langle\phi^{1}_{\\_}1\otimes\ldots\otimes\phi^{n}_{\\_}1|+\ldots+x^{1}_{\\_}m\ldots
x^{n}_{\\_}m|\phi^{1}_{\\_}m\otimes\ldots\otimes\phi^{n}_{\\_}m\rangle\langle\phi^{1}_{\\_}m\otimes\ldots\otimes\phi^{n}_{\\_}m|$.
It then follows that $Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq
i}(|\phi^{1}_{\\_}j\otimes\ldots\otimes\phi^{n}_{\\_}j\rangle\langle|\phi^{1}_{\\_}j\otimes\ldots\otimes\phi^{n}_{\\_}j|)))=1$
for all $i$. Then we know that $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(|\phi^{1}_{\\_}j\otimes\ldots\otimes\phi^{n}_{\\_}j\rangle\langle|\phi^{1}_{\\_}j\otimes\ldots\otimes\phi^{n}_{\\_}j|)))=1$
because
$|\phi^{1}_{\\_}j\otimes\ldots\otimes\phi^{n}_{\\_}j\rangle\langle|\phi^{1}_{\\_}j\otimes\ldots\otimes\phi^{n}_{\\_}j|$
is a basis quantum ballot profile. Therefore, we have $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(\rho)))=x^{1}_{\\_}1\ldots
x^{n}_{\\_}1+\ldots+x^{1}_{\\_}m\ldots x^{n}_{\\_}m=1$. $\square$

###### Theorem 3

The Quantum Condorcet Voting $\mathcal{E}_{\\_}{qcv}$ satisfies unsharp
unanimity.

Proof: Let $\rho=\rho_{\\_}1\otimes\ldots\otimes\rho_{\\_}n$ be a basis
quantum ballot profile where each $\rho_{\\_}i$ is a basis vector of
$\mathsf{H}_{\\_}i$. If $Tr(\Pi^{x\succ y}(\rho_{\\_}i))>0$ for each voter
$v_{\\_}i$, then each $\rho_{\\_}i$ encodes $x\succ y$ since $\rho_{\\_}i$ is
a basis ballot. Then the projector $\Pi^{x\succ y}$ will be applied in the
step of enforcing unanimity. Therefore, $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qpv}(\rho)))=1>0$.

Now let $\rho$ be a quantum ballot profile such that $Tr(\Pi^{x\succ
y}(Tr_{\\_}{\neq i}(\rho)))>0$ for each voter $v_{\\_}i$. Note that
$Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))>0$ implies that $Tr_{\\_}{\neq
i}(\rho)=x_{\\_}i|\phi_{\\_}i\rangle\langle\phi_{\\_}i|+\ldots$ for some basis
vector $|\phi_{\\_}i\rangle$ which encodes $x\succ y$ and $0<x_{\\_}i\leq 1$.
Hence $\rho=x_{\\_}1\ldots
x_{\\_}n|\phi_{\\_}1\otimes\ldots\otimes\phi_{\\_}n\rangle\langle\phi_{\\_}1\otimes\ldots\otimes\phi_{\\_}n|+\ldots$.
Note that
$|\phi_{\\_}1\otimes\ldots\otimes\phi_{\\_}n\rangle\langle\phi_{\\_}1\otimes\ldots\otimes\phi_{\\_}n|$
is a basis quantum ballot profile in which each $\phi_{\\_}i$ encode $x\succ
y$. It then follows that $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(|\phi_{\\_}1\otimes\ldots\otimes\phi_{\\_}n\rangle\langle\phi_{\\_}1\otimes\ldots\otimes\phi_{\\_}n|)))=1$.
From $0<x_{\\_}1\ldots x_{\\_}n\leq 1$ we now know that $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qpv}(\rho)))>0$. $\square$

###### Theorem 4

The Quantum Condorcet Voting $\mathcal{E}_{\\_}{qcv}$ satisfies sharp IIA.

