$\displaystyle\mathcal{T}_{*}^{(1)}\leqslant\int_{\mathbb{R}^{3}}\mathrm{d}y\int_{\mathbb{R}^{3}}\mathrm{d}z\int_{\mathbb{R}^{3}}\mathrm{d}z^{\prime}\;|\alpha_{*}(z)\,\alpha_{*}(z^{\prime})\,\alpha_{*}(z-y)\,\alpha_{*}(z^{\prime}-y)|$
(5.95)
$\displaystyle\times\fint_{Q_{B}}\mathrm{d}X\;\Bigl{|}\bigl{[}\mathrm{e}^{\mathrm{i}\frac{z}{2}\Pi_{X}}\Psi(X)\bigr{]}\bigl{[}\mathrm{e}^{\mathrm{i}\frac{z^{\prime}}{2}\Pi_{X}}\Psi(X)\bigr{]}\bigl{[}\bigl{(}\mathrm{e}^{\mathrm{i}\frac{z+y}{2}\Pi_{X}}-1\bigr{)}\Psi(X)\bigr{]}\bigl{[}\bigl{(}\mathrm{e}^{\mathrm{i}\frac{z^{\prime}+y}{2}\Pi_{X}}-1\bigr{)}\Psi(X)\bigr{]}\Bigr{|}.$

When we use Hölder’s inequality, (4.123), (4.98), and (2.7), we see that the
integral in the second line can be bounded by

$\displaystyle\|\mathrm{e}^{\mathrm{i}\frac{z}{2}\Pi}\Psi\|_{6}^{2}\;\|(\mathrm{e}^{\mathrm{i}\frac{z+y}{2}\Pi}-1)\Psi\|_{6}\;\|(\mathrm{e}^{\mathrm{i}\frac{z+y}{2}\Pi}-1)\Psi\|_{2}\leqslant
C\;\bigl{|}\frac{z+y}{2}\bigr{|}\;B^{-\nicefrac{{3}}{{2}}}\;\|\Pi\Psi\|_{2}^{4}.$
(5.96)

Proposition 5.7 provides us with a bound for $\|\Pi\Psi\|_{2}$. In combination
with (5.95), (5.96), Young’s inequality, and the bound $|z+y|\leqslant
2|z|+|z-y|$, this implies

$\displaystyle\mathcal{T}_{*}^{(1)}$ $\displaystyle\leqslant
C\,B^{-\nicefrac{{3}}{{2}}}\,\left(B^{2}\|\Psi\|_{2}^{4}+D_{1}B^{4}\right)\int_{\mathbb{R}^{3}}\mathrm{d}y\;\Bigl{|}\int_{\mathbb{R}^{3}}\mathrm{d}z\;|z+y|\;|\alpha_{*}(z)\alpha_{*}(y-z)|\Bigr{|}^{2}$
$\displaystyle\leqslant
C\left(B^{-\nicefrac{{1}}{{2}}}\|\Psi\|_{2}^{4}+D_{1}B^{\nicefrac{{5}}{{2}}}\right)\,\|\alpha_{*}\|_{\nicefrac{{4}}{{3}}}\,\|\,|\cdot|\alpha_{*}\|_{\nicefrac{{4}}{{3}}},$
(5.97)

where the right side is finite by (A.1). Similarly, we see that
$\mathcal{T}_{*}^{(2)}$ is bounded by the right side of (5.97).

_Upper bound on $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ in (5.91)._ Bounds for
$\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ can be obtain along the same lines as
the bound for $\mathcal{T}_{*}^{(1)}$. We apply the same change of variables
as above and use estimates similar to the ones in (5.95). In case of
$\mathcal{T}_{1}$, the bound in (5.96) needs to be replaced by

$\displaystyle\|\Psi\|_{6}^{2}\,\bigl{\|}\sin^{2}\bigl{(}\frac{z-y}{2}\Pi\bigr{)}\Psi\bigr{\|}_{6}\,\bigl{\|}\sin^{2}\bigl{(}\frac{z-y}{2}\Pi\bigr{)}\Psi\bigr{\|}_{2}\leqslant
C\;\frac{|z-y|}{2}\;B^{-\nicefrac{{3}}{{2}}}\,\|\Pi\Psi\|_{2}^{4}.$ (5.98)

Here, we used $\sin^{2}(x)\leqslant|x|$ and the operator inequality in (4.98)
to estimate the third factor. For the first and the second factor, we used

$\displaystyle\sin^{2}\bigl{(}\frac{z-y}{4}\Pi\bigr{)}=-\frac{1}{4}\bigl{(}2+\mathrm{e}^{\mathrm{i}\frac{z-y}{2}\Pi}+\mathrm{e}^{-\mathrm{i}\frac{z-y}{2}\Pi}\bigr{)}$

and (2.7) or (4.123), respectively. A bound for $\mathcal{T}_{2}$ can be
proved analogously. The final estimate we obtain in this way reads

$\mathcal{T}_{1}+\mathcal{T}_{2}\leqslant
C\left(B^{-\nicefrac{{1}}{{2}}}\|\Psi\|_{2}^{4}+D_{1}B^{\nicefrac{{5}}{{2}}}\right).$
(5.99)

In combination with (5.90), (5.93), and (5.97), this proves (5.89).

_Step 3._ We denote $c:=\frac{1}{2}\|\widehat{\alpha}_{*}\|_{4}$, insert
(5.87) and (5.89) into (5.86) and obtain

$\displaystyle CB^{\frac{1}{4}}\|\Psi\|_{2}^{\nicefrac{{1}}{{2}}}\geqslant
c\|\Psi\|_{4}-CB^{\frac{1}{8}}\|\Psi\|_{2}-CB^{\frac{1}{2}},$ (5.100)

which holds for $B$ small enough. For $\eta>0$ the left side is bounded from
above by a constant times $\eta\|\Psi\|_{2}+\eta^{-1}B^{\frac{1}{2}}$ and
Hölder’s inequality implies $\|\Psi\|_{4}\geqslant\|\Psi\|_{2}$. Accordingly,

$C\left(\eta\|\Psi\|_{2}+\eta^{-1}B^{\frac{1}{2}}\right)\geqslant(c-CB^{\frac{1}{8}})\|\Psi\|_{2}-CB^{\frac{1}{2}}.$
(5.101)

When we choose $\eta$ and $B$ in (5.101) small enough, this proves the claim.
∎

###  5.4 Proof of Theorem 5.1

The assumption (5.1) in Theorem 5.1, Corollary 5.6, Proposition 5.7 and Lemma
5.14 imply the decomposition $\alpha=A^{*}\Psi+\xi_{0}$, where $\Psi$ and
$\xi_{0}$ in (5.29) obey

$\displaystyle\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}=B^{-1}\|\Psi\|_{2}^{2}+B^{-2}\|\Pi\Psi\|_{2}^{2}\leqslant
C$ (5.102)

and

$\displaystyle\|\xi_{0}\|_{{H^{1}(Q_{B}\times\mathbb{R}_{\mathrm{s}}^{3})}}^{2}\leqslant
CB^{2}\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}+D_{1}\bigr{)}.$
(5.103)

Define

$\displaystyle\xi$
$\displaystyle:=\xi_{0}+\bigl{(}\cos\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}-1\bigr{)}\alpha_{*}(r)\Psi(X).$
(5.104)

Then, (5.2) holds and we claim that $\xi$ satisfies (5.3). To prove this, we
estimate the second term in (5.104). We use $1-\cos(x)\leqslant|x|$ and (4.98)
to bound

$\displaystyle\bigl{\|}\bigl{(}\cos\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}-1\bigr{)}\Psi\alpha_{*}\bigr{\|}_{2}^{2}$
$\displaystyle\leqslant
C\;\||\cdot|\alpha_{*}\|_{2}^{2}\;\|\Pi\Psi\|_{2}^{2}\leqslant
CB^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2},$

where the right side is finite by the decay properties of $\alpha_{*}$ in
(A.1). Using additionally (5.40), we also see that

$\displaystyle\bigl{\|}\Pi_{X}\bigl{(}\cos\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}-1\bigr{)}\alpha_{*}\Psi\bigr{\|}_{2}^{2}$
$\displaystyle\leqslant\bigl{\|}\bigl{[}\cos\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}-1\bigr{]}\Pi_{X}\alpha_{*}\Psi\bigr{\|}_{2}^{2}+\bigl{\|}\sin\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}\mathbf{B}\wedge
r\alpha_{*}\Psi\bigr{\|}_{2}^{2}$ $\displaystyle\leqslant
C\bigl{(}\|\Pi\Psi\|_{2}^{2}+B^{2}\|\Psi\|_{2}^{2}\bigr{)}\leqslant
CB^{3}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}$

holds. Finally, an application of (5.43) and (4.98) allows us to estimate

$\displaystyle\bigl{\|}\pi_{r}\bigl{(}\cos\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}-1\bigr{)}\Psi\alpha_{*}\bigr{\|}_{2}^{2}$
$\displaystyle=\bigl{\|}\bigl{[}\bigl{(}\cos\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}-1\bigr{)}\widetilde{\pi}_{r}+\frac{\mathrm{i}}{2}\sin\bigl{(}\frac{r}{2}\Pi_{X}\bigr{)}\Pi_{X}\bigr{]}\Psi\alpha_{*}\bigr{\|}_{2}^{2}$
$\displaystyle\leqslant
C\bigl{(}\|\Pi\Psi\|_{2}^{2}+B^{2}\|\Psi\|_{2}^{2}\bigr{)}\leqslant
CB^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}.$

This proves that $\xi$ obeys (5.3) and ends the proof of Theorem 5.1.

