TITLE: $n$-distant permutations more than not
QUESTION [2 upvotes]: Let $\mathfrak{S}_{2n}$ be the permutation group of the letters $[2n]=\{1,2,\dots,2n\}$. Call a permutation $\pi\in\mathfrak{S}_{2n}$ has an $n$-distant pair if there is some $j\in [2n-1]$ such that $\vert\pi_j-\pi_{j+1}\vert=n$.
Define $A=$ set of $\pi\in\mathfrak{S}_{2n}$ with exactly one $n$-distant pair and $B=$ set of permutations without $n$-distant pairs.

Question. Which inequality is true in general: $\vert A\vert>\vert B\vert$ or $\vert A\vert<\vert B\vert$? And your proof?

REPLY [3 votes]: There is an injective mapping from $B$ to $A$, which can be constructed as follows:
for a permutation $\pi=(p_1,p_2,\dots,p_{2n})\in B$, find the element $p_k$ that pairs with $p_1$ (i.e., $|p_k-p_1|=n$); clearly we have $k>2$. Map $\pi$ to the permutation
$\pi'=(p_2,\dots,p_{k-1},p_1,p_k,p_{k+1},\dots,p_{2n})$ (i.e., remove $p_1$ from $\pi$ and insert it right before $p_k$).
It can be easily seen that $\pi'\in A$, and $\pi'$ has only one pre-image in $B$. Hence, $|B|\leq |A|$.
Furthermore, for $n>2$, there exist permutations in $A$ that have no pre-images; namely, these are the permutations that start with an $n$-distant pair. It follows that $|B|<|A|$.
In fact, we can infer from the above argument that $|A_n| = |B_n| + 2n\cdot |B_{n-1}|$.
UPDATE. Inclusion-exclusion gives
$$|B_n| = \sum_{j=0}^n (-1)^j \binom{n}{j} 2^j (2n-j)! ,$$
which matches sequence A007060 and can be further used to derive an explicit formula for $|A_n|$ (added as A285850).


TITLE: find the limit of a function with 0 in the divisor
QUESTION [0 upvotes]: Given the expression below I need to find the limit for $x \to 1$:
$$
\lim_{x \to 1}\frac{\frac{1 + \frac{4}{x}}{1+x} -2.5}{1+x-\frac{2}{x}} \div \frac{x^2+x^3+4}{13x + (x-1)^5}
$$
If we start substituting $x$'s for $1$'s we get $0$ in the divisor which means we can't really do that. I suppose the expression can somehow be transformed and/or simplified such that the divisor will not equal $0$ but I don't see how. I tried proceeding with the expression as it is and eventually received $0$ in the multiple ($\frac{1 + \frac{4}{1}}{1+1} -2.5 = 0$). Wolfram Alpha says the limit is $\approx -2.35$ (after my rounding).

REPLY [1 votes]: Without l'Hospital:
Simplify things (until some factor(s) $\;x-1\;$ disappear, if possible), and remember that $\;\cfrac ab\div\cfrac cd=\cfrac ab\cdot\cfrac dc\;$:
$$\frac{\frac{1+\frac4x}{1+x}-\frac52}{1+x-\frac2x}\div\frac{x^3+x^2+4}{13x+(x-1)^5}\stackrel{\text{mult. left factor by}\;\frac xx=1}=\frac{\frac{x+4}{1+x}-\frac{5x}2}{x^2+x-2}\cdot\frac{13x+(x-1)^5}{x^3+x^2+4}=$$
$$=\frac{2x+8-5x-5x^2}{2(x+1)(x+2)(x-1)}\cdot\frac{13x+(x-1)^5}{(x+2)(x^2-x+2)}=$$
$$=-\frac{\overbrace{5x^2+3x-8}^{=(5x+8)(x-1)}}{2(x+1)(x+2)(x-1)}\cdot\frac{13x+(x-1)^5}{(x+2)(x^2-x+2)}=$$
$$-\frac{5x+8}{2(x+1)(x+2)}\cdot\frac{13x+(x-1)^5}{(x+2)(x^2-x+2)}\color{red}{\xrightarrow[x\to1]{}-\frac{13}{12}\cdot\frac{13}{3\cdot2}=-\frac{169}{72}}$$


text:
\section{Leaf Subnetworks} \label{sec:leaf}
In this section, we consider the particular situation in which the subnetworks consist of leaves. We derive a concise expression for the norm of $P^{\vec{\omega}}_{G_i}$ restricted to the subspace $\mathcal{H}_i := \text{span}\{\vec{a}_u\,:\,u \in G_i\}$ and provide sufficient upper-bounds on the relaxation parameters for the vertices in $G_i$ to guarantee admissibility. We recall that the Gram matrix $\mathcal{G}(\vec{x}_1,\vec{x}_2,...,\vec{x}_t)$ is the $t \times t$ matrix of inner-products,
\[
\mathcal{G}(\vec{x}_1,\vec{x}_2,...,\vec{x}_t) = \begin{pmatrix}
\langle\vec{x}_1,\vec{x}_1\rangle & \langle\vec{x}_1,\vec{x}_2\rangle & ... & \langle\vec{x}_1,\vec{x}_t\rangle \\
\langle\vec{x}_2,\vec{x}_1\rangle & \langle\vec{x}_2,\vec{x}_2\rangle & ... & \langle\vec{x}_2,\vec{x}_t\rangle \\
\vdots & \vdots & \ddots & \vdots \\
\langle\vec{x}_t,\vec{x}_1\rangle & \langle\vec{x}_t,\vec{x}_2\rangle & ... & \langle\vec{x}_t,\vec{x}_t\rangle \\
\end{pmatrix}.
\]
We further denote the diagonal matrix $\mathcal{D}_i$ associated with the leaf subnetwork $G_i = \{\ell_{i,1},\ell_{i,2},...,\ell_{i,t_i}\}$ by
\[
\mathcal{D}_i = \begin{pmatrix}
\dfrac{w(g_i,\ell_{i,1})\omega_{\ell_{i,1}}}{\|\vec{a}_{\ell_{i,1}}\|^2} & 0 & \hdots & 0 \\
0 & \dfrac{w(g_i,\ell_{i,2})\omega_{\ell_{i,2}}}{\|\vec{a}_{\ell_{i,2}}\|^2} & \hdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \hdots & \dfrac{w(g_i,\ell_{i,t_i})\omega_{\ell_{i,t_i}}}{\|\vec{a}_{\ell_{i,t_i}}\|^2}
\end{pmatrix}.
\]

We denote the spectrum (collection of eigenvalues) of a matrix $A$ by $\sigma(A)$, and we denote its spectral radius by $\rho(A) = \max\{|\lambda|\,:\,\lambda \in \sigma(A)\}$.

\begin{theorem}\label{PGnorm}
Suppose $G_i = \{\ell_{i,1},\ell_{i,2},...,\ell_{i,t_i}\}$. Then
\[
\|P_{G_i}^{\vec{\omega}}|_{\mathcal{H}_i}\| = \max\{|1 - \lambda|\,:\,\lambda \in \sigma(\mathcal{D}_i\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}}))\smallsetminus\{0\}\}.
\]
\end{theorem}

\begin{proof}
From Equations \ref{operatorP}, \ref{operatorPw} and
\ref{eq15}, we have
\begin{align}\label{PGmatrix}
P^{\vec{\omega}}_{G_i} = I - \sum_{j=1}^{t_i}\dfrac{w(g_i,\ell_{i,j})\omega_{\ell_{i,j}}}{\|\vec{a}_{\ell_{i,j}}\|^2}\vec{a}_{\ell_{i,j}}\vec{a}_{\ell_{i,j}}^*.
\end{align}
Now let $K_{G_i} := \sqrt{\mathcal{D}_i}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})^*$. Then, Equation \ref{PGmatrix} may be expressed as $P^{\vec{\omega}}_{G_i} = I - K_{G_i}^*K_{G_i}$. Note that $\mathcal{H}_i$ is an invariant subspace for $K_{G_i}^*K_{G_i}$. Hence, from the spectral mapping theorem, we find
\[
\sigma(P_{G_i}^{\vec{\omega}}|_{\mathcal{H}_i}) = 1 - \sigma(K_{G_i}^*K_{G_i}|_{\mathcal{H}_i}).
\]
We claim that $\sigma(K_{G_i}^*K_{G_i}|_{\mathcal{H}_i})$ is precisely the collection of all of the nonzero eigenvalues of $K_{G_i}^*K_{G_i}$. Suppose, to the contrary, that there exists a nonzero vector $\vec{x} \in \mathcal{H}_i$ such that $K_{G_i}^*K_{G_i}\vec{x} = 0$. Then $K_{G_i}\vec{x} \in R(K_{G_i})\cap\mathcal{N}(K_{G_i}^*)$ implying $K_{G_i}\vec{x} = 0$, yet this leads to the contradiction that $\vec{x} \in \mathcal{H}_i \cap \mathcal{H}_i^\perp$ or $\vec{x} = 0$. It is well-known that $K_{G_i}^*K_{G_i}$ and $K_{G_i}K_{G_i}^*$ have the same nonzero eigenvalues and
\begin{align*}
\sigma(K_{G_i}K_{G_i}^*) &= \sigma\left(\sqrt{\mathcal{D}_i}\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})^T\sqrt{\mathcal{D}_i}\right) \\
&= \sigma(\mathcal{D}_i\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})),
\end{align*}
concluding the proof.

\end{proof}

\begin{corollary}\label{PGnormcor}
Suppose $G_i = \{\ell_{i,1},\ell_{i,2},...,\ell_{i,t_i}\}$. If
\[
0 < \omega_{\ell_{i,j}} < \dfrac{2\|\vec{a}_{\ell_{i,j}}\|^2}{w(g_i,\ell_{i,j})\rho(\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}}))} \qquad\hbox{for all}\qquad 1 \leq j \leq t_i,
\]
then $\|P_{G_i}^{\vec{\omega}}|_{\mathcal{H}_i}\| < 1$.
\end{corollary}

\begin{proof}
Since $\mathcal{D}_i$ and $\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})$ are positive semi-definite matrices, the eigenvalues of $\mathcal{D}_i\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})$ are nonnegative. Therefore, by Theorem \ref{PGnorm}, it suffices to show $\lambda < 2$ for $\lambda \in \sigma(\mathcal{D}_i\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}}))$. Let $j$ be the index for the largest diagonal entry in $\mathcal{D}_i$. By Theorem 8.12 in \cite{Zhang}, we have
\begin{align*}
\rho(\mathcal{D}_i\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})) \leq \dfrac{w(g_i,\ell_{i,j})\omega_{\ell_{i,j}}}{\|\vec{a}_{\ell_{i,j}}\|^2}\rho(\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})) < 2,
\end{align*}
as desired.

\end{proof}

\begin{remark}\label{spectralnormone}
It is not unusual to require that the rows of $A$ are normalized (i.e., $\|\vec{a}_u\| = 1$ for all $u$). Further, for the case that $\rho(\mathcal{G}(\vec{a}_{\ell_{i,1}},\vec{a}_{\ell_{i,2}},...,\vec{a}_{\ell_{i,t_i}})) \approx 1$, the relaxation parameters for the vertices in $G_i$ are admissible if
\[
\omega_{\ell_{i,j}} \lesssim \dfrac{2}{w(g_i,\ell_{i,j})} \qquad\hbox{for all}\qquad 1 \leq j \leq t_i.
\]
This upper-bound is greater than the usual bound in literature and can be drastically larger than $2$, depending on the associated weights. For example, if the weights are uniformly distributed, then the upper-bound is $2t_i \geq 2$.
\end{remark}

We end this section by observing that it is necessary and sufficient to check that $\Omega_1$ satisfies the admissibility conditions in Theorem \ref{mainthm} when the subnetwork consists of only leaves. We note that this need not hold for other subnetworks.

\begin{theorem}
Suppose $G_i = \{\ell_{i,1},\ell_{i,2},...,\ell_{i,t_i}\}$. Let $\Omega = s\Omega_1$ for some $s \in (0,1]$ as in Theorem \ref{mainthm}. If $\Omega_1$ satisfies the admissibility conditions, then $\Omega$ satisfies the admissibility conditions.
\end{theorem}

\begin{proof}
We check condition (2) in Definition \ref{D:admissible}. Let $\vec{x} \in \mathcal{H}_i$. Then
\begin{align*}
\|P^{\Omega}_{G_i}\vec{x}\| &= \left\|\sum_{j=1}^{t_i}w(g_i,\ell_{i,j})P^{\Omega}_{\ell_{i,j}}\vec{x}\right\| \\
&= \left\|\sum_{j=1}^{t_i}w(g_i,\ell_{i,j})\left[(1 - s)I + sP^{\Omega_1}_{\ell_{i,j}}\right]\vec{x}\right\| \\
&= \left\|(1 - s)\vec{x} + sP^{\Omega_1}_{G_i}\vec{x}\right\| \\
&\leq (1 - s)\|\vec{x}\| + s\alpha_i\|\vec{x}\| \\
&= [(1 - s)1 + s\alpha_i]\|\vec{x}\|,
\end{align*}
where the coefficient is strictly less than one as it is a convex sum of $1$ and $\alpha_i$.
\end{proof}

config: arxiv


TITLE: Is this determinant identity correct?
QUESTION [3 upvotes]: For complex valued matrices $A,B$ where $B$ is invertible, does $$\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})=\det(I+AB^{-1}A^*)=\det(I+A^*B^{-1}A)?$$
Here $A^*$ is the conjugate transform. I guess $\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})$ holds by Sylvester's identity.
Correction
$$\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})=\det(I+A^*B^{-1}A)?$$