Proof: Let $\rho=\rho_{\\_}1\otimes\ldots\otimes\rho_{\\_}n$ and
$\rho^{\prime}=\rho^{\prime}_{\\_}1\otimes\ldots\otimes\rho^{\prime}_{\\_}n$
be two basis quantum ballot profiles. Assume $Tr(\Pi^{x\succ
y}(\rho_{\\_}i))=Tr(\Pi^{x\succ y}(\rho^{\prime}_{\\_}i))$ and $Tr(\Pi^{y\succ
x}(\rho_{\\_}i))=Tr(\Pi^{y\succ x}(\rho_{\\_}i^{\prime}))$ for each voter
$v_{\\_}i$. If $Tr(\Pi^{x\succ y}(\mathcal{E}_{\\_}{qcv}(\rho)))=1$, then we
know $x\succ y$ is encoded by all $\rho_{\\_}i$. For otherwise $\Pi^{y\succ
x}$ will appear in $\sigma^{2}$ in the step of giving the minority a shot,
making $Tr(\Pi^{x\succ y}(\mathcal{E}_{\\_}{qcv}(\rho)))<1$. Since
$Tr(\Pi^{x\succ y}(\rho_{\\_}i))=Tr(\Pi^{x\succ y}(\rho^{\prime}_{\\_}i))$ and
$Tr(\Pi^{y\succ x}(\rho_{\\_}i))=Tr(\Pi^{y\succ x}(\rho_{\\_}i^{\prime}))$ for
each voter $v_{\\_}i$, we know that $x\succ y$ is encoded by all
$\rho^{\prime}_{\\_}i$. Hence $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(\rho^{\prime})))=1$.

Now, let $\rho$ and $\rho^{\prime}$ be quantum ballot profiles such that
$Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))=Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq
i}(\rho^{\prime})))$ and $Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq
i}(\rho)))=Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq i}(\rho^{\prime})))$ for each voter
$v_{\\_}i$. If $Tr(\Pi^{x\succ y}(\mathcal{E}_{\\_}{qcv}(\rho)))=1$, then we
know $x\succ y$ is encoded by all $\rho_{\\_}i$. For otherwise $\Pi^{y\succ
x}$ will appear in $\sigma^{2}$ in the step of giving the minority a shot and
$\Pi^{y\succ x}$ will not appear in $\sigma^{3}$ in the step of enforcing
unanimity, making $Tr(\Pi^{x\succ y}(\mathcal{E}_{\\_}{qcv}(\rho)))<1$. Since
$Tr(\Pi^{x\succ y}(\rho_{\\_}i))=Tr(\Pi^{x\succ y}(\rho^{\prime}_{\\_}i))$ and
$Tr(\Pi^{y\succ x}(\rho_{\\_}i))=Tr(\Pi^{y\succ x}(\rho_{\\_}i^{\prime}))$ for
each voter $v_{\\_}i$, we know that $x\succ y$ is encoded by all
$\rho^{\prime}_{\\_}i$. Hence $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(\rho^{\prime})))=1$. $\square$

###### Theorem 5

The Quantum Condorcet Voting $\mathcal{E}_{\\_}{qcv}$ satisfies unsharp IIA.

Proof: Let $\rho=\rho_{\\_}1\otimes\ldots\otimes\rho_{\\_}n$ and
$\rho^{\prime}=\rho^{\prime}_{\\_}1\otimes\ldots\otimes\rho^{\prime}_{\\_}n$
be two basis quantum ballot profiles. Assume $Tr(\Pi^{x\succ
y}(\rho_{\\_}i))=Tr(\Pi^{x\succ y}(\rho^{\prime}_{\\_}i))$ and $Tr(\Pi^{y\succ
x}(\rho_{\\_}i))=Tr(\Pi^{y\succ x}(\rho_{\\_}i^{\prime}))$ for each voter
$v_{\\_}i$. Now, if $Tr(\Pi^{x\succ y}(\mathcal{E}(\rho)))>0$, then $y\succ x$
is not encoded by all candidates. Without loss of generality, let’s assume
$\rho_{\\_}1$ encodes $x\succ y$ but not $y\succ x$. Then
$\rho^{\prime}_{\\_}1$ also encodes $x\succ y$ but not $y\succ x$. Hence
$Tr(\Pi^{x\succ y}(\mathcal{E}_{\\_}{qcv}(\rho^{\prime})))>0$.