##  6 The Lower Bound on (1.21) and Proof of Theorem 2 (b)

###  6.1 The BCS energy of low-energy states

In this section, we provide the lower bound on (1.21) and the proof of Theorem
2 (b), and thereby complete the proof of Theorems 1 and 2. Let $D_{1}\geqslant
0$ and $D\in\mathbb{R}$ be given and assume that $\Gamma$ is a gauge-periodic
state at temperature $T=T_{\mathrm{c}}(1-DB)$ that satisfies (5.1). Corollary
5.2 provides us with a decomposition of the Cooper pair wave function
$\alpha=[\Gamma]_{12}$ in terms of $\Psi_{\leqslant}$ in (5.4) and $\sigma$ in
(5.9), where $\|\Psi_{\leqslant}\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\leqslant C$
and where the bound

$\displaystyle\|\Psi_{\leqslant}\|_{H_{\mathrm{mag}}^{2}(Q_{B})}^{2}$
$\displaystyle\leqslant C\,\varepsilon
B^{-1}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}$ (6.1)

holds in terms of the function $\Psi$ in Theorem 5.1. With the function
$\Psi_{\leqslant}$ we construct a Gibbs state $\Gamma_{\Delta}$ with the gap
function $\Delta\equiv\Delta_{\Psi_{\leqslant}}$ as in (3.1). Using
Proposition 3.4, we write the BCS free energy of $\Gamma$ as

$\displaystyle\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma)-\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma_{0})$
$\displaystyle=-\frac{1}{4}\langle\Delta,L_{T,B}\Delta\rangle+\frac{1}{8}\langle\Delta,N_{T,B}(\Delta)\rangle+\|\Psi_{\leqslant}\|_{2}^{2}\;\langle\alpha_{*},V\alpha_{*}\rangle$
$\displaystyle\hskip
10.0pt+\operatorname{Tr}\bigl{[}\mathcal{R}_{T,B}(\Delta)\bigr{]}+\frac{T}{2}\mathcal{H}_{0}(\Gamma,\Gamma_{\Delta})-\fint_{Q_{B}}\mathrm{d}X\int_{\mathbb{R}^{3}}\mathrm{d}r\;V(r)\,|\sigma(X,r)|^{2},$

where

$\|\mathcal{R}_{T,B}(\Delta)\|_{1}\leqslant
C\;B^{3}\;\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{6}.$

We also apply Theorem 3.5 to compute the terms in the first line on the right
side, and find the lower bound

$\displaystyle\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma)-\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma_{0})$
$\displaystyle\geqslant
B^{2}\;\mathcal{E}^{\mathrm{GL}}_{D,B}(\Psi_{\leqslant})-C\bigl{(}B^{3}+\varepsilon
B^{2}\bigr{)}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}$ $\displaystyle\hskip
30.0pt+\frac{T}{2}\mathcal{H}_{0}(\Gamma,\Gamma_{\Delta})-\fint_{Q_{B}}\mathrm{d}X\int_{\mathbb{R}^{3}}\mathrm{d}r\;V(r)\,|\sigma(X,r)|^{2}.$
(6.2)

The relative entropy is nonnegative and the last term on the right side is
nonpositive. In the next section we show that their sum is negligible.

###  6.2 Estimate on the relative entropy

In this section we prove a lower bound for the second line in (6.2), showing
that it is negligible. We start with the following lower bound for the
relative entropy.

######  Lemma 6.1.

For all admissible BCS states $\Gamma$, we have

$\displaystyle T\mathcal{H}_{0}(\Gamma,\Gamma_{\Delta})$
$\displaystyle\geqslant\operatorname{Tr}\Bigl{[}(\Gamma-\Gamma_{\Delta})\frac{H_{\Delta}}{\tanh(\frac{\beta}{2}H_{\Delta})}(\Gamma-\Gamma_{\Delta})\Bigr{]}.$
(6.3)

###### Proof.

The proof is given in [17, Lemma 5] and uses the fact that $\Gamma_{\Delta}$
is admissible, which follows from Lemma 3.1. ∎

To be able to combine the term on the right side of (6.3) and the last term on
the right side of (6.2), we first need to replace the operator $H_{\Delta}$ in
the second factor on the right side of (6.3) by $H_{0}$. To that end, we note
that the estimate
$H_{\Delta}^{2}\geqslant(1-\delta)H_{0}^{2}-\delta^{-1}\|\Delta\|_{\infty}^{2}$
holds for $0<\delta<1$ and we rewrite it as

$\displaystyle
H_{0}\leqslant(1-\delta)^{-1}\bigl{(}H_{\Delta}^{2}+\delta^{-1}\|\Delta\|_{\infty}^{2}\bigr{)}.$
(6.4)

Furthermore, we note that the series expansion

$\displaystyle\frac{x}{\tanh(\frac{x}{2})}=2+\sum_{k=1}^{\infty}\bigl{(}2-\frac{8k^{2}\pi^{2}}{x^{2}+4k^{2}\pi^{2}}\bigr{)},$

see [17, Eq. (5.14)], shows that the function
$x\mapsto\frac{\sqrt{x}}{\tanh(\frac{\beta}{2}\sqrt{x})}$ is operator
monotone. We use this together with (6.4) and the monotonicity of the map
$x\mapsto\tanh(x)$, which yields

$\displaystyle\begin{pmatrix}K_{T,B}&0\\\
0&\overline{K_{T,B}}\end{pmatrix}=\frac{H_{0}}{\tanh(\frac{\beta}{2}H_{0})}$
$\displaystyle\leqslant(1-\delta)^{-\nicefrac{{1}}{{2}}}\frac{\sqrt{H_{\Delta}^{2}+\delta^{-1}\|\Delta\|_{\infty}^{2}}}{\tanh(\frac{\beta}{2}\sqrt{H_{\Delta}^{2}+\delta^{-1}\|\Delta\|_{\infty}^{2}})}.$

When we apply a first order Taylor expansion on the right side, the above
inequality can be written as

$\displaystyle\begin{pmatrix}K_{T,B}&0\\\ 0&\overline{K_{T,B}}\end{pmatrix}$
$\displaystyle\leqslant(1-\delta)^{-\nicefrac{{1}}{{2}}}\Bigl{[}\frac{H_{\Delta}}{\tanh(\frac{\beta}{2}H_{\Delta})}+\frac{\beta}{4}\int_{0}^{\delta^{-1}\|\Delta\|_{\infty}^{2}}\mathrm{d}t\;g\bigl{(}\frac{\beta}{2}\sqrt{H_{\Delta}^{2}+t}\bigr{)}\Bigr{]}$

with the nonnegative function

$\displaystyle g(x)$
$\displaystyle:=\frac{1}{x}\frac{1}{\tanh(x)}-\frac{1}{\tanh^{2}(x)}\frac{1}{\cosh^{2}(x)}.$

Using $\sup_{x\geqslant 0}g(x)\leqslant 1$ and $1\leqslant\frac{x}{\tanh(x)}$,
we conclude that

$\displaystyle\begin{pmatrix}K_{T,B}&0\\\
0&\overline{K_{T,B}}\end{pmatrix}\leqslant(1-\delta)^{-\nicefrac{{1}}{{2}}}\bigl{(}1+\frac{\delta^{-1}\beta^{2}}{8}\;\|\Delta\|_{\infty}^{2}\bigr{)}\frac{H_{\Delta}}{\tanh(\frac{\beta}{2}H_{\Delta})}$

holds. We choose $\delta:=\|\Delta\|_{\infty}$ and note that Lemma 4.1 and
(2.7) imply

$\|\Delta\|_{\infty}\leqslant
C\;B^{\nicefrac{{1}}{{4}}}\;\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}.$ (6.5)

In particular, $\delta<1$ as long as $B>0$ is sufficiently small, and we have

$\displaystyle\frac{H_{\Delta}}{\tanh(\frac{\beta}{2}H_{\Delta})}\geqslant\frac{\sqrt{1-\|\Delta\|_{\infty}}}{1+\frac{\beta^{2}}{8}\|\Delta\|_{\infty}}\;\begin{pmatrix}K_{T,B}&0\\\
0&\overline{K_{T,B}}\end{pmatrix}\geqslant(1-C\|\Delta\|_{\infty})\;\begin{pmatrix}K_{T,B}&0\\\
0&\overline{K_{T,B}}\end{pmatrix}.$ (6.6)

In combination, (6.3) and (6.6) prove

$\displaystyle\frac{1}{2}\operatorname{Tr}\Bigl{[}(\Gamma-\Gamma_{\Delta})\frac{H_{\Delta}}{\tanh(\frac{\beta}{2}H_{\Delta})}(\Gamma-\Gamma_{\Delta})\Bigr{]}$
$\displaystyle\geqslant(1-C\|\Delta\|_{\infty})\langle\alpha-\alpha_{\Delta},K_{T,B}(\alpha-\alpha_{\Delta})\rangle+(1-C\|\Delta\|_{\infty})\operatorname{Tr}[(\gamma-\gamma_{\Delta})K_{T,B}(\gamma-\gamma_{\Delta})],$

where we can drop the last term for a lower bound because it is nonnegative if
$B$ is sufficiently small. This is the lower bound for the relative entropy of
$\Gamma$ with respect to $\Gamma_{\Delta}$ we were looking for.