REPLY [3 votes]: Indeed, $\det(I+B^{-1}AA^*)=\det(I+AA^*B^{-1})$ holds by Sylvester's Identity.
The other equalities don't hold. Let
$$
A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ B=\begin{bmatrix}1&0\\0&1/2\end{bmatrix}.
$$
Then
$$
\det(I+AB^{-1}A^*)=3, \ \ \det(I+A^*B^{-1}A)=2.
$$

REPLY [1 votes]: $\begin{align*}
\det(I + B^{-1}AA^*) &= \det(B)\det(I + B^{-1}AA^*)\det(B^{-1}) \\
&= \det(B(I + B^{-1}AA^*)B^{-1}) \\
&= \det(BB^{-1} + BB^{-1}AA^*B^{-1}) \\
&= \det(I + AA^*B^{-1})
\end{align*}$
That shows the first equality.  I don't see why the other two should be correct.  I only mention this because the identity is much easier than Sylvester's identity.
Sylvester's identity says
$$\det(I + CD) = \det(I + DC)$$
Letting $C = A^*B^{-1}$ and $D  = A$ we get
$$\det(I + A^*B^{-1}A) = \det(I + AA^*B^{-1}).$$
Similarly, if we let $C = B^{-1}A$ and $D = A^*$ we get
$$\det(I + B^{-1}AA^*) = \det(I + A^*B^{-1}A).$$


TITLE: Confusion between conditional and joint probability
QUESTION [2 upvotes]: This question might seem very trivial but I want to understand it in an intuitive way.
Let's say I have three options in a question and one of then is correct. I am randomly guessing the answer.
Now if I give the wrong answer in the first attempt I get a second chance
Then what is the probability that I will answer correctly?
I know that this answer should be P(first attempt correct) + P(second attempt correct)
Now what should be expression of P(second attempt correct)? It should be P(correct answer and first attempt wrong). But I am not understanding why it should not be P(correct answer|first attempt wrong)?
Can someone explain it in some intuitive way? And also what is the sample space and what are the correct events?

REPLY [1 votes]: I know that this answer should be P(first attempt correct) + P(second attempt correct) Now what should be expression of P(second attempt correct)? It should be P(correct answer and first attempt wrong). But I am not understanding why it should not be P(correct answer|first attempt wrong)? Can someone explain it in some intuitive way? And also what is the sample space and what are the correct events?

You only get the second attempt if the first fails. (This is why the events are disjoint.)
Therefore the answer should be: P(first attempt correct) + P(second attempt correct & first attempt wrong)
Which is: P(first attempt correct) + P(first attempt wrong) P(second attempt correct | first attempt wrong)  by definition of conditional probability.

And also what is the sample space and what are the correct events?

It depends on how you are guessing.  Let us say, there are $n$ possible (but only one correct), and you are selecting up to two without bias or repetition (basically: drawing without replacement).  Then the probability of obtaining the correct answer is:
$$\tfrac {1}{~n~}+\tfrac{~n-1~}{n}\cdot\tfrac{1}{~n-1~} ~=~ \tfrac {2}{~n~}$$


TITLE: Distance between closed set $A$ and compact set $B$ s.t. $A\cap B=\emptyset.$
QUESTION [0 upvotes]: I have to prove this statement.

Let A and B be nonempty disjoint subsets of $\mathbb{R}$ where A is closed and B is compact. Then there exist $a \in A$ and $b \in B$ such that $\inf\{|a-b| | a \in A, b \in B\}=|a-b|$.

This is the question in Distance between sets with one closed set and one compact set.
I'm trying to complete the proof referring to the questioner's idea and the two answers, but I'm having difficulty in completing the proof.

Let $E=\{|a-b| | a \in A, b \in B\}$.
And I can find the sequence $\{x_n\}\subset E$ s.t. $\lim x_n=\inf E$, since $E$ is not empty. (If $E$ is not bounded below, I consider $\inf E$ as $-\infty$.)
Since each $x_n$ is in $E$, there exist $\{a_n\}\subset A, \{b_n\}\subset B$ s.t. $x_n=|a_n-b_n|$
(Up to here, the questioner's idea. And from here, I refer to
the Hagen's answer.)
From the compactness of $B$, there exists a subsequence $\{b_{n_k}\}$ of $\{b_n\}$ and exists $b\in B$ s.t. $\lim_k b_{n_k}=b.$ ($b$ is called a limit point of $\{b_n\}$.)
Then, the zwim's answer says

$(\ast)$ we can work locally in the disk of center $b$ and radius $r$ (large enough so it intersects $A$ but more importantly contains the convergent part of $(a_n)$)

I have not fully understood this.
Since $\lim_k b_{n_k}=b,$ for all $r>0$, there exists $k_0\in \mathbb N$ s.t. $k\geqq k_0 \Rightarrow |b_{n_k}-b|<r,$ i.e., $k\geqq k_0 \Rightarrow b_{n_k}\in B_r(b)$.
According to $(\ast)$, it appears that I have to choose $r$ so that $B_r(b)$ can interact $A$ and contain the convergent part of $\{a_n\}$.
However, I don't know what "the convergent part of $\{a_n\}$" means.
$\{a_n\}$ is in $A$ and $A$ is not necessarily bounded, so this doesn't necessarily have the convergent subsequence, and I don't know how I should choose $r$ specifically.

REPLY [1 votes]: Your $x:=\inf E=\lim x_n$ is finite since your $E$ is a non-empty subset of $[0,+\infty).$
There exists $k_0\in\mathbb N$ such that
$$k\geq k_0\Rightarrow |b_{n_k}-b|<1.$$
Then, there exists $k_1\ge k_0$ such that $$k\ge k_1\Rightarrow x_{n_k}<x+1,$$
hence
$$k\ge k_1\Rightarrow |a_{n_k}-b|<r:=x+2.$$


TITLE: Upper bound on a vector norm
QUESTION [0 upvotes]: Let $A\subseteq\mathbb{R}^n$, s.t $\|a\|_p<\varepsilon, \forall a\in A$. Can we have an upper bound on $\|a-a'\|_p$.

REPLY [1 votes]: Think about this geometrically, somehow all these $a$'s lie in a ball with radius $\varepsilon$(under the $p$-norm). Then of course the maximum distance any of these points can be apart is $2\varepsilon$.


TITLE: Prove Limit: $x^2\to 4$, as $x\to 2$
QUESTION [2 upvotes]: I have been shown the proof, but I am stuck, have therefore marked out what I do not understand. The first red mark I do not see where five has come from. Could someone please take me through this step by step?

REPLY [3 votes]: First, some informal motivation. We want to choose $\delta$ so that if $0\lt |x-2|\lt \delta$, then $|x-2||x+2|$ is "small."
We have control over $|x-2|$, we can choose to make it small by choosing $\delta$ appropriately. The problem is that conceivably, $|x+2|$ could be large, and spoil things. That is why we insist that $\delta$ will be $\le 1$.
If $\delta\le 1$, then since $x$ is within $\delta$ of $2$, we have $1\le x\le 3$. It follows that $3\le\t x+2\lt 5$, and in particular $|x+2|\lt 5$.
Now the rest is straightforward. We want to make sure that $|x-2||x+2|\lt \epsilon$. The term $|x+2|$ is under control, it cannot get bigger than $5$. So if we make $|x-2|\lt \epsilon/5$, we will be OK.
But recall we want $\delta\le 1$. So we choose $\delta=\min(1,\epsilon/5)$.


TITLE: Asteroid orientation possible?
QUESTION [6 upvotes]: I’m writing a SF story with a question about an element of my plot. I have a Manhattan-sized NiFe asteroid, with little or no spin, in orbit at 2.5 astronomical units from the Sun.
Would it be reasonable to assume that the potato-shaped Ni-Fe asteroid, undisturbed for millennia, would settle into an orbit with its long axis (through the c.g.) parallel to its velocity vector?

REPLY [1 votes]: No.  Not only is there no reason to expect this, but it is especially unlikely to occur.  For the asteroid to remain "pointing" along it's velocity vector, that means that it actually completes a rotation with the exact same period as it's orbit.  This is called "tidal locking" and requires tidal forces to be important.  Tidal forces scale like,
$$F_\mathrm{tidal} \propto \frac{L}{r^3},$$
For an object of size $L$ (in this case very very small for an asteroid) and a distance between objects of $r$ (where here, $r = 2.5AU$).  Even Mercury's orbit isn't tidally locked (corrected by @rob, thanks!).  So the only reason this asteroid would continue to point in the same direction is if it coincidentally happened to have the exact right spin angular momentum.


text:
\begin{document}

\title{\textbf{The solution to the BCS gap equation \\
and the second-order phase transition \\
in superconductivity}}

\author{Shuji Watanabe\\
Division of Mathematical Sciences\\
Graduate School of Engineering, Gunma University\\
4-2 Aramaki-machi, Maebashi 371-8510, Japan\\
Email: shuwatanabe@gunma-u.ac.jp}

\date{}

\maketitle

\begin{abstract}

The existence and the uniqueness of the solution to the BCS gap equation of superconductivity is established in previous papers, but the temperature dependence of the solution is not discussed. In this paper, in order to show how the solution varies with the temperature, we first give another proof of the existence and the uniqueness of the solution and point out that the unique solution belongs to a certain set. Here this set depends on the temperature $T$. We define another certain subset of a Banach space consisting of continuous functions of both $T$ and $x$. Here, $x$ stands for the kinetic energy of an electron minus the chemical potential. Let the solution be approximated by an element of the subset of the Banach space above. We second show, under this approximation, that the transition to a superconducting state is a second-order phase transition.

\medskip

\noindent Mathematics Subject Classification (2000): 45G05, 46N50, 47H10, 82B26, 82D55

\medskip

\noindent Keywords: BCS gap equation, second-order phase transition, superconductivity, nonlinear integral equation, Schauder fixed-point theorem
\end{abstract}

\section{Introduction and preliminaries}

We use the unit $k_B=1$, where $k_B$ stands for the Boltzmann constant. Let $\omega_D>0$ and $k\in\mathbb{R}^3$ stand for the Debye frequency and the wave vector of an electron, respectively. We denote Planck's constant by $h>0$ and set $\hslash=h/(2\pi)$. Let $m>0$ and $\mu>0$ stand for the electron mass and the chemical potential, respectively. We denote by $T (\geq 0)$ the temperature, and by $x$ the kinetic energy of an electron minus the chemical potential, i.e., $x=\hslash^2|k|^2/(2m)-\mu \in [ -\mu, \, \infty)$. Note that $0<\hslash\omega_D<<\mu$.

In the BCS model \cite{bcs, bogoliubov} of superconductivity, the solution to the BCS gap equation \eqref{eq:gapequation} below is called the gap function. We regard the gap function as a function of both $T$ and $x$, and denote it by $u$, i.e., $u: \, (T,\, x) \mapsto u(T,\, x)$ $(\geq 0)$. The BCS gap equation is the following nonlinear integral equation:
\begin{equation}\label{eq:gapequation}
u(T,\, x)=\int_{\varepsilon}^{\hslash\omega_D}
\frac{U(x,\,\xi)\, u(T,\,\xi)}{\,\sqrt{\,\xi^2+u(T,\,\xi)^2\,}\,}\,
\tanh \frac{\,\sqrt{\,\xi^2+u(T,\,\xi)^2\,}\,}{2T}\, d\xi,
\end{equation}
where $U(x,\,\xi)>0$ is the potential multiplied by the density of states per unit energy at the Fermi surface and is a function of $x$ and $\xi$. In \eqref{eq:gapequation} we introduce $\varepsilon>0$, which is small enough and fixed $(0<\varepsilon<<\hslash\omega_D)$. It is known that the BCS gap equation \eqref{eq:gapequation} is based on a superconducting state called the BCS state. In this connection, see \cite[(6.1)]{watanabe} for a new gap equation based on a superconducting state having a lower energy than the BCS state.

The integral with respect to $\xi$ in \eqref{eq:gapequation} is sometimes replaced by the integral over $\mathbb{R}^3$ with respect to the wave vector $k$. Odeh \cite{odeh}, and Billard and Fano \cite{billardfano} established the existence and the uniqueness of the positive solution to the BCS gap equation in the case $T=0$. In the case $T\geq 0$, Vansevenant \cite{vansevesant} determined the transition temperature (the critical temperature) and showed that there is a unique positive solution to the BCS gap equation. Recently, Hainzl, Hamza, Seiringer and Solovej \cite{hhss} proved that the existence of a positive solution to the BCS gap equation is equivalent to the existence of a negative eigenvalue of a certain linear operator to show the existence of a transition temperature. Hainzl and Seiringer \cite{haizlseiringer} also derived upper and lower bounds on the transition temperature and the energy gap for the BCS gap equation. Moreover, Frank, Hainzl, Naboko and Seiringer \cite{fhns} gave a rigorous analysis of the asymptotic behavior of the transition temperature at weak coupling.

However, the solution mentioned above belongs to a function space consisting of functions of the wave vector only, and the temperature is regarded as a parameter, i.e., in previous papers the existence and the uniqueness of the solution is established for $T$ fixed. So the temperature dependence of the solution is not discussed. For example, it is not shown that the solution is continuous for $T \geq 0$. Moreover, how the solution varies with the temperature is not studied. Studying the temperature dependence of the solution to the BCS gap equation is very important. This is because, by dealing with the thermodynamical potential, this study leads to the mathematical challenge of showing that the transition to a superconducting state is a second-order phase transition. So it is highly desirable to study the temperature dependence of the solution.