Now, let $\rho$ and $\rho^{\prime}$ be quantum ballot profiles such that
$Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq i}(\rho)))=Tr(\Pi^{x\succ y}(Tr_{\\_}{\neq
i}(\rho^{\prime})))$ and $Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq
i}(\rho)))=Tr(\Pi^{y\succ x}(Tr_{\\_}{\neq i}(\rho^{\prime})))$ for each voter
$v_{\\_}i$. If $Tr(\Pi^{x\succ y}(\mathcal{E}_{\\_}{qcv}(\rho)))>0$, then we
know $y\succ x$ is not encoded by all $\rho_{\\_}i$. Since $Tr(\Pi^{x\succ
y}(\rho_{\\_}i))=Tr(\Pi^{x\succ y}(\rho^{\prime}_{\\_}i))$ and $Tr(\Pi^{y\succ
x}(\rho_{\\_}i))=Tr(\Pi^{y\succ x}(\rho_{\\_}i^{\prime}))$ for each voter
$v_{\\_}i$, we know that $y\succ x$ is not encoded by all
$\rho^{\prime}_{\\_}i$. Hence $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(\rho^{\prime})))>0$. $\square$

###### Theorem 6

The Quantum Condorcet Voting $\mathcal{E}_{\\_}{qcv}$ does not satisfy sharp
dictatorship.

Proof: We will construct a ballot profile in which no candidate is a dictator.
Let $\\{x,y,z\\}$ be the set of candidates. Let
$\rho=\rho_{\\_}1\otimes\rho_{\\_}2\otimes\rho_{\\_}3$ be a quantum ballot
profile where $\rho_{\\_}1=|x\succ y\succ z\rangle\langle x\succ y\succ
z|,\rho_{\\_}2=|y\succ z\succ x\rangle\langle y\succ z\succ
x|,\rho_{\\_}3=|z\succ x\succ y\rangle\langle z\succ x\succ y|$. Then the weak
order generated by $\mathcal{E}_{\\_}{qcv}$ according to the Condorcet score
is $x\equiv y\equiv z$. The completion of $x\equiv y\equiv z$ is $\\{x\succ
y\succ z,x\succ z\succ y,y\succ x\succ z,y\succ z\succ x,z\succ x\succ
y,z\succ y\succ x\\}$. Therefore the quantum state generated in step 4 of
quantum Condorcet voting is $\sigma^{1}=\frac{1}{6}(|x\succ y\succ
z\rangle\langle x\succ y\succ z|+|x\succ z\succ y\rangle\langle x\succ z\succ
y|+|y\succ x\succ z\rangle\langle y\succ x\succ z|+|y\succ z\succ
x\rangle\langle y\succ z\succ x|+|z\succ x\succ y\rangle\langle z\succ x\succ
y|+|z\succ y\succ x\rangle\langle z\succ y\succ x|)$. Therefore, we have
$Tr(\Pi^{x\succ y}\rho_{\\_}1)=1$ but $Tr(\Pi^{x\succ
y}(\mathcal{E}_{\\_}{qcv}(\rho)))<1$, $Tr(\Pi^{y\succ z}\rho_{\\_}2)=1$ but
$Tr(\Pi^{y\succ z}(\mathcal{E}_{\\_}{qcv}(\rho)))<1$, $Tr(\Pi^{z\succ
x}\rho_{\\_}3)=1$ but $Tr(\Pi^{z\succ x}(\mathcal{E}_{\\_}{qcv}(\rho)))<1$.
This violates sharp dictatorship. $\square$

###### Theorem 7

The Quantum Condorcet Voting $\mathcal{E}_{\\_}{qcv}$ does not satisfy unsharp
dictatorship.

Proof: Consider again the profile constructed in the above proof. We have
$Tr(\Pi^{z\succ x}(\mathcal{E}_{\\_}{qcv}(\rho)))>0$ but $Tr(\Pi^{z\succ
x}\rho_{\\_}1)\ngtr 0$, $Tr(\Pi^{x\succ y}(\mathcal{E}_{\\_}{qcv}(\rho)))>0$
but $Tr(\Pi^{x\succ y}\rho_{\\_}2)\ngtr 0$, $Tr(\Pi^{y\succ
z}(\mathcal{E}_{\\_}{qcv}(\rho)))>0$ but $Tr(\Pi^{y\succ z}\rho_{\\_}3)\ngtr
0$. This violates unsharp dictatorship. $\square$

By combining theorems 2-7 we conclude that Quantum Condorcet Voting satisfies
Quantum Unanimity and the QIIA but does not Quantum Dictatorship. In other
words, we can infer the following corollary:

###### Corollary 1

Arrow’s impossibility theory is not valid in quantum voting.