It remains to combine the first term on the right side and the interaction
term on the right side of (6.2). Let us define the function
$\eta:=\alpha_{*}\Psi_{\leqslant}-\alpha_{\Delta}$. By Corollary 5.2 we have
$\alpha-\alpha_{\Delta}=\sigma+\eta$ as well as

$\displaystyle\frac{T}{2}\mathcal{H}_{0}(\Gamma,\Gamma_{\Delta})-\fint_{Q_{B}}\mathrm{d}X\int_{\mathbb{R}^{3}}\mathrm{d}r\;V(r)|\sigma(X,r)|^{2}$
$\displaystyle\hskip
20.0pt\geqslant(1-C\|\Delta\|_{\infty})\langle\sigma+\eta,K_{T,B}(\sigma+\eta)\rangle-\langle\sigma,V\sigma\rangle$
$\displaystyle\hskip
20.0pt\geqslant(1-C\|\Delta\|_{\infty})\langle\sigma,(K_{T,B}-V)\sigma\rangle-C\|\Delta\|_{\infty}\|V\|_{\infty}\|\sigma\|_{2}^{2}-2\,|\langle\eta,K_{T,B}\sigma\rangle|.$
(6.7)

From (5.14) we know that the lowest eigenvalue of $K_{T,B}-V$ is bounded from
below by $-CB$. In combination with (5.10) and (6.5), this implies that the
first two terms on the right side of (6.7) are bounded from below by

$\displaystyle-C\varepsilon^{-1}B^{\nicefrac{{9}}{{4}}}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\;\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}+D_{1}\bigr{)}^{\nicefrac{{1}}{{2}}}.$
(6.8)

To estimate the last term on the right side of (6.7), we use (5.15) to replace
$K_{T,B}$ by $K_{T_{\mathrm{c}},B}$, which yields the estimate

$\displaystyle|\langle\eta,(K_{T,B}-K_{T_{\mathrm{c}},B})\sigma\rangle|$
$\displaystyle\leqslant 2D_{0}B\,\|\sigma\|_{2}\,\|\eta\|_{2}\leqslant
C\,B^{3}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\;\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}+D_{1}\bigr{)}^{\nicefrac{{1}}{{2}}}.$

To obtain this result we also used (5.10), Proposition 3.2 and (6.1). Next, we
decompose $\eta=\eta_{0}+\eta_{\perp}$ with $\eta_{0}(\Delta)$ and
$\eta_{\perp}(\Delta)$ as in Proposition 3.2 and write

$\displaystyle\langle\eta,K_{T_{\mathrm{c}},B}\sigma\rangle$
$\displaystyle=\langle\eta_{0},K_{T_{\mathrm{c}},B}\sigma\rangle+\langle\eta_{\perp},K_{T_{\mathrm{c}},B}(\sigma-\sigma_{0})\rangle+\langle\eta_{\perp},K_{T_{\mathrm{c}},B}\sigma_{0}\rangle.$
(6.9)

Using (3.7) and (5.10), we see that the first term on the right side of (6.9)
is bounded by

$\displaystyle|\langle\eta_{0},K_{T_{\mathrm{c}},B}\sigma\rangle|$
$\displaystyle\leqslant\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}}\,\eta_{0}\bigr{\|}_{2}\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}}\,\sigma\bigr{\|}_{2}$
$\displaystyle\leqslant
C\varepsilon^{-\nicefrac{{1}}{{2}}}B^{\nicefrac{{5}}{{2}}}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\;\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}+D_{1}\bigr{)}^{\nicefrac{{1}}{{2}}}.$
(6.10)

We note that $\sigma-\sigma_{0}=\xi$ and use (3.8), (5.3), and (6.1) to
estimate

$\displaystyle|\langle\eta_{\perp},K_{T_{\mathrm{c}},B}\xi\rangle|$
$\displaystyle\leqslant\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}}\,\eta_{\perp}\bigr{\|}_{2}\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}}\,\xi\bigr{\|}_{2}$
$\displaystyle\leqslant
C\varepsilon^{\nicefrac{{1}}{{2}}}B^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\;\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}+D_{1}\bigr{)}^{\nicefrac{{1}}{{2}}}.$
(6.11)

It remains to estimate the last term on the right side of (6.9), which we
write as

$\displaystyle\langle\eta_{\perp},K_{T_{\mathrm{c}},B}\sigma_{0}\rangle$
$\displaystyle=\langle\eta_{\perp},K_{T_{\mathrm{c}}}^{r}\sigma_{0}\rangle+\langle\eta_{\perp},[K_{T_{\mathrm{c}},B}^{r}-K_{T_{\mathrm{c}}}^{r}]\sigma_{0}\rangle+\langle\eta_{\perp},(U-1)K_{T_{\mathrm{c}},B}^{r}\sigma_{0}\rangle$
$\displaystyle\hskip
190.0pt+\langle\eta_{\perp},UK_{T_{\mathrm{c}},B}^{r}(U^{*}-1)\sigma_{0}\rangle.$
(6.12)

with the unitary operator $U$ in (5.20). We recall that the operators
$K_{T_{\mathrm{c}},B}^{r}$ and $K_{T_{\mathrm{c}}}^{r}$ act on the relative
coordinate $r=x-y$.

Since $\Delta(X,r)=-2V(r)\alpha_{*}(r)\Psi_{\leqslant}(X)$ and
$\sigma_{0}(X,r)=\alpha_{*}(r)\Psi_{>}(X)$ we know from Proposition 3.2 (c)
that the first term on the right side of (6.12) vanishes. A bound for the
remaining terms is provided by the following lemma. Its proof will be given in
Section 6.4 below.

######  Lemma 6.2.

We have the following estimates on the remainder terms of (6.12):

  1. (a)

$|\langle\eta_{\perp},[K_{T_{\mathrm{c}},B}^{r}-K_{T_{\mathrm{c}}}^{r}]\sigma_{0}\rangle|\leqslant
CB^{3}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}$,

  2. (b)

$|\langle\eta_{\perp},(U-1)K_{T_{\mathrm{c}},B}^{r}\sigma_{0}\rangle|\leqslant
C\varepsilon^{\nicefrac{{1}}{{2}}}B^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}$,

  3. (c)

$|\langle\eta_{\perp},UK_{T_{\mathrm{c}},B}^{r}(U^{*}-1)\sigma_{0}\rangle|\leqslant
C\varepsilon^{\nicefrac{{1}}{{2}}}B^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}$.

Accordingly, we have

$\displaystyle|\langle\sigma,K_{T,B}\eta\rangle|$ $\displaystyle\leqslant
C\bigl{(}\varepsilon^{-\nicefrac{{1}}{{2}}}B^{\nicefrac{{5}}{{2}}}+\varepsilon^{\nicefrac{{1}}{{2}}}B^{2}\bigr{)}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\;\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}+D_{1}\bigr{)}^{\nicefrac{{1}}{{2}}}.$

We combine this with (6.2), (6.7), and (6.8) to see that

$\displaystyle\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma)-\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma_{0})$
$\displaystyle\geqslant\mathcal{E}^{\mathrm{GL}}_{D,B}(\Psi_{\leqslant})-C\bigl{(}\varepsilon^{-\nicefrac{{1}}{{2}}}B^{\nicefrac{{5}}{{2}}}+\varepsilon^{\nicefrac{{1}}{{2}}}B^{2}+\varepsilon^{-1}B^{\nicefrac{{9}}{{4}}}\bigr{)}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}+D_{1}\bigr{)}^{\nicefrac{{1}}{{2}}}.$
(6.13)

The optimal choice $\varepsilon=B^{\nicefrac{{1}}{{6}}}$ in (6.13) yields

$\displaystyle\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma)-\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma_{0})$
$\displaystyle\hskip 56.9055pt\geqslant
B^{2}\,\bigl{(}\mathcal{E}^{\mathrm{GL}}_{D,B}(\Psi_{\leqslant})-C\,B^{\nicefrac{{1}}{{12}}}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\;\bigl{(}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}+D_{1}\bigr{)}^{\nicefrac{{1}}{{2}}}\bigr{)}.$
(6.14)

The proves the lower bound for the BCS free energy in Theorem 1.

###  6.3 Conclusion

Using (6.14), we now finish the proofs of Theorem 1 and Theorem 2, and we
start with the former. Let $\Gamma$ be an approximate minimizer of the BCS
functional, i.e., let (5.1) hold with

$\displaystyle D_{1}:=E^{\mathrm{GL}}(D)+\rho$ (6.15)

and $\rho\geqslant 0$. Since $\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}\leqslant
C$ by (5.3), (6.14) implies

$\displaystyle
B^{2}\bigl{(}E^{\mathrm{GL}}(D)+\rho\bigr{)}\geqslant\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma)-\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma_{0})\geqslant
B^{2}\bigl{(}\mathcal{E}^{\mathrm{GL}}_{D,B}(\Psi_{\leqslant})-C\,B^{\nicefrac{{1}}{{12}}}\bigr{)}.$

This proves the claimed bound for the Cooper pair wave function of an
approximate minimizer of the BCS functional in Theorem 1.

We turn to the proof of Theorem 2. Let the temperature $T$ obey

$\displaystyle
T_{\mathrm{c}}(1-B(D_{\mathrm{c}}-D_{0}B^{\nicefrac{{1}}{{12}}}))<T\leqslant
T_{\mathrm{c}}(1+CB)$ (6.16)

with $D_{\mathrm{c}}$ in (1.20) and $D_{0}>0$. We claim that the normal state
$\Gamma_{0}$ minimizes the BCS functional for such temperatures $T$ if $D_{0}$
is chosen sufficiently large. Since Corollary 5.5 takes care of the remaining
temperature range, this implies part (b) of Theorem 2 and completes its proof.