In this paper, in order to show how the solution varies with the temperature, we first give another proof of the existence and the uniqueness of the solution to the BCS gap equation \eqref{eq:gapequation}. More precisely, we show that the unique solution belongs to $V_T$ (see \eqref{eq:vt} below). Note that the set $V_T$ depends on $T$. We define a certain subset $W$ (see \eqref{eq:spacew} below) of a Banach space consisting of continuous functions of both $T$ and $x$. Let the solution be approximated by an element of $W$. We second show, under this approximation, that the transition to a superconducting state is a second-order phase transition. In other words, we show that the condition that the solution belongs to $W$ is a sufficient condition for the second-order phase transition in superconductivity.

In this connection, on the basis of the Banach fixed-point theorem, the author \cite[Theorem 2.3]{watanabe3} recently gave a proof of the statement that the solution to the BCS gap equation \eqref{eq:gapequation} is continuous for both  $T$ and $x$ when $T$ $(\geq 0)$ is small enough.

Let
\begin{equation}\label{eq:potentialu}
U(x,\,\xi)=U_1 \qquad \mbox{at all} \quad (x,\,\xi) \in [\varepsilon, \, \hslash\omega_D]^2,
\end{equation}
where $U_1>0$ is a constant. Then the gap function depends on the temperature $T$ only. We therefore denote the gap function by $\Delta_1$ in this case, i.e., $\Delta_1:\, T \mapsto \Delta_1(T)$. Then \eqref{eq:gapequation} leads to the simplified gap equation
\begin{equation}\label{eq:smplgapequation}
1=U_1\int_{\varepsilon}^{\hslash\omega_D}
 \frac{1}{\,\sqrt{\,\xi^2+\Delta_1(T)^2\,}\,}\,
 \tanh \frac{\, \sqrt{\,\xi^2+\Delta_1(T)^2\,}\,}{2T}\,d\xi.
\end{equation}
It is known that superconductivity occurs at temperatures below the transition temperature. The following is the definition of the transition temperature, which originates from the simplified gap equation \eqref{eq:smplgapequation}.
\begin{definition}[\cite{bcs}]\label{dfn:tcsmpl}
The transition temperature originating from the simplified gap equation \eqref{eq:smplgapequation} is the temperature $\tau_1>0$ satisfying
\[
1=U_1\int_{\varepsilon}^{\hslash\omega_D}
\frac{1}{\,\xi\,}\,\tanh \frac{\xi}{\,2\tau_1\,}\,d\xi.
\]
\end{definition}

The BCS model makes the assumption that there is a unique solution $\Delta_1: T \mapsto \Delta_1(T)$ to the simplified gap equation \eqref{eq:smplgapequation} and that it is of class $C^2$ with respect to the temperature $T$ (see e.g. \cite{bcs} and \cite[(11.45), p.392]{ziman}). The author \cite{watanabe2} has given a mathematical proof of this assumption on the basis of the implicit function theorem. Set
\begin{equation}\label{eq:delta0}
\Delta=\frac{\,
\sqrt{ \left( \hslash\omega_D-\varepsilon\, e^{1/U_1} \right)
\left( \hslash\omega_D-\varepsilon\, e^{-1/U_1} \right) }\,}
{\,\sinh\frac{1}{\,U_1\,}\,}.
\end{equation}

\begin{proposition}[{\cite[Proposition 2.2]{watanabe2}}]\label{prp:simplegap}
Let $\Delta$ be as in \eqref{eq:delta0}. Then there is a unique nonnegative solution $\Delta_1: [\,0,\,\tau_1\,] \to [0,\,\infty)$ to the simplified gap equation \eqref{eq:smplgapequation} such that the solution $\Delta_1$ is continuous and strictly decreasing on the closed interval $[\,0,\,\tau_1\,]$:
\[
\Delta_1(0)=\Delta>\Delta_1(T_1)>\Delta_1(T_2)>\Delta_1(\tau_1)=0, \qquad 0<T_1<T_2<\tau_1.
\]
Moreover, it is of class $C^2$ on the interval $[\,0,\,\tau_1\,)$ and satisfies
\[
\Delta_1'(0)=\Delta_1''(0)=0 \quad \mbox{and} \quad \lim_{T\uparrow \tau_1} \Delta_1'(T)=-\infty.
\]
\end{proposition}

\begin{remark}
We set $\Delta_1(T)=0$ for $T>\tau_1$.
\end{remark}

\begin{remark}
In Proposition \ref{prp:simplegap}, $\Delta_1(T)$ is nothing but $\sqrt{f(T)}$ in \cite[Proposition 2.2]{watanabe2}.
\end{remark}

Let $0<U_1<U_2$ , where $U_2>0$ is a constant. We assume the following condition on $U(\cdot,\,\cdot)$ throughout this paper:
\begin{equation}\label{eq:condition}
U_1 \leq U(x,\,\xi) \leq U_2 \quad \mbox{at all} \quad (x,\,\xi) \in [\varepsilon,\, \hslash\omega_D]^2, \qquad U(\cdot,\,\cdot) \in C^2([\varepsilon,\, \hslash\omega_D]^2).
\end{equation}
When $U(x,\,\xi)=U_2$ at all $(x,\,\xi) \in [\varepsilon,\, \hslash\omega_D]^2$, an argument similar to that in Proposition \ref{prp:simplegap} gives that there is a unique nonnegative solution $\Delta_2: [\,0,\,\tau_2\,] \to [0,\,\infty)$ to the simplified gap equation
\begin{equation}\label{eq:smplgapequation2}
1=U_2\int_{\varepsilon}^{\hslash\omega_D}
 \frac{1}{\,\sqrt{\,\xi^2+\Delta_2(T)^2\,}\,}\,
 \tanh \frac{\, \sqrt{\,\xi^2+\Delta_2(T)^2\,}\,}{2T}\,d\xi, \qquad
0\leq T\leq \tau_2.
\end{equation}
Here the transition temperature $\tau_2>0$ is defined by
\begin{equation}\label{dfn:tcsmpl2}
1=U_2\int_{\varepsilon}^{\hslash\omega_D}
\frac{1}{\,\xi\,}\,\tanh \frac{\xi}{\,2\tau_2\,}\,d\xi.
\end{equation}
We again set $\Delta_2(T)=0$ for $T>\tau_2$. A straightforward calculation gives the following.

\begin{lemma}\label{lm:tautau}\quad {\rm (a)} The inequality $\tau_1<\tau_2$ holds.

\noindent {\rm (b)} If \   $0\leq T<\tau_2$, then $\Delta_1(T)<\Delta_2(T)$. If \  $T\geq \tau_2$, then $\Delta_1(T)=\Delta_2(T)=0$.
\end{lemma}

On the basis of Proposition \ref{prp:simplegap}, the author \cite[Theorem 2.3]{watanabe2} proved that the transition to a superconducting state is a second-order phase transition under the restriction \eqref{eq:potentialu}.

As is mentioned above, we now introduce the thermodynamical potential $\Omega$ to study the phase transition in superconductivity. The thermodynamical potential $\Omega$ is of the form
\[
\Omega=-k_BT \,\ln Z,
\]
where $Z$ is the function called the partition function. For more details on the thermodynamical potential, see e.g. \cite[sec.~III]{bcs} or Niwa \cite[sec.~7.7.3]{niwa}. Let $N(x)\geq 0$ stand for the density of states per unit energy at the energy $x$ \quad $(-\mu \leq x < \infty)$ and set $N_0=N(0)>0$. Here, $N_0$ stands for the density of states per unit energy at the Fermi surface $(x=0)$. Note that the function $x \mapsto N(x)$ is continuous on $[-\mu , \, \infty)$. For the gap function $u$, set
\[
\Omega_S(T)=\Omega_N(T)+\Psi(T),
\]
where
\begin{eqnarray}\label{eq:omegan}
\Omega_N(T)&=&
-2N_0\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}} x\,dx
-4N_0T\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
 \ln\left( 1+e^{\displaystyle{-x/T}} \right)\,dx \\ \nonumber
& &+\Phi(T),\qquad T>0,
\end{eqnarray}
\begin{eqnarray}\label{eq:v}
\Phi(T)&=&
2\int_{\displaystyle{-\mu}}^{\displaystyle{-\hslash\omega_D}} x\, N(x)\,dx
 -2T\int_{\displaystyle{-\mu}}^{\displaystyle{-\hslash\omega_D}}
 N(x)\ln\left( 1+e^{\displaystyle{\,x/T}} \right)\,dx \\ \nonumber
& &-2T\int_{\displaystyle{\hslash\omega_D}}^{\infty} N(x)
 \ln\left( 1+e^{\displaystyle{-x/T}} \right)\,dx,\qquad T>0,
\end{eqnarray}
\begin{eqnarray}\label{eq:delta}
\Psi(T)&=&
-2N_0\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
 \left\{ \sqrt{x^2+u(T,\,x)^2}-x\right\}\,dx  \\ \nonumber
& &+N_0\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
\frac{u(T,\, x)^2}{\,\sqrt{\,x^2+u(T,\, x)^2\,}\,}\,
\tanh \frac{\,\sqrt{\,x^2+u(T,\, x)^2\,}\,}{2T}\, dx \\ \nonumber
& &-4N_0T
\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
\ln \frac{1+e^{-\displaystyle{\sqrt{x^2+u(T,\, x)^2}/T}}}
{1+e^{-\displaystyle{x/T}}} \,dx,\quad 0<T\leq T_c\,.
\end{eqnarray}
Here, $T_c$ is the transition temperature defined by Definition \ref{dfn:tcgeneral} below and originates from the BCS gap equation \eqref{eq:gapequation}.

\begin{remark}
The integral $\displaystyle{ \int_{\displaystyle{\hslash\omega_D}}^{\infty} N(x) \ln\left( 1+e^{\displaystyle{-x/T}} \right)\,dx }$ on the right side of (\ref{eq:v}) is well defined for $T>0$, since $N(x)=O(\sqrt{x})$ as $x \to \infty$.
\end{remark}

\begin{definition}\label{dfn:thpo}
Let $\Omega_S(T)$ and $\Omega_N(T)$ be as above. The thermodynamical potential $\Omega$ in the BCS model is defined by (see e.g. Niwa \cite[sec.~7.7.3]{niwa})
\[
\Omega(T)=\left\{ \begin{array}{ll}\displaystyle{
 \Omega_S(T)} \qquad &(0<T\leq T_c),\\
\noalign{\vskip0.3cm} \displaystyle{
 \Omega_N(T)} \qquad &(T>T_c).
\end{array}\right.
\]
\end{definition}

\begin{remark}
Generally speaking, the thermodynamical potential $\Omega$ is a function of the temperature $T$, the chemical potential $\mu$ and the volume of our physical system. Fixing both $\mu$ and the volume of our physical system, we deal with the dependence of $\Omega$ on the temperature $T$ only.
\end{remark}

\begin{remark}
Hainzl, Hamza, Seiringer and Solovej \cite{hhss} studied the BCS gap equation with a more general potential examining the thermodynamic pressure.
\end{remark}

\begin{remark}\label{rmk:omegan-v}
It is shown in \cite[Lemmas 6.1 and 6.2]{watanabe2} that both of the functions $\Omega_N$ (see \eqref{eq:omegan}) and $\Phi$ (see \eqref{eq:v}), regarded as functions of $T$, are of class $C^2$ on $(0,\,\infty)$.
\end{remark}

\begin{definition}
We say that the transition to a superconducting state at the transition temperature $T_c$ is a second-order phase transition if the following conditions are fulfilled.

{\rm (a)}\   The thermodynamical potential $\Omega$, regarded as a function of $T$, is of class $C^1$ on $(0,\,\infty)$.

{\rm (b)}\   The thermodynamical potential $\Omega$, regarded as a function of $T$, is of class $C^2$ on $(0,\,\infty) \setminus \{ T_c\}$, and the second-order partial derivative $\left( \partial^2\Omega/\partial T^2\right)$ is discontinuous at $T=T_c$.
\end{definition}

\begin{remark}
As is known in condensed matter physics, condition (a) implies that the entropy $\displaystyle{ S=-\left( \partial\Omega/\partial T\right) }$ is continuous on $(0,\,\infty)$ and that, as a result, no latent heat is observed
at $T=T_c$ . On the other hand, (b) implies that the specific heat at constant volume, $\displaystyle{C_V=-T\left( \partial^2\Omega/\partial T^2\right)}$, is discontinuous at $T=T_c$ and that the gap $\Delta C_V$ in the specific heat at constant volume is observed at $T=T_c$ . For more details on the entropy well as the specific heat at constant volume, see e.g. \cite[sec.~III]{bcs} or Niwa \cite[sec.~7.7.3]{niwa}.
\end{remark}

The paper proceeds as follows. In section 2 we state our main results without proof. In sections 3, 4 and 5 we prove our main results.