##  4 Related work

Most of the related work on quantum voting focus on the security of voting.
The first quantum voting protocol was proposed by Hillery et al. [9]. They
proposed two voting modes, namely traveling ballot and distributed ballot to
ensure the security of voting. The protocol designed by Vaccaro et al. [25]
uses entangled states to ensure that the votes are anonymous and to allow the
votes to be tallied. The entanglement is distributed over separated sites; the
physical inaccessibility of any site is sufficient to guarantee the anonymity
of the votes. Horoshko and Kilin [10] proposed a quantum anonymous voting
scheme based on Bell-state. Their protocol protects both the voters from a
curious tallyman and all the participants from a dishonest voter in an
unconditionally secure way. Wang et al. [26] proposed a quantum anonymous
voting protocol assisted by two kinds of entangled quantum states. They
provided a mechanism of opening and permuting the ordered votes of all the
voters in an anonymous manner; any party who is interested in the voting
results can obtain the voting result through a simple calculation. Their
protocol possesses the properties of privacy, self-tallying, nonreusability,
verifiability, and fairness at the same time.

In our previous work [21] a simple voting protocol based on Quantum Blockchain
was proposed. Despite its simplicity, our protocol satisfies the most
important properties of secure voting protocols: is anonymous, binding, non-
reusable, verifiable, eligible, fair and self-tallying. The protocol could
also be implemented using presently available technology. One limitation of
this protocol is that it works for only 2 candidates. In a recent paper [20]
we overcame this limitation by realizing classical Condorcet voting on Quantum
Blockchain.

##  5 Conclusion and future work

In this paper, we study Arrow’s Impossibility Theorem in the quantum setting.
We first modify the definition of QIIA such that it precisely captures the
idea of independence of irrelevant alternatives. Then we present a detailed
proof of the violation of Arrow’s Impossibility Theorem with our modified
definition.

In [20], we have demonstrated that Condorcet voting on Quantum Blockchain
significantly simplifies the task of electronic voting and in the same time
ensures many desired security properties. In the future, we will further
improve Quantum Condorcet Voting such that it has advantages for both security
and the quality of the democratic processes. We will also investigate the
validity of other theorems of classical social choice theory in the quantum
setting. Those theorems include Sen’s Theorem on the impossibility of a
Paretian Liberal [19], the Muller-Satterthwaite Theorem on surjective
monotonicity [14] and the Gibbard-Satterthwaite Theorem on strategic
manipulation [8]. The third direction of research we are interested in is
quantum logic for social choice. Modal logic has been used as a powerful tool
to model and reason about social choice [24, 2, 7, 17]. It is both natural and
valuable to develop a quantum logic to model and reason about quantum social
choice.

## Appendix: basics of quantum information

Some primitives of quantum information which are used in this paper are
collected in this appendix. The readers who are interested in quantum
information are recommended to textbooks such as Yanofsky and Mannucci [28],
Scherer [18], Nielsen and Chuang [15] and Watrous [27].

######  Definition 8 (Hilbert space)

A (finite-dimensional) Hilbert space $\mathsf{H}$ is a

  1. 1.

complex vector space, that is,

$\phi,\psi\in\mathsf{H}$ and $a,b\in\mathbb{C}$ $\Rightarrow$
$a\phi+b\psi\in\mathsf{H}$,

  2. 2.

with a (positive-definite) scalar product
$\langle\cdot|\cdot\rangle:\mathsf{H}\times\mathsf{H}\mapsto\mathbb{C}$ such
that for all $\phi,\psi,\phi_{\\_}1,\phi_{\\_}2\in\mathsf{H}$ and
$a,b\in\mathbb{C}$

    1. (a)

$\langle\phi|\psi\rangle=\overline{\langle\psi|\phi\rangle}$

    2. (b)

$\langle\phi|\phi\rangle\geq 0$

    3. (c)

$\langle\phi|\phi\rangle=0$ iff $\phi=0$

    4. (d)

$\langle\psi|a\phi_{\\_}1+b\phi_{\\_}2\rangle=a\langle\psi|\phi_{\\_}1\rangle+b\langle\psi|\phi_{\\_}2\rangle$

In quantum computation and quantum information, we only consider finite-
dimensional Hilbert spaces. A vector $\phi$ of a Hilbert space is usually
represented in the Dirac notion as $|\phi\rangle$ in quantum computing. A
Hilbert space $\mathsf{H}$ induces a norm $\lVert.\rVert$ defined by
$\lVert\phi\rVert=\sqrt{\langle\phi|\phi\rangle}$ for any $\phi\in\mathsf{H}$.