To see that the above claim is true, we start with the lower bound in (6.14)
and assume that (5.1) holds with $D_{1}=0$. We drop the nonnegative quartic
term in the Ginzburg–Landau functional for a lower bound and obtain

$\displaystyle\mathcal{E}^{\mathrm{GL}}_{D,B}(\Psi_{\leqslant})$
$\displaystyle\geqslant
B^{-2}\langle\Psi_{\leqslant},(\Lambda_{0}\,\Pi^{2}-DB\Lambda_{2})\Psi_{\leqslant}\rangle\geqslant\Lambda_{2}\,(D_{\mathrm{c}}-D)\;B^{-1}\|\Psi_{\leqslant}\|_{2}^{2},$

with $\Lambda_{0}$ in (3.22), $\Lambda_{2}$ in (3.23), and with
$D\in\mathbb{R}$ defined by $T=T_{\mathrm{c}}(1-DB)$. We combine (5.6) and
(5.26) and estimate

$\displaystyle\|\Psi_{\leqslant}\|_{2}\geqslant\|\Psi\|_{2}-\|\Psi_{>}\|_{2}\geqslant
c\;B^{\nicefrac{{1}}{{2}}}\;\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}(1-C\,B^{\nicefrac{{5}}{{12}}}).$

When we insert our findings in the lower bound for the BCS energy in (6.14),
this gives

$0\geqslant\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma)-\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}(\Gamma_{0})\geqslant
c\;B^{2}\;\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}\bigl{(}(D_{\mathrm{c}}-D)-CB^{\nicefrac{{1}}{{12}}}\bigr{)},$
(6.17)

We note that the lower bound in (6.16) is equivalent to

$\displaystyle D_{\mathrm{c}}-D>D_{0}B^{\nicefrac{{1}}{{12}}}.$ (6.18)

When we choose $D_{0}>C$ with $C>0$ in (6.17) and use (6.18) to obtain a lower
bound for the right side of (6.17), we conclude that $\Psi=0$. By (5.2) and
(5.3), this implies that $\alpha=0$ whence $\Gamma$ is a diagonal state.
Therefore, $\Gamma_{0}$ is the unique minimizer of
$\mathcal{F}^{\mathrm{BCS}}_{\mathbf{B},T}$ if $T$ satisfies (6.16) with our
choice of $D_{0}$. As explained below (6.16), this proves Theorem 2.

###  6.4 Proof of Lemma 6.2

In this section we prove Lemma 6.2. Our proof of part (a) uses a Cauchy
integral representation for the operator $K_{T_{\mathrm{c}},B}-(\pi^{2}-\mu)$,
which is provided in Lemma 6.4 below. Let us start by defining the contour for
the Cauchy integral.

######  Definition 6.3 (Speaker path).

Let $R>0$, assume that $\mu\leqslant 1$ and define the following complex paths

$\displaystyle\begin{split}u_{1}(t)&:=\frac{\pi\mathrm{i}}{2\beta_{\mathrm{c}}}+(1+\mathrm{i})t\\\
u_{2}(t)&:=\frac{\pi\mathrm{i}}{2\beta_{\mathrm{c}}}-(\mu+1)t\\\
u_{3}(t)&:=-\frac{\pi\mathrm{i}}{2\beta_{\mathrm{c}}}t-(\mu+1)\\\
u_{4}(t)&:=-\frac{\pi\mathrm{i}}{2\beta_{\mathrm{c}}}-(\mu+1)(1-t)\\\
u_{5}(t)&:=-\frac{\pi\mathrm{i}}{2\beta_{\mathrm{c}}}+(1-\mathrm{i})t\end{split}$
$\displaystyle\begin{split}\phantom{\frac{\mathrm{i}}{\beta_{\mathrm{c}}}}t&\in[0,R],\\\
\phantom{\frac{\mathrm{i}}{\beta_{\mathrm{c}}}}t&\in[0,1],\\\
\phantom{\frac{\mathrm{i}}{\beta_{\mathrm{c}}}}t&\in[-1,1],\\\
\phantom{\frac{\mathrm{i}}{\beta_{\mathrm{c}}}}t&\in[0,1],\\\
\phantom{\frac{\mathrm{i}}{\beta_{\mathrm{c}}}}t&\in[0,R].\end{split}$

The speaker path is defined as the union of paths $u_{i}$, $i=1,\ldots,5$,
with $u_{1}$ taken in reverse direction, i.e.,

$\displaystyle{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}_{R}:=\mathop{\dot{-}}u_{1}\mathop{\dot{+}}u_{2}\mathop{\dot{+}}u_{3}\mathop{\dot{+}}u_{4}\mathop{\dot{+}}u_{5}.$

If $\mu>1$ we choose the same path as in the case $\mu=1$.

This path has the property that certain norms of the resolvent kernel of
$\pi^{2}$ are uniformly bounded for $z\in{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}_{R}$
and $R>0$. More precisely, Lemma 4.5 implies

$\displaystyle\sup_{0\leqslant B\leqslant B_{0}}\sup_{R>0}\sup_{w\in{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}_{R}}\bigl{[}\left\|\,|\cdot|^{a}g_{B}^{w}\right\|_{1}+\left\|\,|\cdot|^{a}\nabla
g_{B}^{w}\right\|_{1}\bigr{]}<\infty.$ (6.19)

We could also choose a path parallel to the real axis in Lemma 6.4 below. In
this case the above norms would depend on $R$. Although our analysis also
works in this case, we decided to use the path ${\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}_{R}$
because of the more elegant bound in (6.19). With the above definition at
hand, we are prepared to state the following lemma.

######  Lemma 6.4.

Let $H\colon\mathcal{D}(H)\rightarrow\mathcal{H}$ be a self-adjoint operator
on a separable Hilbert space $\mathcal{H}$ with $H\geqslant-\mu$ and let
$\beta>0$. Then, we have

$\displaystyle\frac{H}{\tanh(\frac{\beta}{2}H)}=H+\lim_{R\to\infty}\int_{{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}_{R}}\frac{\mathrm{d}z}{2\pi\mathrm{i}}\Bigl{(}\frac{z}{\tanh(\frac{\beta}{2}z)}-z\Bigr{)}\frac{1}{z-H},$
(6.20)

with the speaker path ${\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}_{R}$
in Definition 6.3. The above integral including the limit is understood as an
improper Riemann integral with respect to the uniform operator topology.

###### Proof.

We have that the function
$f(z)=\frac{z}{\tanh(\frac{\beta}{2}z)}-z=\frac{2z}{\mathrm{e}^{\beta z}-1}$
is analytic in the open domain $\mathbb{C}\setminus 2\pi
T\mathrm{i}\mathbb{Z}_{\neq 0}$. The construction of the Riemann integral over
the path ${\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}_{R}$
with respect to the uniform operator topology is standard. The fact that the
limit $R\to\infty$ exists in the same topology follows from the exponential
decay of the function $f(z)$ along the speaker path. To check the equality in
(6.20), we evaluate both sides in the inner product with two vectors in
$\operatorname{ran}\mathbbs 1_{(-\infty,K]}(H)$ for $K>0$, use the functional
calculus, the Cauchy integral formula, and the fact that
$\bigcup_{K>0}\operatorname{ran}\mathbbs 1_{(-\infty,K]}(H)$ is a dense subset
of $\mathcal{H}$. This proves the claim. ∎

Henceforth, we use the symbol $\int_{{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}}$ to
denote the integral on the right side of (6.20) including the limit and we
denote ${\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}=\bigcup_{R>0}{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}_{R}$.

###### Proof of Lemma 6.2.

We apply Cauchy-Schwarz to estimate

$\displaystyle|\langle\eta_{\perp},(K_{T_{\mathrm{c}},B}^{r}-K_{T_{\mathrm{c}}}^{r})\sigma_{0}\rangle|\leqslant\|\eta_{\perp}\|_{2}\,\|(K_{T_{\mathrm{c}},B}^{r}-K_{T_{\mathrm{c}}}^{r})\sigma_{0}\|_{2}$
(6.21)

and claim that

$\displaystyle\|[K_{{T_{\mathrm{c}}},B}^{r}-K_{T_{\mathrm{c}}}^{r}]\sigma_{0}\|_{2}\leqslant
C\varepsilon^{-\nicefrac{{1}}{{2}}}B^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}$
(6.22)

holds. To see this, we apply Lemma 6.4 and write

$\displaystyle
K_{{T_{\mathrm{c}}},B}^{r}-K_{T_{\mathrm{c}}}^{r}=\pi_{r}^{2}-p_{r}^{2}+\int_{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}\frac{\mathrm{d}w}{2\pi\mathrm{i}}\;f(w)\;\frac{1}{w+\mu-\pi_{r}^{2}}[\pi_{r}^{2}-p_{r}^{2}]\frac{1}{w+\mu-
p_{r}^{2}},$ (6.23)

where $\pi_{r}^{2}-p_{r}^{2}=\mathrm{i}\,\mathbf{B}\wedge
r\;p_{r}+\frac{1}{4}|\mathbf{B}\wedge r|^{2}$. Using (5.6) and (A.1), we
estimate the first term on the right side of (6.23) by

$\displaystyle\|[\pi_{r}^{2}-p_{r}^{2}]\sigma_{0}\|_{2}$
$\displaystyle\leqslant
B\,\|\,|\cdot|\nabla\alpha_{*}\|_{2}\|\Psi_{>}\|_{2}+B^{2}\|\,|\cdot|^{2}\alpha_{*}\|_{2}\|\Psi_{>}\|_{2}$
$\displaystyle\leqslant
C\varepsilon^{-\nicefrac{{1}}{{2}}}B^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}.$
(6.24)