\section{Main results}

Let $0 \leq T \leq \tau_2$ and fix $T$, where $\tau_2$ is that in \eqref{dfn:tcsmpl2}. We first consider the Banach space $C([\varepsilon,\, \hslash\omega_D])$ consisting of continuous functions of $x$ only, and deal with the following subset $V_T$:
\begin{equation}\label{eq:vt}
V_T=\left\{ u(T,\,\cdot) \in C([\varepsilon,\, \hslash\omega_D]): \; \Delta_1(T) \leq u(T,\,x) \leq \Delta_2(T) \;\; \mbox{at} \;\; x \in [\varepsilon,\, \hslash\omega_D] \right\}.
\end{equation}

\begin{remark}
The set $V_T$ depends on $T$. So we denote each element of $V_T$ by $u(T,\,\cdot)$.
\end{remark}

As is mentioned in the introduction, the existence and the uniqueness of the solution to the BCS gap equation was established in previous papers \cite{billardfano, fhns, hhss, haizlseiringer, odeh, vansevesant} and the uniqueness only holds for a nonnegative $U(\cdot,\,\cdot)$. However the temperature dependence of the solution is not discussed, and so we give another proof of the existence and the uniqueness of the solution to the BCS gap equation \eqref{eq:gapequation} so as to show how the solution varies with the temperature. More precisely, we show that for $T$ fixed, the unique solution belongs to $V_T$. Note that Proposition \ref{prp:simplegap} and Lemma \ref{lm:tautau} point out how $\Delta_1$ and $\Delta_2$ depend on the temperature and how $\Delta_1$ and $\Delta_2$ vary with the temperature.

\begin{theorem}\label{thm:gap}
Assume condition \eqref{eq:condition} on $U(\cdot,\,\cdot)$. Let $T \in [0,\, \tau_2]$ be fixed. Then there is a unique nonnegative solution $u_0(T,\,\cdot) \in V_T$ to the BCS gap equation \eqref{eq:gapequation}:
\[
u_0(T,\, x)=\int_{\varepsilon}^{\hslash\omega_D}
\frac{U(x,\,\xi)\, u_0(T,\, \xi)}{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}\,
\tanh \frac{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}{2T}\, d\xi, \quad
x \in [\varepsilon,\, \hslash\omega_D].
\]
Consequently, the solution is continuous with respect to $x$ and varies with the temperature as follows:
\[
\Delta_1(T) \leq u_0(T,\, x) \leq \Delta_2(T) \quad \mbox{at} \quad
(T,\,x) \in [0,\, \tau_2] \times [\varepsilon,\, \hslash\omega_D].
\]
\end{theorem}

\begin{remark}
In fact, Theorem \ref{thm:gap} holds true under
\[
U_1 \leq U(x,\,\xi) \leq U_2 \quad \mbox{at all} \quad (x,\,\xi) \in [\varepsilon,\, \hslash\omega_D]^2, \qquad U(\cdot,\,\cdot) \in C([\varepsilon,\, \hslash\omega_D]^2).
\]
But we assume condition \eqref{eq:condition} on $U(\cdot,\,\cdot)$ instead. This is because we deal with the subset $W$ (see \eqref{eq:spacew} below) so as to prove Theorem \ref{thm:phasetransition}.
\end{remark}

\begin{proposition}\label{prp:uzerotzero}
Let $T \in [\tau_1,\, \tau_2]$ be fixed and let $u_0(T,\,\cdot)$ be as in Theorem \ref{thm:gap}. If there is a point $x_1 \in [\varepsilon,\,\hslash\omega_D]$ satisfying $u_0(T,\,x_1)=0$, then $u_0(T,\,x)=0$ at all $x \in [\varepsilon,\,\hslash\omega_D]$.
\end{proposition}.

The existence of the transition temperature $T_c$ is pointed out in previous papers \cite{fhns, hhss, haizlseiringer, vansevesant}. In our case, it is defined as follows.

\begin{definition}\label{dfn:tcgeneral}\quad  Let $u_0(T,\,\cdot) \in V_T$ be as in Theorem \ref{thm:gap}. The transition temperature $T_c$ originating from the BCS gap equation \eqref{eq:gapequation} is defined by
\[
T_c=\inf \left\{ T>0:\, u_0(T,\, x)=0 \quad \mbox{at all} \; x \in [\varepsilon,\, \hslash\omega_D] \right\}.
\]
\end{definition}

\begin{remark}\label{rmk:tc1tctc2}
Combining Definition \ref{dfn:tcgeneral} with Theorem \ref{thm:gap} implies that $\tau_1\leq T_c \leq \tau_2$. For $T>T_c$, we set $u_0(T,\, x)=0$ at all $x \in [\varepsilon,\, \hslash\omega_D]$.
\end{remark}

\begin{proposition}\label{prp:u0delta}\quad Let $u_0(T,\,\cdot)$ be as in Theorem \ref{thm:gap}. If $U(x,\,\xi)=U_1$ at all $(x,\,\xi)\in [\varepsilon,\, \hslash\omega_D]^2$, then $u_0(T,\, x)=\Delta_1(T)$ and $T_c=\tau_1$.
\end{proposition}

We next consider the Banach space $C([0,\, T_c] \times [\varepsilon,\,\hslash\omega_D])$ consisting of continuous functions of both $T$ and $x$. Let us consider the following condition, which gives the behavior of functions as $T \to T_c$. We assume condition \eqref{eq:condition} on $U(\cdot,\,\cdot)$. Let $T_c$ be as in Definition \ref{dfn:tcgeneral} and let $\varepsilon_1>0$ be arbitrary.

\noindent \textbf{Condition (C).} \  For $u \in C([0,\, T_c] \times [\varepsilon,\, \hslash\omega_D]) \cap C^2\left( (0,\, T_c) \times [\varepsilon,\, \hslash\omega_D] \right)$, there are a unique $v \in C([\varepsilon,\, \hslash\omega_D])$ and a unique $w \in C([\varepsilon,\, \hslash\omega_D])$ satisfying the following.

(C1) \   $v(x)>0$ at all $x \in [\varepsilon,\, \hslash\omega_D]$.

(C2) \   For $\varepsilon_1>0$, there is a $\delta>0$ such that
$\left| T_c-T \right|<\delta$ implies
\[
\left| v(x)-\frac{\,u(T,\, x)^2\,}{\,T_c-T\,} \right|<T_c \,\varepsilon_1
\quad \mbox{and} \quad
\left| v(x)+2\,u(T,\, x)\frac{\partial u}{\,\partial T\,}(T,\, x) \right|<T_c \, \varepsilon_1\,,
\]
where $\delta$ does not depend on $x \in [\varepsilon,\, \hslash\omega_D]$.

(C3) \   Set $f(T,\, x)=u(T,\, x)^2$. Then, for $\varepsilon_1>0$, there is a $\delta>0$ such that $\left| T_c-T \right|<\delta$ implies
\[
\left| \frac{\, w(x)\,}{2}+
\frac{\,f(T,\, x)+(T_c-T)\,\displaystyle{ \frac{\,\partial f\,}{\,\partial T\,} }(T,\, x)\,}{\,(T_c-T)^2\,} \right|<\varepsilon_1 \quad \mbox{and} \quad
\left| w(x)-\frac{\,\partial^2 f\,}{\,\partial T^2\,}(T,\, x) \right|<\varepsilon_1\,,
\]
where $\delta$ does not depend on $x \in [\varepsilon,\, \hslash\omega_D]$.
\begin{remark}\label{rmk:utc} \   If $u \in C([0,\, T_c] \times [\varepsilon,\, \hslash\omega_D]) \cap C^2\left( (0,\, T_c) \times [\varepsilon,\, \hslash\omega_D] \right)$ satisfies condition (C), then \  $\displaystyle{ u(T_c\,,\, x)=0 }$ at all $x \in [\varepsilon,\, \hslash\omega_D]$.
\end{remark}

We deal with the following subset $W$ of the Banach space $C([0,\, T_c] \times [\varepsilon,\,\hslash\omega_D])$. Dealing with $W$ is important both in studying smoothness of the thermodynamical potential with respsect to $T$ and in showing that the transition to a superconducting state is a second-order phase transition.
\begin{eqnarray}\label{eq:spacew}
& & \qquad \\ \nonumber
W&=&\left\{ u \in C([0,\, T_c] \times [\varepsilon,\,\hslash\omega_D]) \cap C^2((0,\, T_c) \times [\varepsilon,\,\hslash\omega_D]):
\Delta_1(T)\leq u(T,\,x) \leq \Delta_2(T) \right. \\ \nonumber
& & \quad \left. \;
\mbox{at}\; (T,\,x)\in [0,\, T_c] \times [\varepsilon,\,\hslash\omega_D], \quad u \; \mbox{satisfies condition (C)} \right\}.
\end{eqnarray}

\begin{remark}
Let $u \in W$. Then, for $T \geq T_c$, we set $u(T,\,x)=0$ at all $x \in [\varepsilon,\,\hslash\omega_D]$.
\end{remark}

The set $W$ is not empty. Let $U_3 > 0$ be a constant satisfying
\[
1=U_3\int_{\varepsilon}^{\hslash\omega_D}
\frac{1}{\,\xi\,}\,\tanh \frac{\xi}{\,2T_c\,}\,d\xi.
\]
Note that $(0<)\, U_1\leq U_3\leq U_2$ (see Remark \ref{rmk:tc1tctc2}). An argument similar to that in Proposition \ref{prp:simplegap} gives that there is a unique nonnegative solution $\Delta_3: [\,0,\,T_c\,] \to [0,\,\infty)$ to the simplified gap equation
\[
1=U_3\int_{\varepsilon}^{\hslash\omega_D}
 \frac{1}{\,\sqrt{\,\xi^2+\Delta_3(T)^2\,}\,}\,
 \tanh \frac{\, \sqrt{\,\xi^2+\Delta_3(T)^2\,}\,}{2T}\,d\xi.
\]
Indeed, the function $\Delta_3$ just above is an element of $W$ (see \cite{watanabe2}). Therefore, $W \not= \emptyset$.

Define a mapping $A$ by
\begin{equation}\label{eq:mapping}
Au(T,\, x)=\int_{\varepsilon}^{\hslash\omega_D}
\frac{U(x,\,\xi)\, u(T,\, \xi)}{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}\,
\tanh \frac{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}{2T}\, d\xi,\qquad
u \in W.
\end{equation}

\begin{proposition}\label{prp:uauu0} \   Assume condition \eqref{eq:condition} on $U(\cdot,\,\cdot)$. Let $W$, a subset  of the Banach space $C([0,\, T_c] \times [\varepsilon,\,\hslash\omega_D])$, be as above. Let $u_0(T,\,\cdot) \in V_T$ be as in Theorem \ref{thm:gap}.

{\rm (a)}\   The mapping $\displaystyle{ A: \, W \longrightarrow W }$ is continuous with respect to the norm of the Banach space $C([0,\, T_c] \times [\varepsilon,\,\hslash\omega_D])$.

{\rm (b)}\   Let $u \in W$. Let $0 \leq T \leq \tau_2$ and fix $T$. Then all of $u(T,\, \cdot)$, $Au(T,\, \cdot)$ and $u_0(T,\,\cdot)$ belong to $V_T$. Consequently, at all $(T,\,x) \in [0,\, \tau_2] \times [\varepsilon,\,\hslash\omega_D]$,
\[
\Delta_1(T) \leq u(T,\,x), \, Au(T,\, x), \, u_0(T,\, x) \leq \Delta_2(T).
\]
\end{proposition}

We choose $U_1$ and $U_2$ (see \eqref{eq:condition}) such that the following inequality holds:
\begin{equation}\label{eq:inequality}
\sup_{0 \leq T \leq \tau_2} \left| \Delta_2(T)-\Delta_1(T) \right| < \varepsilon_2,
\end{equation}
where $\varepsilon_2>0$ is small enough. Then it follows from Proposition \ref{prp:uauu0} (b) that for $u \in W$,
\begin{equation}\label{eq:gapfunction}
\left| u(T,\,x)-u_0(T,\, x) \right|< \varepsilon_2, \quad
\left| Au(T,\, x)-u_0(T,\, x) \right|< \varepsilon_2, \quad
\left| Au(T,\,x)-u(T,\, x) \right|< \varepsilon_2.
\end{equation}
at all $(T,\,x) \in [0,\, \tau_2] \times [\varepsilon,\,\hslash\omega_D]$.

\noindent \textbf{Approximation (A).} \   The gap function on the right side of \eqref{eq:delta} is the solution $u_0$ of Theorem \ref{thm:gap}, i.e., the solution to the BCS gap equation \eqref{eq:gapequation}. But no one gives the proof of the statement that there is a unique solution in $W$ to the BCS gap equation \eqref{eq:gapequation}. In view of \eqref{eq:gapfunction}, we then let $u_0$ be approximated by a $u \in W$, and replace the gap function on the right side of \eqref{eq:delta} by this $u \in W$.

Let $g: \, [0,\, \infty) \rightarrow \mathbb{R}$ be given by
\begin{equation}\label{eq:fng}
g(\eta)= \left\{ \begin{array}{ll}\displaystyle{
\frac{1}{\,\eta^2\,}\left( \frac{1}{\,\cosh^2\eta \,}
 -\frac{\,\tanh\eta\,}{\eta}\right) } \qquad &(\eta>0),\\
\noalign{\vskip0.3cm} \displaystyle{
-\frac{\,2\,}{\,3\,} } &(\eta=0).
\end{array}\right.
\end{equation}
Note that $g(\eta)<0$. See Lemma \ref{lm:gproperty} below for some properties of $g$.

\begin{theorem}\label{thm:phasetransition} \   Assume condition \eqref{eq:condition} on $U(\cdot,\,\cdot)$. Let $U_1$ and $U_2$ be chosen such that \eqref{eq:inequality} holds, and let the solution $u_0$ of Theorem \ref{thm:gap} be approximated by a $u \in W$ as stated in approximation (A) above. Let $v \in C([\varepsilon,\,\hslash\omega_D])$ be as in condition (C). Then the following hold.