######  Definition 9 (orthonormal basis and dimension)

An orthonormal basis $\\{|\phi_{\\_}i\rangle\\}$ for a Hilbert space
$\mathsf{H}$ is a basis of $\mathsf{H}$ whose vectors are unit vectors and are
orthogonal to each other, that is, for any
$|\phi_{\\_}i\rangle,|\phi_{\\_}j\rangle$, $\lVert\phi_{\\_}i\rVert=1$ and
$\langle\phi_{\\_}i|\phi_{\\_}j\rangle=0$. The dimension of a Hilbert space is
the number of vectors of an orthonormal basis.

######  Definition 10 (tensor product)

Given Hilbert spaces $V$ and $W$ of dimension $m$ and $n$ respectively, their
tensor product, denoted $V\otimes W$, is a $mn$-dimensional space consisting
of linear combinations of outer products $|v\rangle\otimes|w\rangle$ of
vectors $|v\rangle=(v_{\\_}1,v_{\\_}2,\dots,v_{\\_}m)^{T}\in V$ and
$|w\rangle=(w_{\\_}1,w_{\\_}2,\dots,w_{\\_}n)^{T}\in W$, where

$|v\rangle\otimes|w\rangle=\begin{bmatrix}v_{\\_}1w_{\\_}1\\\
v_{\\_}1w_{\\_}2\\\ \vdots\\\ v_{\\_}mw_{\\_}n\end{bmatrix}$ (1)

######  Definition 11 (subspace)

A subspace of a Hilbert space $V$ is a subset $W$ of $V$ such that $W$ is also
a Hilbert space.

######  Definition 12 (operator)

A linear map $A:\mathsf{H}\mapsto\mathsf{H}$ is called an operator on
$\mathsf{H}$.

We use $L(\mathsf{H})$ to denote the set of all operators on $\mathsf{H}$.

######  Definition 13 (adjoint)

The operator $A^{*}:\mathsf{H}\mapsto\mathsf{H}$ that satisfies $\langle
A^{*}\phi|\psi\rangle=\langle\phi|A\psi\rangle$ for all
$\phi,\psi\in\mathsf{H}$ is called the adjoint operator to $A$.

######  Definition 14 (projector)

A projector of a Hilbert space $\mathsf{H}$ is a linear map
$P:\mathsf{H}\mapsto\mathsf{H}$ such that $P^{2}=P$ and $P^{*}=P$.

Projectors are related to projective measurements in quantum mechanics. We use
an operator $M$ to represent an observable of the quantum system being
observed, with a decomposition $M=\sum_{\\_}{m}P_{\\_}{m}$, where $P_{\\_}m$
is the projector onto the eigenspace of $M$ with eigenvalue $m$. The result of
measuring the state $|\psi\rangle$ will be one of $M$’s eigenvalues, and the
probability of getting result $m$ is $p(m)=\langle\psi|P_{\\_}m|\psi\rangle$.

######  Definition 15 (trace)

Let $\mathsf{H}$ be a Hilbert space and $\rho$ be an operator on $\mathsf{H}$.
The trace of $\rho$ is defined by

$Tr(\rho)=\sum_{\\_}{i}\langle i|\rho|i\rangle$

where $\\{|i\rangle\\}$ is an orthonormal basis of $\mathsf{H}$.

######  Definition 16 (positive semidefinite operator)

An operator $A:\mathsf{H}\mapsto\mathsf{H}$ is positive semidefinite if it
holds that $A=B^{*}B$ for some operator $B\in L(\mathsf{H})$.

######  Definition 17 (density operator)

A positive semidefinite operator $\rho$ on $\mathsf{H}$ is a density operator
if it holds that $\rho=\rho^{*}$ and $Tr(\rho)=1$.

######  Definition 18 (partial trace)

Suppose the composite system of two subsystems $A$ and $B$ is described by the
density operator $\rho_{\\_}{AB}$. The partial trace over $B$ is defined by

$\rho_{\\_}A=Tr_{\\_}B(\rho_{\\_}{AB})=\sum_{\\_}{i}(I_{\\_}A\otimes\langle
i|)\rho_{\\_}{AB}(I_{\\_}A\otimes|i\rangle)$

where $\\{|i\rangle\\}$ is an orthonormal basis of the Hilbert space
$\mathsf{H}_{\\_}B$. $\rho_{\\_}A$ is called the reduced density operator of
the subsystem $A$. The partial trace over $A$ can be defined in a similar way.

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