To estimate the second term in (6.23), we use Hölder’s inequality in (2.1) and
find

$\displaystyle\Bigl{\|}\int_{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}\frac{\mathrm{d}w}{2\pi\mathrm{i}}\,f(w)\;\frac{1}{w+\mu-\pi_{r}^{2}}[\pi_{r}^{2}-p_{r}^{2}]\frac{1}{w+\mu-
p_{r}^{2}}\sigma_{0}\Bigr{\|}_{2}$ $\displaystyle\leqslant\int_{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}\frac{\mathrm{d}|w|}{2\pi}\,|f(w)|\,\Bigl{\|}\frac{1}{w+\mu-\pi_{r}^{2}}\Bigr{\|}_{\infty}\Bigl{\|}[\pi_{r}^{2}-p_{r}^{2}]\frac{1}{w+\mu-
p_{r}^{2}}\sigma_{0}\Bigr{\|}_{2},$

where $\mathrm{d}|w|=\mathrm{d}t\;|w^{\prime}(t)|$. Eq. (6.19) implies that
the operator norm of the magnetic resolvent is uniformly bounded for
$w\in{\hskip 0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}$. Since the function $f$ is exponentially decaying along the speaker
path it suffices to prove a bound on the last factor that is uniform for
$w\in{\hskip 0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip
0.7pt}}$. We have

$\displaystyle[\pi_{r}^{2}-p_{r}^{2}]\frac{1}{w+\mu-p_{r}^{2}}\sigma_{0}(X,r)$
$\displaystyle=\int_{\mathbb{R}^{3}}\mathrm{d}s\;[\pi_{r}^{2}-p_{r}^{2}]g_{0}^{w}(r-s)\alpha_{*}(s)\Psi_{>}(X),$

which implies

$\displaystyle\Bigl{\|}[\pi_{r}^{2}-p_{r}^{2}]\frac{1}{w+\mu-
p_{r}^{2}}\sigma_{0}\Bigr{\|}_{2}^{2}$
$\displaystyle\leqslant\|\Psi_{>}\|_{2}^{2}\int_{\mathbb{R}^{3}}\mathrm{d}r\,\Bigl{|}\int_{\mathbb{R}^{3}}\mathrm{d}s\;|[\pi_{r}^{2}-p_{r}^{2}]g_{0}^{w}(r-s)\alpha_{*}(s)|\Bigr{|}^{2}.$
(6.25)

Moreover,

$\displaystyle\int_{\mathbb{R}^{3}}\mathrm{d}r\;\Bigl{|}\int_{\mathbb{R}^{3}}\mathrm{d}s\;|[\pi_{r}^{2}-p_{r}^{2}]g_{0}^{w}(r-s)\alpha_{*}(s)|\Bigr{|}^{2}$
$\displaystyle\leqslant CB^{2}\bigl{(}\|\,|\cdot|\nabla
g_{0}^{w}\|_{1}^{2}\;\|\alpha_{*}\|_{2}^{2}+\|\nabla
g_{0}^{w}\|_{1}^{2}\;\|\,|\cdot|\alpha_{*}\|_{2}^{2}+\|\,|\cdot|^{2}g_{0}^{w}\|_{1}^{2}\;\|\alpha_{*}\|_{2}^{2}+\|g_{0}^{w}\|_{1}^{2}\;\|\,|\cdot|^{2}\alpha_{*}\|_{2}^{2}\bigr{)}.$

The right side is uniformly bounded for $w\in{\hskip
0.7pt\text{\raisebox{-1.7pt}{\scalebox{1.4}{\faVolumeOff}}\hskip 0.7pt}}$ by
(6.19) and (A.1). In combination with (5.6) and (6.25), this implies

$\displaystyle\Bigl{\|}[\pi_{r}^{2}-p_{r}^{2}]\frac{1}{w+\mu-
p_{r}^{2}}\sigma_{0}\Bigr{\|}_{2}^{2}$ $\displaystyle\leqslant
CB^{2}\;\|\Psi_{>}\|_{2}^{2}\leqslant
C\varepsilon^{-1}B^{4}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}.$

Using this and (6.24), we read off (6.22). Finally, we apply Proposition 3.2
to estimate $\|\eta_{\perp}\|_{2}$ in (6.21), which proves part (a).

To prove part (b), we start by noting that

$\displaystyle|\langle\eta_{\perp},(U-1)K_{T_{\mathrm{c}},B}^{r}\sigma_{0}\rangle|$
$\displaystyle\leqslant\|\,|r|\eta_{\perp}\|_{2}\;\|\,|r|^{-1}(U-1)K_{T_{\mathrm{c}},B}^{r}\,\sigma_{0}\|_{2}.$

A bound for the left factor on the right side is provided by Proposition 3.2.
To estimate the right factor, we use (5.6), (A.1) and the operator inequality
in (4.98), which implies $|U-1|^{2}\leqslant 3r^{2}\Pi_{X}^{2}$, and find

$\displaystyle\|\,|r|^{-1}(U-1)K_{T_{\mathrm{c}},B}^{r}\,\sigma_{0}\|_{2}$
$\displaystyle\leqslant
C\|K_{T_{\mathrm{c}},B}^{r}\alpha_{*}\|_{2}\;\|\Pi\Psi_{>}\|_{2}\leqslant
CB\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}.$

This proves part (b).

For part (c), we estimate

$\displaystyle|\langle\eta_{\perp},UK_{T_{\mathrm{c}},B}^{r}(U^{*}-1)\sigma_{0}\rangle|$
$\displaystyle\leqslant\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}^{r}}\,U^{*}\eta_{\perp}\bigr{\|}_{2}\;\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}^{r}}\,(U^{*}-1)\sigma_{0}\bigr{\|}_{2}$
(6.26)

and note that $K_{T_{\mathrm{c}},B}^{r}\leqslant C(1+\pi_{r}^{2})$ implies

$\displaystyle\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}^{r}}(U^{*}-1)\sigma_{0}\bigr{\|}_{2}^{2}$
$\displaystyle=\langle\sigma_{0},(U-1)K_{T_{\mathrm{c}},B}^{r}(U^{*}-1)\sigma_{0}\rangle$
$\displaystyle\leqslant
C\|(U^{*}-1)\sigma_{0}\|_{2}^{2}+C\|\pi_{r}(U^{*}-1)\sigma_{0}\|_{2}^{2}.$
(6.27)

Using the bound for $|U-1|^{2}$ in part (b), (5.6) and (A.1), we see that the
first term is bounded by $C\||r|\alpha_{*}\Pi_{X}\Psi_{>}\|^{2}\leqslant
CB^{2}$. Lemma 5.11 allows us to write

$\displaystyle\pi_{r}(U^{*}-1)$
$\displaystyle=(U^{*}-1)\widetilde{\pi}_{r}+\frac{1}{2}U^{*}\Pi_{X}-\frac{1}{4}\mathbf{B}\wedge
r.$

Accordingly, we have

$\displaystyle\|\pi_{r}(U^{*}-1)\sigma_{0}\|_{2}^{2}$ $\displaystyle\leqslant
C\bigl{(}\|\,|r|p_{r}\alpha_{*}\Pi_{X}\Psi_{>}\|_{2}^{2}+B\,\|\,|r|^{2}\alpha_{*}\Pi_{X}\Psi_{>}\|_{2}^{2}+\|\alpha_{*}\Pi_{X}\Psi_{>}\|_{2}^{2}+B\|\,|r|\alpha_{*}\Psi_{>}\|_{2}^{2}\bigr{)}$
$\displaystyle\leqslant C\bigl{(}B^{2}+\varepsilon^{-1}B^{3}\bigr{)}\leqslant
CB^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}.$

We conclude that the right side of (6.27) is bounded by
$CB^{2}\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}$.

With $K_{T}(p)\leqslant C(1+p^{2})$ we see that the first factor on the right
side of (6.26) is bounded by

$\displaystyle\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}^{r}}\,U^{*}\eta_{\perp}\bigr{\|}_{2}^{2}$
$\displaystyle=\langle\eta_{\perp},UK_{T_{\mathrm{c}},B}^{r}U^{*}\eta_{\perp}\rangle\leqslant
C\|\eta_{\perp}\|_{2}^{2}+C\|\pi_{r}U^{*}\eta_{\perp}\|_{2}^{2}.$

From Lemma 5.11 we know that
$\pi_{r}U^{*}=U^{*}[\widetilde{\pi}_{r}+\frac{1}{2}\Pi_{X}]$, and hence

$\displaystyle\bigl{\|}\sqrt{K_{T_{\mathrm{c}},B}^{r}}\,U^{*}\eta_{\perp}\bigr{\|}_{2}^{2}$
$\displaystyle\leqslant
C\bigl{(}\|\eta_{\perp}\|_{2}^{2}+\|\widetilde{\pi}_{r}\eta_{\perp}\|_{2}^{2}+\|\Pi_{X}\eta_{\perp}\|_{2}^{2}\bigr{)}\leqslant
C\varepsilon B^{2}\,\|\Psi\|_{H_{\mathrm{mag}}^{1}(Q_{B})}^{2}.$

This proves part (c) and ends the proof of the Lemma 6.2. ∎

— Appendix —

##  Appendix A Estimates on Eigenvalues and Eigenfunctions of
$K_{T_{\mathrm{c}},B}-V$

In this section, we investigate the low lying eigenvalues of
$K_{T_{\mathrm{c}},B}-V$ and its ground state wave function. Our analysis is
carried out at $T=T_{\mathrm{c}}$ and we omit $T_{\mathrm{c}}$ from the
notation throughout the appendix. The goal is to prove the following result.