{\rm (a)}\   The transition to a superconducting state at the transition temperature $T_c$ is a second-order phase transition. Consequently, the condition that the solution to the BCS gap equation \eqref{eq:gapequation} belongs to $W$ is a sufficient condition for the second-order phase transition in superconductivity.

{\rm (b)}\   The gap $\Delta C_V$ in the specific heat at constant volume at the transition temperature $T_c$ is given by the form
\[
\Delta C_V=-\frac{N_0}{\, 8\, T_c\,}
 \int_{\varepsilon/(2\, T_c)}^{\hslash\omega_D/(2\, T_c)} v(2T_c\,\eta)^2
 \, g(\eta)\, d\eta \quad (>0).
\]
\end{theorem}

\begin{remark}
Suppose that $U(x,\,\xi)=U_1$ at all $(x, \,\xi) \in [\varepsilon,\, \hslash\omega_D]^2$. By Proposition \ref{prp:u0delta}, $u_0(T,\, x)=\Delta_1(T)$ and $T_c=\tau_1$. Therefore the function $\Delta_1$ of Proposition \ref{prp:simplegap} becomes an element of $W$ (see \cite{watanabe2}), and hence, $u$ of Theorem \ref{thm:phasetransition} can be replaced by  $\Delta_1$. So, setting $\varepsilon=0$, we find that the gap $\Delta C_V$ reduces to the form (see \cite[Proposition 2.4]{watanabe2})
\begin{equation}\label{eq:smplform}
\Delta C_V=-N_0f'(T_c)\tanh \frac{\,\hslash\omega_D\,}{2T_c} \quad (>0),
\end{equation}
where \  $\displaystyle{ f'(T_c)=-\lim_{T \uparrow T_c}
\frac{\,\Delta_1(T)^2\,}{\,T_c-T\,} }$.
\end{remark}

\section{Proof of Theorem \ref{thm:gap}}

Let $0 \leq T \leq \tau_2$ and fix $T$. Let $V_T$ be as in \eqref{eq:vt}. A straightforward calculation gives the following.

\begin{lemma}
The set $V_T$ is bounded, closed and convex.
\end{lemma}

Let $x \in [\varepsilon,\,\hslash\omega_D]$ and define a mapping $B$ by
\begin{equation}\label{eq:mappingb}
Bu(T,\, x)=\int_{\varepsilon}^{\hslash\omega_D}
\frac{U(x,\,\xi)\, u(T,\, \xi)}{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}\,
\tanh \frac{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}{2T}\, d\xi,\qquad
u(T,\,\cdot) \in V_T \, .
\end{equation}

\begin{lemma}\label{lm:aucd} \quad $\displaystyle{ Bu(T,\,\cdot) \in C([\varepsilon,\,\hslash\omega_D]) }$ for $u(T,\,\cdot) \in V_T$.
\end{lemma}

\begin{proof} \  For $\varepsilon_1>0$, let \
$\displaystyle{ \delta=
\frac{\varepsilon_1}{\, \displaystyle{ \hslash\omega_D\,\sup_{(x,\,\xi) \in [\varepsilon,\,\hslash\omega_D]^2} \left| \frac{\,\partial U\,}{\partial x}(x,\,\xi) \right| }
\,} }$. Then $\left| \, x-x_0\,\right|<\delta$ implies
\begin{eqnarray*}
\left| \, Bu(T,\, x)-Bu(T,\, x_0)\,\right|
&\leq& \int_{\varepsilon}^{\hslash\omega_D}
\left| U(x,\,\xi)-U(x_0,\,\xi) \right| \, d\xi \cr
&\leq& \hslash\omega_D\,\left| \, x-x_0\,\right|
 \sup_{(x,\,\xi) \in [\varepsilon,\,\hslash\omega_D]^2}
 \left| \frac{\,\partial U\,}{\partial x}(x,\,\xi) \right| \cr
&<& \varepsilon_1\,.
\end{eqnarray*}
\end{proof}

Since $\delta$ in the proof just above does not depend on $u(T,\,\cdot) \in V_T$, we immediately have the following.

\begin{lemma}\label{lm:equic}
The set $\displaystyle{ BV_T=\left\{ Bu(T,\,\cdot):\; u(T,\,\cdot) \in V_T \right\} }$  $\left( \subset C([\varepsilon,\,\hslash\omega_D]) \right)$ is equicontinuous.
\end{lemma}

\begin{lemma}\label{lm:d1aud2} \quad Let $u(T,\,\cdot) \in V_T$. Then \quad$\displaystyle{ \Delta_1(T) \leq Bu(T,\, x) \leq \Delta_2(T) }$ at $x \in [\varepsilon,\,\hslash\omega_D]$.
\end{lemma}

\begin{proof}\quad We show $Bu(T,\, x) \leq \Delta_2(T)$.
Since
\[
\frac{u(T,\, \xi)}{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}\leq
\frac{\Delta_2(T)}{\,\sqrt{\,\xi^2+\Delta_2(T)^2\,}\,}\,,
\]
it follows from \eqref{eq:smplgapequation2} that
\begin{eqnarray*}
Bu(T,\, x) \leq \int_{\varepsilon}^{\hslash\omega_D}
\frac{U_2\, \Delta_2(T)}{\,\sqrt{\,\xi^2+\Delta_2(T)^2\,}\,}\,
\tanh \frac{\,\sqrt{\,\xi^2+\Delta_2(T)^2\,}\,}{2T}\, d\xi
=\Delta_2(T).
\end{eqnarray*}
The rest can be shown similarly by \eqref{eq:smplgapequation}.
\end{proof}

Combining Lemma \ref{lm:aucd} with Lemma \ref{lm:d1aud2} immediately implies the following.

\begin{lemma} \quad $\displaystyle{BV_T \subset V_T}$.
\end{lemma}

By Lemma \ref{lm:d1aud2}, the set $BV_T$ is uniformly bounded since
\[
Bu(T,\, x) \leq \Delta_2(0)=\frac{\,
\sqrt{ \left( \hslash\omega_D-\varepsilon\, e^{1/U_2} \right)
\left( \hslash\omega_D-\varepsilon\, e^{-1/U_2} \right) }\,}
{\,\sinh\frac{1}{\,U_2\,}\,} \qquad \mbox{for} \quad u(T,\,\cdot) \in V_T.
\]
Combining Lemma \ref{lm:equic} with the Ascoli--Arzel$\grave{\mbox{a}}$ theorem thus yields the following.

\begin{lemma}
The set $BV_T$, a subset of the Banach space $C([\varepsilon,\,\hslash\omega_D])$, is relatively compact.
\end{lemma}

\begin{lemma}\label{lm:Bconti} \   The mapping $\displaystyle{ B: \, V_T \longrightarrow V_T }$ is continuous with respect to the norm of the Banach space $C([\varepsilon,\,\hslash\omega_D])$.
\end{lemma}

\begin{proof} \   Let $u(T,\,\cdot),\,v(T,\,\cdot) \in V_T$. Then
\begin{eqnarray*}
& & \left| \frac{u(T,\, \xi)}{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}\,
 \tanh \frac{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}{2T}
 -\frac{v(T,\, \xi)}{\,\sqrt{\,\xi^2+v(T,\, \xi)^2\,}\,}\,
 \tanh \frac{\,\sqrt{\,\xi^2+v(T,\, \xi)^2\,}\,}{2T} \right| \cr
\noalign{\vskip2mm}
& & \leq \frac{\,\left| u(T,\, \xi)-v(T,\, \xi)\right|\,}
 {\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}
 +v(T,\, \xi)\left| \frac{1}{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}
    -\frac{1}{\,\sqrt{\,\xi^2+v(T,\, \xi)^2\,}\,} \right| \cr
\noalign{\vskip2mm}
& & \quad +\frac{v(T,\, \xi)}{\,\sqrt{\,\xi^2+v(T,\, \xi)^2\,}\,}\,
   \left| \tanh \frac{\,\sqrt{\,\xi^2+u(T,\, \xi)^2\,}\,}{2T}
   -\tanh \frac{\,\sqrt{\,\xi^2+v(T,\, \xi)^2\,}\,}{2T} \right| \cr
\noalign{\vskip2mm}
& & \leq 3\frac{\,\left| u(T,\, \xi)-v(T,\, \xi)\right|\,}{\xi}\,.
\end{eqnarray*}
Thus \  $\displaystyle{
\left\| \, Bu(T,\,\cdot)-Bv(T,\,\cdot)\,\right\| \leq 3\, U_2
\ln \frac{\,\hslash\omega_D\,}{\,\varepsilon\,}\cdot
\left\| \, u(T,\,\cdot)-v(T,\,\cdot)\,\right\| }$, \quad
where $\displaystyle{ \left\| \cdot \right\| }$ stands for the norm of the Banach space $C([\varepsilon,\,\hslash\omega_D])$.
\end{proof}

We now have the following.

\begin{lemma}\quad The mapping $\displaystyle{ B: \, V_T \longrightarrow V_T}$ is compact, i.e., the mapping $\displaystyle{ B: \, V_T \longrightarrow V_T}$ is continuous and transforms bounded sets into relatively compact sets.
\end{lemma}

See Zeidler \cite[pp.39--40]{zeidler} for (nonlinear) compact operators. The Schauder fixed-point theorem (see e.g. Zeidler \cite[p.61]{zeidler}) thus implies the following.

\begin{lemma}\label{lm:u0}\quad Let $T \in [0,\, \tau_2]$ be fixed. Then the mapping $\displaystyle{ B: \, V_T \longrightarrow V_T}$ has at least one fixed point $u_0(T,\,\cdot) \in V_T$, i.e.,
\[
u_0(T,\,\cdot)=Bu_0(T,\,\cdot)\,,\qquad u_0(T,\,\cdot) \in V_T.
\]
\end{lemma}

Let us prove the uniqueness of $u_0(T,\,\cdot) \in V_T$.

\begin{lemma}\label{lm:unique} \quad Let $T \in [0,\, \tau_2]$ be fixed. Then the mapping $\displaystyle{ B: \, V_T \longrightarrow V_T}$ has a unique fixed point $u_0(T,\,\cdot) \in V_T$.
\end{lemma}

\begin{proof}
We give a proof similar to that of Thereom 24.2 given by Amann \cite{amann}. Let $v_0(T,\,\cdot) \in V_T$ be another fixed point of $B$, i.e., $v_0(T,\,\cdot)=Bv_0(T,\,\cdot)$. We deal with the case where $u_0(T,\, x)>0$ and $v_0(T,\, x)>0$ at all $x \in [\varepsilon,\,\hslash\omega_D]$ (see Proposition \ref{prp:uzerotzero}).

\textit{Step 1. The case where $\{ x \in [\varepsilon,\,\hslash\omega_D]: u_0(T,\, x) \geq v_0(T,\, x) \} \not= \emptyset$ and $\{ x \in [\varepsilon,\,\hslash\omega_D]: u_0(T,\, x) < v_0(T,\, x) \} \not= \emptyset$.}

Then there are a number $t$ \  $(0<t<1)$ and a point $x_0 \in [\varepsilon,\,\hslash\omega_D]$ such that
\begin{equation}\label{eq:T_0x_0}
u_0(T,\, x)\geq t\,v_0(T,\, x) \quad (x \in [\varepsilon,\,\hslash\omega_D]) \quad \mbox{and} \quad u_0(T,\, x_0)=t\, v_0(T,\, x_0).
\end{equation}
Hence
\begin{eqnarray*}
u_0(T,\, x_0) &=& \int_{\varepsilon}^{\hslash\omega_D}
 \frac{U(x_0,\,\xi)\, u_0(T,\, \xi)}{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}
 \,\tanh \frac{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}{2T}\, d\xi \cr
& & \cr
&\geq& \int_{\varepsilon}^{\hslash\omega_D}
 \frac{U(x_0,\,\xi)\, t\,v_0(T,\, \xi)}{\,\sqrt{\,\xi^2+t^2\,v_0(T,\, \xi)^2 \,}\,}\,
 \tanh \frac{\,\sqrt{\,\xi^2+t^2\,v_0(T,\, \xi)^2\,}\,}{2T}\, d\xi \cr
& & \cr
&>& t\,\int_{\varepsilon}^{\hslash\omega_D}
\frac{U(x_0,\,\xi)\, v_0(T,\, \xi)}{\,\sqrt{\,\xi^2+v_0(T,\, \xi)^2\,}\,}\,\tanh \frac{\,\sqrt{\,\xi^2+v_0(T,\, \xi)^2\,}\,}{2T}\, d\xi \cr
& & \cr
&=& t\,v_0(T,\, x_0),
\end{eqnarray*}
which contradicts \eqref{eq:T_0x_0}.

\textit{Step 2. The case where $u_0(T,\, x) \leq v_0(T,\, x)$ at all $x \in [\varepsilon,\,\hslash\omega_D]$.}

We again have \eqref{eq:T_0x_0}. Hence the same reasoning applies.

Thus $u_0(T,\,\cdot)=v_0(T,\,\cdot)$.
\end{proof}

The proof of Lemma \ref{lm:unique} is based on the condition that $U(\cdot,\,\cdot)$ is nonnegative. The proof of Theorem \ref{thm:gap} is complete.

We assumed also, in the proof of Lemma \ref{lm:unique}, that $u_0(T,\, x)>0$ and $v_0(T,\, x)>0$ at all $x \in [\varepsilon,\,\hslash\omega_D]$. We now show that if there is a point $x_1 \in [\varepsilon,\,\hslash\omega_D]$ satisfying $u_0(T,\,x_1)=0$, then $u_0(T,\,x)=0$ at all $x \in [\varepsilon,\,\hslash\omega_D]$ (see Proposition \ref{prp:uzerotzero}).

\noindent \textit{Proof of Proposition \ref{prp:uzerotzero}}.

Since $u_0(T,\,x_1)=0$, Theorem \ref{thm:gap} implies
\[
\int_{\varepsilon}^{\hslash\omega_D}
 \frac{U(x_1,\,\xi)\, u_0(T,\, \xi)}{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}
 \,\tanh \frac{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}{2T}\, d\xi=0.
\]
Since the integrand is nonnegative, it follows that $u_0(T,\, \xi)=0$ at all $\xi \in [\varepsilon,\,\hslash\omega_D]$.

\begin{remark}\quad Let $\tau_1 \leq T < \tau_2$ and fix $T$. Clearly, $0 \in V_T$ is a fixed point of $B$. If, for such a $T$, there are the two fixed points $u_0(T,\,\cdot) \in V_T$ mentioned in Lemma \ref{lm:unique} and $0 \in V_T$, then the fixed point $0 \in V_T$ is disregarded.
\end{remark}

Let us give a proof of Proposition \ref{prp:u0delta}.

\noindent \textit{Proof of Proposition \ref{prp:u0delta}}.

By Theorem \ref{thm:gap},
\[
u_0(T,\, x)=U_1 \int_{\varepsilon}^{\hslash\omega_D}
 \frac{u_0(T,\, \xi)}{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}
 \,\tanh \frac{\,\sqrt{\,\xi^2+u_0(T,\, \xi)^2\,}\,}{2T}\, d\xi, \qquad
 x \in D.
\]
Hence $u_0(T,\, x)$ does not depend on $x$. Therefore we denote $u_0(T,\,\cdot)$ by $u_0(T)$. Then
\[
u_0(T)\left\{ 1-U_1\int_{\varepsilon}^{\hslash\omega_D}
 \frac{1}{\,\sqrt{\,\xi^2+u_0(T)^2\,}\,}
 \,\tanh \frac{\,\sqrt{\,\xi^2+u_0(T)^2\,}\,}{2T}\, d\xi \right\}=0.
\]
Since $u_0(T) \not= 0$, it follows from \eqref{eq:smplgapequation} and Proposition \ref{prp:simplegap} that $u_0(T)=\Delta_1(T)$ and $T_c=\tau_1$.

\section{Proof of Proposition \ref{prp:uauu0}}

Let $A$ be as in \eqref{eq:mapping}. A straightforward calculation gives the following.

\begin{lemma} \   Let $u\in W$. Then $\displaystyle{ Au \in C([0,\, T_c] \times [\varepsilon,\,\hslash\omega_D]) \cap C^2( (0,\, T_c) \times [\varepsilon,\,\hslash\omega_D] ) }$.
\end{lemma}

A proof similar to that of Lemma \ref{lm:d1aud2} immediately gives the following.
\begin{lemma} \   Let $u\in W$. Then
\[
\Delta_1(T) \leq Au(T,\, x) \leq \Delta_2(T) \quad \mbox{at} \quad
(T,\,x) \in [0,\, T_c] \times [\varepsilon,\,\hslash\omega_D].
\]
\end{lemma}

We next show that $Au$ \   $(u \in W)$ satisfies condition (C) above. For $u \in W$, let $v$ be as in condition (C). Set
\begin{equation}\label{eq:functionF}
F(x)=\left( \int_{\varepsilon}^{\hslash\omega_D}
\frac{\, U(x,\,\xi)\,\sqrt{v(\xi)}\,}{\xi}
\tanh \frac{\xi}{\,2T_c\,}\, d\xi \right)^2 \quad (>0), \quad
\varepsilon \leq x \leq \hslash\omega_D\,.
\end{equation}

A straightforward calculation gives the following.

\begin{lemma}
Let $F$ be as in \eqref{eq:functionF}. Then $F \in C([\varepsilon,\,\hslash\omega_D])$.
\end{lemma}

\begin{lemma}
Let $F$ be as in \eqref{eq:functionF}. Then, for $\varepsilon_1>0$, there is a $\delta>0$ such that $\left| T_c-T \right|<\delta$ implies
\[
\left| F(x)-\frac{\,\{Au(T,\, x)\}^2\,}{\,T_c-T\,} \right|<T_c \,
\varepsilon_1\,,
\]
where $\delta$ does not depend on $x$ \  $(\varepsilon \leq x \leq \hslash\omega_D)$.
\end{lemma}

\begin{proof} \   Set $\displaystyle{M=\sup_{\varepsilon \leq \xi \leq \hslash\omega_D} v(\xi)}$. Let $\varepsilon_1$ and $\delta$ be as in (C2) of condition (C), and let $0<\varepsilon_1<1$. Then,
\begin{eqnarray*}
& & \; \left| F(x)-\frac{\,\{Au(T,\, x)\}^2\,}{\,T_c-T\,} \right| \cr
&\leq& \left| \int_{\varepsilon}^{\hslash\omega_D}
\frac{\, U_2\,\sqrt{v(\xi)}\,}{\xi}\, d\xi
+\int_{\varepsilon}^{\hslash\omega_D}
\frac{\, U_2\,}{\xi}\sqrt{\frac{\,u(T,\,\xi)^2\,}{\,T_c-T\,}}
\, d\xi \right| \cdot \left| I_1(x)+I_2(x)+I_3(x) \right| \cr
&\leq& 2U_2\sqrt{M+T_c}\,\ln\frac{\,\hslash\omega_D\,}{\varepsilon} \cdot
\left| I_1(x)+I_2(x)+I_3(x) \right|,
\end{eqnarray*}
where
\begin{eqnarray*}
I_1(x) &=& \int_{\varepsilon}^{\hslash\omega_D}
\frac{\,U(x,\,\xi)\,}{\xi}\left(
\sqrt{v(\xi)}-\sqrt{ \frac{\,u(T,\,\xi)^2\,}{\,T_c-T\,} }
\right) \tanh \frac{\xi}{\,2T_c\,}\,d\xi,\cr
I_2(x) &=& \int_{\varepsilon}^{\hslash\omega_D}
U(x,\,\xi)\sqrt{ \frac{\,u(T,\,\xi)^2\,}{\,T_c-T\,} } \left(
\frac{1}{\,\xi\,}-\frac{1}{\,\sqrt{\xi^2+u(T,\,\xi)^2}\,}\right)
\tanh \frac{\xi}{\,2T_c\,}\,d\xi,\cr
I_3(x) &=& \int_{\varepsilon}^{\hslash\omega_D}
\frac{U(x,\,\xi)}{\,\sqrt{\xi^2+u(T,\,\xi)^2}\,}
\sqrt{ \frac{\,u(T,\,\xi)^2\,}{\,T_c-T\,} } \left(
\tanh \frac{\xi}{\,2T_c\,}
-\tanh \frac{\,\sqrt{\xi^2+u(T,\,\xi)^2}\,}{\,2T\,} \right)\,d\xi.
\end{eqnarray*}
Let $T>T_c/2$. Then, by (C2), $\left| T_c-T \right|<\delta$ implies
\[
\left| I_1(x) \right| \leq \varepsilon_1\frac{\,U_2\,}{2}
\int_{\varepsilon}^{\hslash\omega_D} \frac{d\xi}{\,\sqrt{v(\xi)}\,}.
\]
Note that $\delta$ does not depend on $x$ \  $(\varepsilon \leq x \leq \hslash\omega_D)$. Since
\[
u(T,\,\xi)^2=\frac{\, u(T,\,\xi)^2\,}{T_c-T}\left( T_c-T \right),
\]
$u(T,\,\xi)$ tends to 0 uniformly with respect to $\xi$ as $T \to T_c$ by (C2). Therefore, $\left| T_c-T \right|<\delta$ implies $u(T,\,\xi) < T_c \, \varepsilon_1$. Recalling $T>T_c/2$ and (C2), we can show similarly that
\[
\left| I_2(x) \right| \leq \varepsilon_1\frac{\,U_2\, T_c\sqrt{M+T_c}\,}{\varepsilon} \quad \hbox{and} \quad
\left| I_3(x) \right| \leq \varepsilon_1\, U_2\left( \frac{\,\sqrt{M+T_c}\,}{2}\ln\frac{\,\hslash\omega_D\,}{\varepsilon}+\frac{\, \hslash\omega_D \,}{\, \sqrt{2T_c} \,} \right).
\]
Note also that $\varepsilon >0$ is fixed.
\end{proof}

A straightforward calculation similar to the above gives the following.

\begin{lemma}
Let $F$ be as in \eqref{eq:functionF}. Then, for $\varepsilon_1>0$, there is a $\delta>0$ such that $\left| T_c-T \right|<\delta$ implies
\[
\left| F(x)+2\, Au(T,\, x)\frac{\partial Au}{\,\partial T\,}(T,\, x)
\right|<T_c \, \varepsilon_1\,,
\]
where $\delta$ does not depend on $x$ \  $(\varepsilon \leq x \leq \hslash\omega_D)$.
\end{lemma}

For $u \in W$, let $v$ and $w$ be as in condition (C). For $\varepsilon \leq x \leq \hslash\omega_D$, set
\begin{eqnarray}\label{eq:functionG}
& &G(x)=\int_{\varepsilon}^{\hslash\omega_D}
\frac{\, U(x,\,\xi)\,\sqrt{v(\xi)}\,}{\xi}
\tanh \frac{\xi}{\,2T_c\,}\, d\xi
\int_{\varepsilon}^{\hslash\omega_D} U(x,\,\eta) \times \\
&\times& \left\{ \left( \frac{w(\eta)}{\,\eta\sqrt{v(\eta)}\,}-
\frac{\,2\sqrt{v(\eta)^3}\,}{\eta^3} \right)
\tanh \frac{\eta}{\,2T_c\,}+\frac{\sqrt{v(\eta)}}{\,\cosh^2
\displaystyle{ \frac{\eta}{\,2T_c\,}\,} }
\left( \frac{v(\eta)}{\,T_c\,\eta^2\,}+\frac{2}{\,T_c^2\,} \right)
\right\}\, d\eta.\nonumber
\end{eqnarray}

A straightforward calculation similar to the above gives the following.

\begin{lemma}\   Let $G$ be as in \eqref{eq:functionG}. Then $G \in C([\varepsilon,\,\hslash\omega_D])$.
\end{lemma}

\begin{lemma}\   Let $G$ be as in \eqref{eq:functionG}. Set $H(T,\, x)=\{ Au(T,\, x)\}^2$. Then, for $\varepsilon_1>0$, there is a $\delta>0$
such that $\left| T_c-T \right|<\delta$ implies
\[
\left| \frac{\, G(x)\,}{2}+\frac{\, H(T,\, x)+(T_c-T)\,\displaystyle{
\frac{\,\partial H\,}{\,\partial T\,} }(T,\, x)\,}{\,(T_c-T)^2\,} \right|<\varepsilon_1 \quad \mbox{and} \quad
\left| G(x)-\frac{\,\partial^2 H\,}{\,\partial T^2\,}(T,\, x) \right|<\varepsilon_1\,,
\]
where $\delta$ does not depend on $x$ \  $(\varepsilon \leq x \leq \hslash\omega_D)$.
\end{lemma}

We therefore have $AW \subset W$. A proof similar to that of Lemma \ref{lm:Bconti} immediately gives the following.
\begin{lemma}\   The mapping $\displaystyle{
A: \, W \longrightarrow W}$ is continuous with respect to the norm of the Banach space $C([0,\, T_c] \times [\varepsilon,\,\hslash\omega_D])$.
\end{lemma}

We have thus proved (a) of Proposition \ref{prp:uauu0}. Part (b) follows immediately from (a).