######  Proposition A.1.

Let Assumptions 1.3 and 1.5 hold. There is a constant $B_{0}>0$ such that for
any $0\leqslant B\leqslant B_{0}$ the following holds. Let $e_{0}^{B}$ and
$e_{1}^{B}$ denote the lowest and next to lowest eigenvalue of
$K_{T_{\mathrm{c}},B}-V$. Then:

  1. (a)

$|e_{0}^{B}|\leqslant CB$,

  2. (b)

$K_{T_{\mathrm{c}},B}-V$ has a uniform spectral gap above $e_{0}^{B}$, i.e.,
$e_{1}^{B}-e_{0}^{B}\geqslant\kappa>0$.

  3. (c)

Let $\alpha_{*}$ be the eigenfunction in (1.15) and let $\alpha_{*}^{B}$ be an
eigenfunction corresponding to $e_{0}^{B}$ such that
$\langle\alpha_{*}^{B},V\alpha_{*}\rangle$ is real and nonnegative for all
$0\leqslant B\leqslant B_{0}$. Then,

$\displaystyle\|\alpha_{*}^{B}-\alpha_{*}\|_{2}+\|\pi^{2}(\alpha_{*}^{B}-\alpha_{*})\|_{2}\leqslant
CB.$

  4. (d)

With $P_{B}:=|\alpha_{*}^{B}\rangle\langle\alpha_{*}^{B}|$ and
$P:=|\alpha_{*}\rangle\langle\alpha_{*}|$ and with $\alpha_{*}^{B}$ and
$\alpha_{*}$ as in part (c), we have

$\displaystyle\|P_{B}-P\|_{\infty}+\|\pi^{2}(P_{B}-P)\|_{\infty}\leqslant CB.$

######  Remark A.2.

We emphasize that this appendix is the only place in the paper where the
assumption $V\geqslant 0$ is used. It simplifies our analysis because it
implies that the Birman–Schwinger operator
$V^{\nicefrac{{1}}{{2}}}[K_{T_{\mathrm{c}},B}-e]^{-1}V^{\nicefrac{{1}}{{2}}}$
is self-adjoint. However, for the statement of Proposition A.1 to be true, it
is not necessary that $V$ has a sign. In fact, with the help of a
Combes–Thomas estimate for the resolvent kernel of $K_{T_{\mathrm{c}}}-V$ it
is possible to show that Proposition A.1 also holds for potentials $V$ without
a definite sign. This approach requires more effort, and we therefore refrain
from giving a general proof here. It can be found in M. Schaub’s PhD thesis.

Let us recall the decay properties of the eigenfunction $\alpha_{*}$
corresponding to the lowest eigenvalue of the operator $K_{T_{\mathrm{c}}}-V$.
Since $\alpha_{*}=K_{T_{\mathrm{c}}}^{-1}V\alpha_{*}$ and $V\in
L^{\infty}(\mathbb{R}^{3})$, we immediately have $\alpha_{*}\in
H^{2}(\mathbb{R}^{3})$. Furthermore, for any $\nu\in\mathbb{N}_{0}^{3}$, by
[17, Proposition 2],

$\displaystyle\int_{\mathbb{R}^{3}}\mathrm{d}x\;\bigl{[}|x^{\nu}\alpha_{*}(x)|^{2}+|x^{\nu}\nabla\alpha_{*}(x)|^{2}\bigr{]}<\infty.$
(A.1)

In fact, more regularity of $\alpha_{*}$ is known, see [17, Appendix A] but
(A.1) is all we use in this paper. Before we give the proof of Proposition A.1
in Section A.2 below, we prove two preparatory statements.

Let $e\in(-\infty,2T_{\mathrm{c}})$ and denote the kernel of the resolvent
$(e-K_{T_{\mathrm{c}}})^{-1}$ by $\mathcal{G}^{e}(x-y)$.

######  Lemma A.3.

For $e\in(-\infty,2T_{\mathrm{c}})$ and $k\in\mathbb{N}_{0}$ the functions
$|\cdot|^{k}\mathcal{G}^{e}$ and $|\cdot|^{k}\nabla\mathcal{G}^{e}$ belong to
$L^{1}(\mathbb{R}^{3})$.

###### Proof.

The function $(e-K_{T_{\mathrm{c}}}(p))^{-1}$ and its derivatives belong to
$L^{2}(\mathbb{R}^{3})$. Therefore, we have

$\displaystyle\|\,|\cdot|^{k}\mathcal{G}^{e}\|_{1}$
$\displaystyle\leqslant\Bigl{(}\int_{\mathbb{R}^{3}}\mathrm{d}x\;\bigl{|}\frac{|x|^{k}}{1+|x|^{k+2}}\bigr{|}^{2}\Bigr{)}^{\nicefrac{{1}}{{2}}}\;\|(1+|\cdot|^{k+2})\mathcal{G}^{e}\|_{2}<\infty.$
(A.2)

This proves the first claim and the second follows from a similar argument. ∎

###  A.1 Phase approximation for $K_{T_{\mathrm{c}},B}$

######  Proposition A.4.

Let $V$ and $|\cdot|^{2}V$ belong to $L^{\infty}(\mathbb{R}^{3})$. There is
$B_{0}>0$ such that for $0\leqslant B\leqslant B_{0}$ and
$e\in(-\infty,2T_{\mathrm{c}})$, we have

$\displaystyle\Bigl{\|}\Bigl{[}\frac{1}{e-K_{T_{\mathrm{c}},B}}-\frac{1}{e-K_{T_{\mathrm{c}}}}\Bigr{]}V^{\nicefrac{{1}}{{2}}}\Bigr{\|}_{\infty}+\Bigl{\|}\pi^{2}\Bigl{[}\frac{1}{e-K_{T_{\mathrm{c}},B}}-\frac{1}{e-K_{T_{\mathrm{c}}}}\Bigr{]}V\Bigr{\|}_{\infty}\leqslant
C_{e}B.$ (A.3)

###### Proof.

To prove this result, we apply a phase approximation to the operator
$K_{T_{\mathrm{c}},B}$. We pursue the strategy that we used in the proof of
Lemma 4.5 and define

$\displaystyle\mathcal{S}_{B}^{e}(x,y):=\mathrm{e}^{\mathrm{i}\frac{\mathbf{B}}{2}\cdot(x\wedge
y)}\;\mathcal{G}^{e}(x-y).$ (A.4)

Let $\mathcal{S}_{B}^{e}$ be the operator defined by the kernel
$\mathcal{S}_{B}^{e}(x,y)$. We claim that

$\displaystyle(e-K_{T_{\mathrm{c}},B})\mathcal{S}_{B}^{e}=\mathbbs
1-\mathcal{T}_{B}^{e}$ (A.5)

with the operator $\mathcal{T}_{B}^{e}$ defined by the kernel

$\displaystyle\mathcal{T}_{B}^{e}(x,y):=\mathrm{e}^{\mathrm{i}\frac{\mathbf{B}}{2}\cdot(x\wedge
y)}\bigl{[}(K_{T_{\mathrm{c}}}(\pi_{x,y})-K_{T_{\mathrm{c}}})\frac{1}{e-K_{T_{\mathrm{c}}}}\bigr{]}(x,y)$
(A.6)

and $\pi_{x,y}=-\mathrm{i}\nabla_{x}+\mathbf{A}(x-y)$. To prove (A.5), it is
sufficient to note that

$\displaystyle
K_{T_{\mathrm{c}},B}\;\mathrm{e}^{\mathrm{i}\frac{\mathbf{B}}{2}\cdot(x\wedge
y)}=\mathrm{e}^{\mathrm{i}\frac{\mathbf{B}}{2}\cdot(x\wedge
y)}\;K_{T_{\mathrm{c}}}(\pi_{x,y}).$

Using Lemma 6.4 and Lemma A.3, a straightforward computation shows that

$\displaystyle\|\mathcal{T}_{B}^{e}\|_{\infty}\leqslant C_{e}\,B$ (A.7)

holds for $B$ small enough.