\section{Proof of Theorem \ref{thm:phasetransition}}

Let $T_1>0$ be arbitrary.

\begin{lemma}\label{lm:deltat}
Let the function $\Psi$ be as in \eqref{eq:delta}. Then the function $\Psi$ is differentiable on $(T_1\,,\,T_c]$, and $(\partial \Psi/\partial T)(T_c)=0$.
\end{lemma}

\begin{proof}
A straightforward calculation gives that $\Psi$ is differentiable on $(T_1\,,\,T_c)$. So it suffices to show that $\Psi$ is differentiable at $T=T_c$ and $(\partial \Psi/\partial T)(T_c)=0$. Note that $\Psi(T_c)=0$ since $u(T_c\,,\,x)=0$ at all $x \in [\varepsilon,\, \hslash\omega_D]$ (see Remark \ref{rmk:utc}). Then, for $T<T_c$,
\begin{eqnarray}\label{eq:deltadifferentiable}
\,\quad -\frac{\Psi(T)}{\,T_c-T\,}&=&
\frac{2N_0}{\,T_c-T\,}
\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
 \left\{ \sqrt{x^2+u(T,\,x)^2}-x\right\}\,dx  \\ \nonumber
& &-\frac{N_0}{\,T_c-T\,}
\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
\frac{u(T,\, x)^2}{\,\sqrt{\,x^2+u(T,\, x)^2\,}\,}\,
\tanh \frac{\,\sqrt{\,x^2+u(T,\, x)^2\,}\,}{2T}\, dx \\ \nonumber
& &+\frac{4N_0T}{\,T_c-T\,}
\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
\ln \frac{1+e^{-\displaystyle{\sqrt{x^2+u(T,\, x)^2}/T}}}
{1+e^{-\displaystyle{x/T}}} \,dx.
\end{eqnarray}
By (C2) in condition (C),
\[
\left| \frac{\,\sqrt{x^2+u(T,\,x)^2}-x\,}{\,T_c-T\,} \right| \leq
\frac{\,v(x)+T_c\,}{x},
\]
where $\varepsilon_1<1$ is assumed. Hence, the Lebesgue dominated convergence theorem implies that as $T \uparrow T_c$ , the first term on the right side of \eqref{eq:deltadifferentiable} tends to
\[
\lim_{T \uparrow T_c} \frac{2N_0}{\,T_c-T\,}
\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
\left\{ \sqrt{x^2+u(T,\,x)^2}-x\right\}\,dx=
N_0 \int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
\frac{\, v(x)\,}{x}\,dx.
\]
Similarly we can deal with the second and the third terms. Thus $\Psi$ is differentiable at $T=T_c$ and
\[
-\lim_{T \uparrow T_c} \frac{\Psi(T)}{\,T_c-T\,}=0.
\]
\end{proof}

A straightforward calculation gives the following.

\begin{lemma}\label{lm:gproperty}
Let $g$ be as in \eqref{eq:fng}. Then the function $g$ is of class $C^1$
on $[0,\,\infty)$ and satisfies
\[
g(\eta)<0 \quad (\eta \geq 0),\qquad g'(0)=0,\qquad
\lim_{\eta\to\infty}g(\eta)=\lim_{\eta\to\infty}g'(\eta)=0.
\]
\end{lemma}

\begin{lemma}\label{lm:deltac2}
Let $\Psi$ be as in Lemma \ref{lm:deltat}. Then $\partial \Psi/\partial T$ is differentiable on $(T_1\,,\,T_c]$, and
\[
\frac{\,\partial^2\Psi\,}{\,\partial T^2\,}(T_c)=\frac{N_0}{\,8T_c^2\,}
\int_{\varepsilon/(2T_c)}^{\hslash\omega_D/(2T_c)}
v(2T_c\,\eta)^2 \, g(\eta)\,d\eta \quad (<0).
\]
\end{lemma}

\begin{proof}
A straightforward calculation gives that $\partial \Psi/\partial T$ is differentiable on $(T_1\,,\,T_c)$. So it suffices to discuss its differentiability at $T=T_c$. Note that $(\partial \Psi/\partial T)(T_c)=0$ by Lemma \ref{lm:deltat}. Then
\begin{eqnarray*}
-\lim_{T \uparrow T_c} \frac{\,\displaystyle{
\frac{\,\partial \Psi\,}{\partial T}(T)}\,}{\,T_c-T\,}&=&-N_0
\lim_{T \uparrow T_c}
\int_{\displaystyle{\varepsilon}}^{\displaystyle{\hslash\omega_D}}
\frac{\, u(T,\,x)^2\,}{\, T_c-T\,}\; u(T,\,x)
\frac{\,\partial u(T,\,x)\,}{\,\partial T\,}\frac{1}{\,x^2+u(T,\,x)^2\,}
\times \\ \noalign{\vskip0.3cm} & &\quad \times
\left( \frac{1}{\,2T\cosh^2 \frac{\,\sqrt{\,x^2+u(T,\, x)^2\,}\,}{2T}\,}
-\frac{\, \tanh \frac{\,\sqrt{\,x^2+u(T,\, x)^2\,}\,}{2T}\,}{\sqrt{\,x^2+u(T,\, x)^2\,}} \right)\,dx.
\end{eqnarray*}
Note that $u(T_c\,,\,x)=0$ at all $x \in [\varepsilon,\, \hslash\omega_D]$ by Remark \ref{rmk:utc}. Hence, combining the Lebesgue dominated convergence theorem with (C2) in condition (C) proves the lemma.
\end{proof}

Note that $T_1>0$ is arbitrary. Therefore, combining Lemmas \ref{lm:deltat} and \ref{lm:deltac2} with Remark \ref{rmk:omegan-v} immediately implies the following.

\begin{lemma}
The thermodynamical potential $\Omega$, regarded as a function of $T$, is of class $C^1$ on $(0,\,\infty)$.
\end{lemma}

A straightforward calculation based on (C3) in condition (C) gives the following.

\begin{lemma}\  The function $\partial^2 \Psi/\partial T^2$ is continuous on $(0,\,T_c]$.
\end{lemma}

Lemma \ref{lm:deltac2} immediately implies the following.

\begin{lemma}\  The thermodynamical potential $\Omega$, regarded as a function of $T$, is of class $C^2$ on $(0,\,\infty) \setminus \{ T_c\}$. Consequently, the transition to a superconducting state at the transition temperature $T_c$ is a second-order phase transition, and hence the condition that the solution to the BCS gap equation \eqref{eq:gapequation} belongs to $W$ is a sufficient condition for the second-order phase transition in superconductivity.
\end{lemma}

Since the specific heat at constant volume is of the form $\displaystyle{C_V=-T\left( \partial^2\Omega/\partial T^2\right)}$, the gap $\Delta C_V$ in $C_V$ at the transition temperature $T_c$ is given by (see Definition \ref{dfn:thpo})
\[
\Delta C_V=-T_c \left\{ \lim_{T\uparrow T_c}
\frac{\,\partial^2\Omega\,}{\,\partial T^2\,}(T)-
\lim_{T\downarrow T_c} \frac{\,\partial^2\Omega\,}{\,\partial T^2\,}(T)
\right\}=-T_c\,\frac{\,\partial^2\Psi\,}{\,\partial T^2\,}(T_c).
\]

We thus have the following by Lemma \ref{lm:deltac2}.

\begin{lemma}
The gap $\Delta C_V$ in the specific heat at constant volume at the transition temperature $T_c$ is given by the form
\[
\Delta C_V=-\frac{N_0}{\, 8\, T_c\,}
 \int_{\varepsilon/(2\, T_c)}^{\hslash\omega_D/(2\, T_c)} v(2T_c\,\eta)^2
 \, g(\eta)\, d\eta \quad (>0).
\]
\end{lemma}

\noindent \textbf{Acknowledgments}

S. Watanabe is supported in part by the JSPS Grant-in-Aid for Scientific Research (C) 21540110.


TITLE: $\aleph_\omega$ and large ordinals?
QUESTION [4 upvotes]: Assuming the GCH, each time you take the powerset of an aleph number, the subscript increases by one. So there is $\aleph_0, \aleph_1, \aleph_2\ldots \aleph_\omega\ldots \aleph_{\epsilon_0}\ldots$ and so on.
This may be a dumb question, but how large of an ordinal has to be present in the subscript below the aleph in order for the resulting aleph number to make it onto the large cardinal hierarchy? For example, how does aleph_church-kleene compare to the smallest inaccessible cardinal?

REPLY [7 votes]: The term "large cardinal" doesn't have an agreed up, concrete definition. When we say that $\kappa$ is a large cardinal we might mean that $\kappa$ is inaccessible and has additional properties; or we might mean that it has properties which imply the consistency of large cardinals.
Some people would refer to the former as large cardinals, and the latter as "large cardinal properties".
If we agree with that convention, then every large cardinal is a weakly inaccessible cardinal at least, which means that it is an $\aleph$-fixed point. Namely $\kappa=\aleph_\kappa$ (and in fact, it is the limit of fixed points, and limits of limits of fixed points and so on).
As to your second question, $\aleph_{\omega_1^{CK}}$ has only countably many $\aleph$'s below it. That's just a tiny fraction of the cardinals you will find below the first $\aleph$ fixed point. If $\kappa$ is inaccessible then it is the $\kappa$-th fixed point, meaning it is unfathomably larger.
Let me add a side remark, $\sf GCH$ has little to do with this. A strongly inaccessible cardinal is also a $\beth$ fixed point (with similar properties as discussed above), where $\beth$ numbers are defined using the power set operation, rather than successor cardinals.

Added:
It may be a bit misleading to think that being a large cardinal is all about having a large index as an $\aleph$ number. Being a large cardinal means having some combinatorial properties which allows you to prove more. Successor of large cardinals are not "large" themselves, because in $\sf ZFC$ successor cardinals are do not posses the basic requirement of a large cardinal, being a regular limit cardinal.


TITLE: Intégral of cumulative distribution function and the density function
QUESTION [0 upvotes]: Let $\Phi: \mathbb{R} \to \mathbb{R}$ be the cumulative distribution function of a standard normal variable $N(0,1)$ and $\phi: \mathbb{R} \to \mathbb{R}$ be its density function. I am interested in evaluating the following integral

$$ \int_{0}^{\infty} \Phi(x) \phi(x) \, dx. $$

Note that if the integral is changed to $\int_{\mathbb{R}} \Phi(x) \phi(x) \,dx $, then by using the facts that $\Phi- \frac{1}{2}$ is an odd function and that $\phi$ is an even function,
\begin{eqnarray}
\int_{\mathbb{R}} \Phi(x) \phi(x) \,dx & = & \int_{\mathbb{R}} \Big( \Phi(x)- \frac{1}{2} \Big) \phi(x) \, dx + \frac{1}{2} \\
& = & \frac{1}{2}.
\end{eqnarray}
However, this trick does not work in this case when the limits of integration are $0$ to $\infty$. Integration by substitution (using $u=-x$) does not work either. Any ideas?
Intuitively, the answer should be $\frac{1}{4}$, by some kind of symmetry. But I can’t show it.

REPLY [1 votes]: Notice that $\phi(x) = \Phi^{\prime}(x)$. Thus if we make the substitution $u = \Phi(x)$, we get that $\mathrm{d}u = \phi(x)\mathrm{d}x$. Since $\Phi(0) = 0.5$ and $\Phi(\infty) = 1$, we get
\begin{align*}
\int_{0}^{\infty}\Phi(x)\phi(x)\mathrm{d}x = \int_{1/2}^{1}u\mathrm{d}u = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}
\end{align*}
At your case, we have $\Phi(-\infty) = 0$ and $\Phi(\infty) = 1$. Consequently, it results that
\begin{align*}
\int_{\textbf{R}}\Phi(x)\phi(x)\mathrm{d}x = \int_{0}^{1}u\mathrm{d}u = \frac{1}{2}
\end{align*}


TITLE: Approximating basic trigonometric functions to a integrable form
QUESTION [0 upvotes]: I wondered if there is a way of approximating trigonometric functions in terms of basic functions (possibly trigonometric functions) so that one can derive the indefinite integral of said function. The function I have in mind is the following:
$$f(x)=sin(\frac{sin(2x)}{2}+x)$$
The presented function is the one currently hussling me; however, I also ask for a general method if one exists.
Sidenote:
The mentioned function was intended to be used as an approximation for Jacobian Elliptic Functions with proper constants dependent on k to match the period and "fatness" of the function.

REPLY [0 votes]: You can calculate the taylor-series of the function. In this case, it is
difficult to do it by hand. PARI/GP gives :
? sin(sin(2*x)/2+x)
%1 = 2*x - 2*x^3 + 26/15*x^5 - 376/315*x^7 + 1982/2835*x^9 - 57212/155925*x^11 +
 1075684/6081075*x^13 - 50590052/638512875*x^15 + O(x^17)


text:
\section{Stratified Morse Lemma}

In this section, we prove an isotopy lemma (Theorem
\ref{isotopy}) adapted to the local study of stratified Morse
functions.  As consequences, we derive some local normal form
statements, one of which (Corollary \ref{ml}) may be called
the stratified Morse lemma.  The results of this section are
very close to those of H. King in \cite{K1} and \cite{K2}.  We
continue with a Whitney stratified smooth manifold $(X, \cS)$.
As in Section 1.2, we denote by $\Lambda^0$ the set of generic
conormal vectors to $\cS$.  We also let $\Lambda^0_S = \Lambda^0
\cap \Lambda_S$, for each $S \in \cS$.

\begin{thm}\label{isotopy}
Let $S \in \cS$ with $\dim S = s$.  Let $B \subset S$ be
a closed $s$-ball smoothly embedded in $S$, with interior
$B^\circ \subset B$ and a fixed point $a \in B^\circ$.  Let
$\cU \subset X$ be an open neighborhood of $B$, let $p : \cU
\to S$ be a smooth submersion restricting to the identity on
$\cU \cap S$, and let $f: \cU \to \R$ be a smooth function
such that $f|^{}_{S \cap \cU} = 0$ and $d_b f \in \Lambda^0_S$
for every $b \in B$.  Write $N = p^{-1} (a)$.  Then there
exist an open set $U \subset \cU$ with $U \cap S = B^\circ$,
control data on $U$ which is $p$-compatible on $U$ and
$f$-compatible on $U \setminus S$, and a controlled
homeomorphism $\phi : (U \cap N) \times B^\circ \to U$
such that:

(i)   $\phi \, (x,a) = x$ for every $x \in U \cap N$;

(ii)  $p \circ \phi \, (x,y) = y$ for every
$x \in U \cap N$ and $y \in B^\circ$.