With the operator $\mathcal{S}_{B}^{e}$ in (A.4) we write

$\displaystyle\frac{1}{e-K_{T_{\mathrm{c}},B}}-\frac{1}{e-K_{T_{\mathrm{c}}}}=\frac{1}{e-K_{T_{\mathrm{c}},B}}-\mathcal{S}_{B}^{e}+\mathcal{S}_{B}^{e}-\frac{1}{e-K_{T_{\mathrm{c}}}}$
(A.8)

For the first term (A.5) implies

$\displaystyle\frac{1}{e-K_{T_{\mathrm{c}},B}}-\mathcal{S}_{B}^{e}=\mathcal{S}_{B}^{e}\;\sum_{n=1}^{\infty}(\mathcal{T}_{B}^{e})^{n}$

and since $\mathcal{S}_{B}^{e}$ is a bounded operator with norm bounded by
$\|\mathcal{G}^{e}\|_{1}$, (A.7) yields

$\displaystyle\Bigl{\|}\frac{1}{e-K_{T_{\mathrm{c}},B}}-\mathcal{S}_{B}^{e}\Bigr{\|}_{\infty}\leqslant
C_{e}B.$

To estimate the second term on the right side of (A.8), we use
$|\mathrm{e}^{\mathrm{i}\frac{\mathbf{B}}{2}\cdot(x\wedge y)}-1|\leqslant
B|x-y||y|$ and bound the kernel of this term by

$\displaystyle\bigl{|}\bigl{[}\mathcal{S}_{B}^{e}-\frac{1}{e-K_{T_{\mathrm{c}}}}\bigr{]}V^{\nicefrac{{1}}{{2}}}(x,y)\bigr{|}$
$\displaystyle\leqslant
B\;|x-y||\mathcal{G}^{e}(x-y)|\;|y||V^{\nicefrac{{1}}{{2}}}(y)|.$

We further bound
$|y||V^{\nicefrac{{1}}{{2}}}(y)|\leqslant\|\,|\cdot|^{2}V\|_{\infty}^{\nicefrac{{1}}{{2}}}$,
which shows that

$\displaystyle\Bigl{\|}\Bigl{[}\mathcal{S}_{B}^{e}-\frac{1}{e-K_{T_{\mathrm{c}}}}\Bigr{]}V^{\nicefrac{{1}}{{2}}}\Bigr{\|}_{\infty}$
$\displaystyle\leqslant
B\;\|\,|\cdot|^{2}V\|_{\infty}^{\nicefrac{{1}}{{2}}}\;\|\,|\cdot|\mathcal{G}^{e}\|_{1}\leqslant
C_{e}B.$

This completes the proof of the first estimate in (A.3).

To prove the second estimate, we note that

$\displaystyle\pi^{2}\Bigl{[}\frac{1}{e-K_{T_{\mathrm{c}},B}}-\frac{1}{e-K_{T_{\mathrm{c}}}}\Bigr{]}V=\pi^{2}\frac{1}{e-K_{T_{\mathrm{c}},B}}[K_{T_{\mathrm{c}},B}-K_{T_{\mathrm{c}}}]\frac{1}{e-K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}.$

Since $\pi^{2}(e-K_{T_{\mathrm{c}},B})^{-1}$ is a bounded function of
$\pi^{2}$, we know that the operator norm of the operator in the above
equation is uniformly bounded in $B$. Thus, it suffices to show that
$[K_{T_{\mathrm{c}},B}-K_{T_{\mathrm{c}}}]\frac{1}{e-K_{T_{\mathrm{c}}}}V$
satisfies the claimed operator norm bound. To this end, we use (6.23) and
obtain two terms. Since $\pi^{2}-p^{2}=\mathbf{B}\wedge x\cdot
p+\frac{1}{4}|\mathbf{B}\wedge x|^{2}$, the estimate for the first term reads

$\displaystyle\bigl{[}(\pi^{2}-p^{2})\frac{1}{e-K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}\bigr{]}(x,y)\leqslant\bigl{[}B\cdot|x||\nabla\mathcal{G}^{e}(x-y)|+B^{2}\;|x|^{2}|\mathcal{G}^{e}(x-y)|\bigr{]}|V(y)|.$

The $L^{1}(\mathbb{R}^{3})$-norm in $x-y$ of the right side is bounded by

$\displaystyle
CB\,\bigl{[}\bigl{(}\|\,|\cdot|\nabla\mathcal{G}^{e}\|_{1}+\|\,|\cdot|^{2}\mathcal{G}^{e}\|_{1}\bigr{)}\|V\|_{\infty}+\|\nabla\mathcal{G}^{e}\|_{1}\,\|\,|\cdot|V\|_{\infty}+\|\mathcal{G}^{e}\|_{1}\,\|\,|\cdot|^{2}V\|_{\infty}\bigr{]},$

which is finite by Lemma A.3. With the help of (6.19) the remaining term can
bounded similarly. This proves the claim. ∎

###  A.2 Asymptotics for eigenvalues and eigenfunctions

We are now prepared to give the proof of Proposition A.1.

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

We start with the upper bound of part (a). By the variational principle for
$e_{0}^{B}$ we have

$\displaystyle
e_{0}^{B}\leqslant\langle\alpha_{*},(K_{T_{\mathrm{c}},B}-V)\alpha_{*}\rangle=\langle\alpha_{*},(K_{T_{\mathrm{c}}}-V)\alpha_{*}\rangle+\langle\alpha_{*},(K_{T_{\mathrm{c}},B}-K_{T_{\mathrm{c}}})\alpha_{*}\rangle,$
(A.9)

where the first term on the right side equals $0$ by the definition of
$\alpha_{*}$. We use (6.19), Lemma 6.4, and (A.1), and argue as in the proof
of (6.24) to see that the second term is bounded by $CB$.

In the next step, we show the lower bound of part (a) and part (b) at the same
time. Thus, for $n=0,1$ we aim to show

$\displaystyle e_{n}^{B}\geqslant e_{n}^{0}-C_{n}B$ (A.10)

for the lowest and next-to-lowest eigenvalue $e_{0}^{B}$ and $e_{1}^{B}$,
respectively. For notational convenience, we give the proof for general
$n\in\mathbb{N}_{0}$ and we order the eigenvalues $e_{n}^{B}$ increasingly.
Let $\alpha_{n}^{B}$ be the eigenfunction to $e_{n}^{B}$ for $n\geqslant 0$
and note that $\alpha_{*}^{B}=\alpha_{0}^{B}$.

Now, we switch to the Birman–Schwinger picture: $e_{n}^{B}$ being the
$(n+1)$-st to smallest eigenvalue of $K_{T_{\mathrm{c}},B}-V$ is equivalent to
1 being the $(n+1)$-st to largest eigenvalue of the Birman–Schwinger operator
$V^{\nicefrac{{1}}{{2}}}(K_{T_{\mathrm{c}},B}-e_{n}^{B})^{-1}V^{\nicefrac{{1}}{{2}}}$
corresponding to $e_{n}^{B}$. Accordingly, the min-max principle, see, e.g.,
[38, Theorem 12.1 (5)], implies

$\displaystyle 1=\max_{\begin{subarray}{c}u_{0},\ldots,u_{n}\in
L^{2}(\mathbb{R}^{3})\\\ u_{i}\perp u_{j},\;i\neq
j\end{subarray}}\min\Bigl{\\{}\Bigl{\langle}\Phi,V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi\Bigr{\rangle}:\Phi\in\mathrm{span}\\{u_{0},\ldots,u_{n}\\},\;\|\Phi\|_{2}=1\Bigr{\\}}.$
(A.11)

We obtain a lower bound on (A.11) by choosing the functions $u_{i}$,
$i=0,\ldots,n$ as the first $n+1$ orthonormal eigenfunctions $\varphi_{i}^{B}$
satisfying

$\displaystyle
V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\varphi_{i}^{B}=\eta_{i}^{B}\,\varphi_{i}^{B},\qquad\qquad
i=0,\ldots,n,$ (A.12)

where $\eta_{i}^{B}\geqslant 1$ denote the first $n$ eigenvalues of the
Birman–Schwinger operator in (A.12) ordered decreasingly. In particular, we
have $\eta_{n}^{B}=1$, as well as the relations
$\varphi_{n}^{B}=V^{\nicefrac{{1}}{{2}}}\alpha_{n}^{B}$ and
$\alpha_{n}^{B}=(K_{T_{\mathrm{c}},B}-e_{n}^{B})^{-1}V^{\nicefrac{{1}}{{2}}}\varphi_{n}^{B}$.

Denote $e_{n}:=e_{n}^{0}$ and apply the resolvent equations to write

$\displaystyle
V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}$
$\displaystyle=V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}}}-e_{n}}V^{\nicefrac{{1}}{{2}}}+(e_{n}^{B}-e_{n})\,\mathcal{Q}_{n}^{B}+\mathcal{R}_{n}^{B}$
(A.13)

with

$\displaystyle\mathcal{Q}_{n}^{B}$
$\displaystyle:=V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}}\,\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}},$
$\displaystyle\mathcal{R}_{n}^{B}$
$\displaystyle:=V^{\nicefrac{{1}}{{2}}}\Bigl{[}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}}-\frac{1}{K_{T_{\mathrm{c}}}-e_{n}}\Bigr{]}V^{\nicefrac{{1}}{{2}}}.$

By Proposition A.4, we have $\|\mathcal{R}_{n}^{B}\|_{\infty}\leqslant
C_{n}B$. Furthermore, we may assume without loss of generality that
$e_{n}^{B}\leqslant e_{n}$, because otherwise there is nothing to prove. We
combine this with (A.11) for $B=0$ and (A.13), which yields

$\displaystyle 1$
$\displaystyle\geqslant\min\Bigl{\\{}\Bigl{\langle}\Phi,V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi\Bigr{\rangle}:\Phi\in\mathrm{span}\\{\varphi_{0}^{B},\ldots,\varphi_{n}^{B}\\},\;\|\Phi\|_{2}=1\Bigr{\\}}$
$\displaystyle\hskip
30.0pt-[e_{n}^{B}-e_{n}]\min\bigl{\\{}\langle\Phi,\mathcal{Q}_{n}^{B}\Phi\rangle:\Phi\in\mathrm{span}\\{\varphi_{0}^{B},\ldots,\varphi_{n}^{B}\\},\;\|\Phi\|_{2}=1\bigr{\\}}-C_{n}B.$
(A.14)

We observe that the first term on the right side equals 1. To be able to
conclude, we therefore need to show that there is a constant $c>0$,
independent of $B$, such that

$\displaystyle\min\bigl{\\{}\langle\Phi,\mathcal{Q}_{n}^{B}\Phi\rangle:\Phi\in\mathrm{span}\\{\varphi_{0}^{B},\ldots,\varphi_{n}^{B}\\}\bigr{\\}}\geqslant
c.$ (A.15)

Then, (A.14) implies $-[e_{n}^{B}-e_{n}]\leqslant C_{n}B$, which proves
(A.10).

We will prove that (A.15) holds with $c=\|V\|_{\infty}^{-1}$. To that end, we
write

$\displaystyle\langle\Phi,\mathcal{Q}_{n}^{B}\Phi\rangle$
$\displaystyle=\Bigl{\langle}\Phi,V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}(K_{T_{\mathrm{c}},B}-e_{n}^{B})\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi\Bigr{\rangle},$

apply $-e_{n}^{B}\geqslant-e_{n}$, and infer

$\displaystyle\langle\Phi,\mathcal{Q}_{n}^{B}\Phi\rangle$
$\displaystyle\geqslant\Bigl{\|}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi\Bigr{\|}_{2}^{2}.$

Since $\Phi\in\mathrm{span}\\{\varphi_{0}^{B},\ldots,\varphi_{n}^{B}\\}$,
there are coefficients $c_{i}^{B}\in\mathbb{C}$, $i=0,\ldots,n$ such that we
have $\Phi=\sum_{i=0}^{n}c_{i}^{B}\varphi_{i}^{B}$. We use the eigenvalue
equation in (A.12) for $\varphi_{i}^{B}$ as well as
$\langle\varphi_{i}^{B},\varphi_{j}^{B}\rangle=\delta_{i,j}$ to see that

$\displaystyle\|V\|_{\infty}\Bigl{\|}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi\Bigr{\|}_{2}^{2}$
$\displaystyle\geqslant\Bigl{\langle}\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi,V\frac{1}{K_{T_{\mathrm{c}},B}-e_{n}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi\Bigr{\rangle}$
$\displaystyle=\sum_{i,j=0}^{n}\overline{c_{i}^{B}}c_{j}^{B}\,\eta_{i}^{B}\eta_{j}^{B}\,\langle\varphi_{i}^{B},\varphi_{j}^{B}\rangle=\sum_{i=0}^{n}|c_{i}^{B}|^{2}\;(\eta_{i}^{B})^{2}\geqslant
1.$

Here, we used that $\eta_{i}\geqslant 1$ and $\|\Phi\|_{2}=1$. This shows
(A.15) and completes the proof of (A.10).

The case $n=0$ in (A.10) yields the lower bound of part (a), since $e_{0}=0$.
For $n=1$, we have $e_{1}=\kappa$, showing part (b) with the help of part (a).

To prove part (c), we write $\varphi_{0}:=\varphi_{0}^{0}$. The chosen phase
of $\varphi_{0}^{B}$ in the lemma is such that
$\langle\varphi_{0},\varphi_{0}^{B}\rangle$ is real and nonnegative. We write
$\varphi_{0}^{B}=a_{B}\varphi_{0}+b_{B}\Phi$ with
$\langle\Phi,\varphi_{0}\rangle=0$, $\|\Phi\|_{2}=1$ and
$|a_{B}|^{2}+|b_{B}|^{2}=1$. By construction,
$a_{B}=\langle\varphi_{0},\varphi_{0}^{B}\rangle$. Furthermore, by (A.13),

$\displaystyle
1=\bigl{\langle}\varphi_{0}^{B},V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}},B}-e_{0}^{B}}V^{\nicefrac{{1}}{{2}}}\varphi_{0}^{B}\bigr{\rangle}=\bigl{\langle}\varphi_{0}^{B},V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}\varphi_{0}^{B}\bigr{\rangle}+\langle\varphi_{0}^{B},\mathcal{T}_{B}\varphi_{0}^{B}\rangle$
(A.16)

with $\mathcal{T}_{B}:=e_{0}^{B}\,Q_{0}^{B}+\mathcal{R}_{0}^{B}$. Thus,

$\displaystyle 1\leqslant
a_{B}^{2}+|b_{B}|^{2}\bigl{\langle}\Phi,V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}\Phi\bigr{\rangle}+2a_{B}\operatorname{\mathrm{Re}}\bigl{[}b_{B}\bigl{\langle}\varphi_{0},V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}\Phi\bigr{\rangle}\bigr{]}+\|\mathcal{T}_{B}\|_{\infty}.$

By parts (a) and (b) and Proposition A.4, we know that
$\|\mathcal{T}_{B}\|_{\infty}\leqslant CB$. Furthermore, the term in square
brackets vanishes, since
$V^{\nicefrac{{1}}{{2}}}K_{T_{\mathrm{c}}}^{-1}V^{\nicefrac{{1}}{{2}}}\varphi_{0}=\varphi_{0}$
and $\langle\varphi_{0},\Phi\rangle=0$. Using the orthogonality of $\Phi$ and
$\varphi_{0}$ once more as well as the fact that 1 is the largest eigenvalue
of $V^{\nicefrac{{1}}{{2}}}K_{T_{\mathrm{c}}}^{-1}V^{\nicefrac{{1}}{{2}}}$, we
see that there is an $\eta<1$ such that
$\langle\Phi,V^{\nicefrac{{1}}{{2}}}K_{T_{\mathrm{c}}}^{-1}V^{\nicefrac{{1}}{{2}}}\Phi\rangle\leqslant\eta$.
It follows that

$\displaystyle 1\leqslant a_{B}^{2}+|b_{B}|^{2}\eta+CB.$

Since $a_{B}^{2}+|b_{B}|^{2}=1$, this implies $|b_{B}|^{2}\leqslant CB$ as
well as $a_{B}^{2}\geqslant 1-CB$. Since $a_{B}\geqslant 0$, we infer
$1-a_{B}\leqslant CB$.

The next step is to improve the estimate on $b_{B}$ to $|b_{B}|\leqslant CB$.
To this end, we combine the two eigenvalue equations of $\varphi_{0}^{B}$ and
$\varphi_{0}$. With $\mathcal{T}_{B}$ as in (A.16), we find

$\displaystyle\varphi_{0}^{B}-\varphi_{0}$
$\displaystyle=V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}(\varphi_{0}^{B}-\varphi_{0})+\mathcal{T}_{B}\varphi_{0}^{B}.$

Testing this against $\Phi$, we obtain

$\displaystyle
b_{B}=\langle\Phi,\varphi_{0}^{B}-\varphi_{0}\rangle=\bigl{\langle}V^{\nicefrac{{1}}{{2}}}\frac{1}{K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}\Phi,\varphi_{0}^{B}-\varphi_{0}\bigr{\rangle}+\langle\Phi,\mathcal{T}_{B}\varphi_{0}\rangle.$

We apply Cauchy-Schwarz on the right side and use that
$\|V^{\nicefrac{{1}}{{2}}}K_{T_{\mathrm{c}}}^{-1}V^{\nicefrac{{1}}{{2}}}\Phi\|_{2}\leqslant\eta$,
which follows from the orthogonality of $\Phi$ and $\varphi_{0}$. We also use
$\|\varphi_{0}^{B}-\varphi_{0}\|_{2}\leqslant(1-a_{B})+|b_{B}|$. This implies

$\displaystyle|b_{B}|\leqslant\eta\,(1-a_{B})+\eta\,|b_{B}|+CB,$

from which we conclude that $|b_{B}|\leqslant CB$.

It remains to use these findings to prove the claimed bounds for
$\alpha_{*}^{B}-\alpha_{*}$. According to the Birman–Schwinger correspondence,
we have
$\alpha_{*}^{B}=(K_{T_{\mathrm{c}},B}-e_{0}^{B})^{-1}V^{\nicefrac{{1}}{{2}}}\varphi_{0}^{B}$.
Thus, since $\varphi_{0}=V^{\nicefrac{{1}}{{2}}}\alpha_{*}$,

$\displaystyle\alpha_{*}^{B}-\alpha_{*}$
$\displaystyle=\frac{1}{K_{T_{\mathrm{c}},B}-e_{0}^{B}}V^{\nicefrac{{1}}{{2}}}\bigl{[}a_{B}\varphi_{0}+b_{B}\Phi\bigr{]}-\frac{1}{K_{T_{\mathrm{c}}}}V^{\nicefrac{{1}}{{2}}}\varphi_{0}$
$\displaystyle=a_{B}\Bigl{[}\frac{1}{K_{T_{\mathrm{c}},B}-e_{0}^{B}}-\frac{1}{K_{T_{\mathrm{c}}}}\Bigr{]}V\alpha_{*}+b_{B}\frac{1}{K_{T_{\mathrm{c}},B}-e_{0}^{B}}V^{\nicefrac{{1}}{{2}}}\Phi-(1-a_{B})\frac{1}{K_{T_{\mathrm{c}}}}V\alpha_{*}.$
(A.17)

The proof of the norm estimates for $\alpha_{*}^{B}-\alpha$ and
$\pi^{2}(\alpha_{*}^{B}-\alpha_{*})$ is obtained from Lemma A.3, Proposition
A.4, and the estimates of part (a) on $e_{0}^{B}$.

Part (d) follows from part (c). This ends the proof of Proposition A.1. ∎

Acknowledgements

A. D. gratefully acknowledges funding from the European Union’s Horizon 2020
research and innovation programme under the Marie Sklodowska-Curie grant
agreement No 836146 and from the Swiss National Science Foundation through the
Ambizione grant PZ00P2 185851. It is a pleasure for A. D. to thank Stefan
Teufel and the Institute of Mathematics at the University of Tübingen for its
warm hospitality.

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