(iii) $f \circ \phi \, (x,y) = f(x)$ for every
$x \in U \cap N$ and $y \in B^\circ$.
\end{thm}

\begin{cor}\label{ml}
Let $f : X \to \R$ be a smooth function with a stratified
Morse critical point $a$, lying in a stratum $S$, let
$\cU \subset X$ be a neighborhood of $a$, and let $p : \cU
\to S$ be a smooth submersion restricting to the identity on
$\cU \cap S$.  Write $N = p^{-1} (a)$.  Then there exist a
smaller neighborhood $U \subset \cU$ of $a$, control data on
$U$ which is $p$-compatible on $U$ and $f$-compatible on
$U \setminus S$, and a controlled homeomorphism
$\phi : (U \cap N) \times (U \cap S) \to U$ such that:

(i)  $\phi \, (x,a) = x$ for every $x \in U \cap N$;

(ii) $p \circ \phi \, (x,y) = y$ for every
$x \in U \cap N$ and $y \in U \cap S$;

(iii) $f \circ \phi \, (x,y) = f(x) + f(y) - f(a)$
for every $x \in U \cap N$ and $y \in U \cap S$.
\end{cor}

\noindent
{\bf Proof:}  Apply Theorem \ref{isotopy} to the function
$f - f \circ p$.
\hfill$\Box$

\begin{rmk}{\em
Corollary \ref{ml} can be used to give a short proof of Goresky and
MacPherson's product theorem for Morse data, as stated in Chapter 1 of
\cite{GM2} (Theorem STM, part B).  This was earlier observed by H. King
in \cite{K1} and \cite{K2}.  We should also mention that a different
short proof of Goresky and MacPherson's result has appeared recently in
\cite{Ha}.}
\end{rmk}

\begin{cor}\label{ns}
Let $S$ be a stratum.  Suppose we have two points $a, \hat a \in S$,
two normal slices $N, \hat N$ to $S$ passing through these points,
and two functions $f : N \to \R$ and $\hat f : \hat N \to \R$ with
$f(a) = \hat f (\hat a) = 0$.  Assume that the differentials $d_a f$
and $d_{\hat a} \hat f$ are both in $\Lambda^0_S$ and, moreover, in
the same path-component of $\Lambda^0_S$.  Then there exist open
neighborhoods $U \subset N$ and $\hat U \subset \hat N$ of $a$ and
$\hat a$, control data on $U$ and $\hat U$, $f$- (resp. $\hat f$-)
compatible on $U \setminus \{ a \}$ (resp. $\hat U \setminus
\{ \hat a \}$), and a controlled homeomorphism $\phi : U \to \hat U$
such that $f = \hat f \circ \phi$.
\end{cor}

\noindent
{\bf Proof:}  Apply Theorem \ref{isotopy} to a suitable
function on the product $X \times \R$.
\hfill$\Box$

\begin{rmk}{\em
As another corollary of Theorem \ref{isotopy}, we note the fact
that stratified Morse functions are topologically locally stable
in the same sense as the ordinary Morse functions.  In other words,
a stratified Morse function is (locally) right-equivalent, by a
controlled homeomorphism, to any nearby function with the same
critical value.  We omit the precise statement.}
\end{rmk}

The proof of Theorem \ref{isotopy} is based on the following
lemma.

\begin{lemma}\label{clever}  In the situation of Theorem
\ref{isotopy}, fix a Riemannian metric $g$ on $X$.  Let
$r : \cU \to \R_{\geq 0}$ be the distance-to-$S$ function.
Let $\alpha = (f,r): \cU \setminus S \to \R \times \R_+$.
Then there exist an open neighborhood $\cU' \subset \cU$ of $B$ and
number $k > 0$, such that $(p, \alpha) : \cU \setminus S \to
S \times \R \times \R_+$ is a stratified submersion on the set
$\{ x \in \cU' \setminus S \; | \; |f(x)| \leq k \cdot r(x) \}$.
\end{lemma}

\noindent
{\bf Proof:}  Suppose the lemma is false.  Let $\Sigma \subset
\cU \setminus S$ be the stratified critical locus of the map
$(p, \alpha)$.  Then there exists a sequence $\{ x_i \in \Sigma \}$,
converging to a point $b \in B$, such that
\begin{equation}\label{inclever}
\lim_{i \to \infty} \frac{f(x_i)}{r(x_i)} = 0.
\end{equation}
By passing if necessary to a subsequence, we can assume that all
the $\{ x_i \}$ lie in the same stratum $R$, and that there
exists a limit $\Delta = \lim T_{x_i} R \subset T_b X$.  But then,
combining equation (\ref{inclever}) with the Whitney conditions
for the pair $(S, R)$, we may conclude that the differential
$d_b f$ annihilates $\Delta$.  This, however, contradicts the
genericity assumption $d_b f \in \Lambda^0_S$.
\hfill$\Box$

\vspace{.1in}

\noindent
{\bf Proof of Theorem \ref{isotopy}:}  Without loss of generality, we
may assume that $X$ is the total space of a vector bundle $X \to M$,
that $S$ is the zero section, and that $f : \cU \to \R$ is the
restriction of a fiber-wise linear function $\tilde f : X \to \R$.
Fix a Euclidean structure in the bundle $X \to M$, and let
$r : X \to \R_{\geq 0}$ be the associated norm.
Let $\alpha = (f,r): \cU \setminus S \to \R \times \R_+$, and apply
Lemma \ref{clever} to produce a neighborhood $\cU' \subset \cU$ of
$B$ and a number $k > 0$.

{\it Step 1.}  Let $R_k = \{ (\xi, \eta) \in \R \times \R_+ \; |
\; |\xi| \leq k \cdot \eta \}$.  There exists a smooth,
$R_+$-equivariant function $\tilde\rho : \R \times \R_+ \to \R_+$
such that $\tilde\rho(\xi, \eta) = |\xi|$ for all $(\xi, \eta)
\notin R_k$.

{\it Step 2.}  Let $\rho : \cU' \setminus S \to \R_+$ be the
composition $\rho = \tilde\rho \circ \alpha$.  Then $\rho$ is a
quasi-distance function for $S$.

{\it Step 3.}  There exist control data $\{ U_T, \Pi_T, \rho_T \}$
on $\cU'$ such that $U_S = \cU'$, $\Pi_S = p|^{}_{\cU'}$,
$\rho_S = \rho$, and $\{ U_T, \Pi_T, \rho_T \}$ is
$\alpha$-compatible on some open set containing
$\cU' \cap \alpha^{-1} (R_k)$.

{\it Step 4.}  Every smooth vector field $V$ on $B$ extends to a
controlled vector field $\tilde V$ on $\cU' \cap p^{-1} (B)$,
compatible with $\{ U_T, \Pi_T, \rho_T \}$ and satisfying
$d \alpha (\tilde V_x) = 0$ for every $x \in \cU' \cap p^{-1} (B)
\cap \alpha^{-1} (R_k)$.

{\it Step 5.}  The homeomorphism $\phi$ is constructed by
integrating the $\tilde V_i$ for a suitable collection
$\{ V_i \}_{i=1}^s$ of vector fields on $B$.
\hfill$\Box$


text:
\begin{document}
\title[Isoperimetry and Symmetrization for Sobolev spaces]{Isoperimetry and Symmetrization for Sobolev spaces on metric spaces}
\author{Joaquim Martin$^{\ast}$}
\address{Department of Mathematics\\
Universitat Aut\`onoma de Barcelona}
\email{jmartin@mat.uab.cat}
\author{Mario Milman}
\address{Department of Mathematics\\
Florida Atlantic University}
\email{extrapol@bellsouth.net}
\urladdr{http://www.math.fau.edu/milman}
\thanks{2000 Mathematics Subject Classification Primary: 46E30, 26D10.}
\thanks{$^{\ast}$ Supported by Grants MTM2007-60500 and by 2005SGR00556.}

\begin{abstract} Using isoperimetry we obtain
new symmetrization inequalities that allow us to provide a unified framework
to study Sobolev inequalities in metric spaces. The applications include
concentration inequalities, as well as metric versions of the P\'{o}
lya-Szeg\"{o} and
Faber-Krahn principles.
\end{abstract}
\maketitle
\section{Introduction}

\label{}

This is a follow up to our recent work \cite{mami0}, where we obtained new
symmetrization inequalities for Sobolev functions that compare the
rearrangement of a function with the rearrangement of its gradient, and
incorporate in their formulation the isoperimetric profile (cf. (\ref{rea})
below). These inequalities imply in a straightforward fashion functional
inequalities for very general rearrangement invariant norms or quasi-norms
(e.g. $L^{p},$ Orlicz, Lorentz, Marcinkiewicz spaces). One remarkable
characteristic of these inequalities is that they preserve their form as we
move from one measure space to another, the only thing that changes are the
corresponding isoperimetric profiles. As a consequence we were able to provide
a unified framework to study the classical Sobolev-Poincar\'{e} inequalities,
logarithmic Sobolev inequalities, as well as concentration inequalities (cf.
\cite{LE} and the references therein). Importantly, if the isoperimetric
profile does not depend on the dimension (like in the Gaussian case) then the
corresponding inequalities are dimension free.

The purpose of this note is to outline the modifications that are necessary to
extend our earlier results to the setting of metric spaces. Indeed, under
relatively weak assumptions, all the tools that we need are available in the
metric setting (cf. \cite{BH}), and our methods can be readily adapted to
provide an almost painless extension. In particular, the results of this note,
when combined with the method developed\footnote{Our method builds on a
variant of Maz'ya's truncation principle, combined with the relevant
isoperimetric inequalities, the co-area formula and classical arguments from
real interpolation theory (cf. Calder\'{o}n \cite{CA}). We call this method to
obtain symmetrization inequalities ``symmetrization via truncation''.} in
\cite{mami0}, produce concentration inequalities in metric spaces, as well as
a Sobolev metric space version of the P\'{o}lya-Szeg\"{o} principle; while our
results combined with the method of [\cite{KM}, Theorem 3] imply metric
Faber-Krahn inequalities.

Let $\left(  \Omega,d,\mu\right)  $ be a metric space equipped with a
separable Borel probability measure $\mu$. For measurable functions
$u:\Omega\rightarrow\mathbb{R},$ the distribution function of $u$ is given by
$\lambda_{u}(t)=\mu(\{x\in{\Omega}:\left|  f(x)\right|  >t\}$ $(t>0)$, the
decreasing rearrangement $u^{\ast}$ of $u$ is defined, as usual, by $u^{\ast
}(s)=\inf\{t\geq0:\lambda_{u}(t)\leq s\}$ $(t\in(0,1)]),$ and we let
$u^{\ast\ast}(t)=\frac{1}{t}\int_{0}^{t}u^{\ast}(s)ds.$ For $A\subset\Omega,$
a Borel set, let $Per(A)=\lim\inf_{\varepsilon\rightarrow0}\frac{\mu\left(
A_{\varepsilon,d}\right)  -\mu\left(  A\right)  }{\varepsilon},$ where
$A_{\varepsilon,d}=\left\{  x\in\Omega:\exists y\in A\mbox{
}d(x,y)<\varepsilon\right\}  $ denotes the $\varepsilon-$extension of $A$ with
respect to the metric $d.$ An isoperimetric inequality measures the relation
between $Per(A)$ and $\mu(A)$ by means of the isoperimetric profile
$I=I_{(\Omega;d;\mu)}$, which is defined as the pointwise maximal function
$I:[0,1]\rightarrow\left[  0,\infty\right)  $, such that $Per(A)\geq
I(\mu(A)),$for all Borel sets $A$. Finally, in this setting for a given
Lipschitz function $f$ (we shall write in what follows $f\in Lip(\Omega)$) the
modulus of the gradient is defined, as usual, by $|\nabla f(x)|=\lim
\sup_{d(x,y)\rightarrow0}\frac{|f(x)-f(y)|}{d(x,y)}.$

\section{Main results}

\begin{theorem}
\label{teomain}Suppose that the isoperimetric profile $I$ is concave,
continuous, increasing on $(0,1/2)$ and symmetric about the point $1/2.$ Then
the following statements hold\footnote{except where indicated all
rearrangements are with respect to the measure $\mu$.} (and in fact are
equivalent):
\[
(i):\text{ }\forall\text{ }f\in Lip(\Omega),\text{ }\int_{0}^{\infty}
I(\lambda_{f}(s))ds\leq\int_{{\Omega}}\left|  \nabla f(x)\right|
d\mu(x)\text{ (Ledoux)}.
\]
\[
(ii):\text{ }\forall\text{ }f\in Lip(\Omega),\text{ }(-f^{\ast})^{\prime
}(s)I(s)\leq\frac{d}{ds}\int_{\{\left|  f\right|  >f^{\ast}(s)\}}\left|
\nabla f(x)\right|  d\mu(x)\text{ (Talenti-Maz'ya).}
\]
\[
(iii):\text{ }\forall\text{ }f\in Lip(\Omega),\text{ }\int_{0}^{t}((-f^{\ast
})^{\prime}(.)I(.))^{\ast}(s)ds\leq\int_{0}^{t}\left|  \nabla f\right|
^{\ast}(s)ds\text{ (P\'{o}lya-Szeg\"{o}).}
\]
(The second rearrangement on the left hand side is with respect to the
Lebesgue measure).
\begin{equation}
(iv):\text{ }\forall\text{ }f\in Lip(\Omega),\text{ }(f^{\ast\ast}(t)-f^{\ast
}(t))\leq\frac{t}{I(t)}\left|  \nabla f\right|  ^{\ast\ast}(t).\label{rea}
\end{equation}
\end{theorem}

Given any rearrangement invariant space\footnote{A Banach lattice of functions
$X(\Omega)$ is called a rearrangement invariant (r.i.) space (cf. \cite{BS})
if $g\in X(\Omega)$ implies that all functions $f$ \ with the same decreasing
rearrangement, $f^{\ast}=g^{\ast},$ also belong to $X(\Omega),$ and, moreover,
$\Vert f\Vert_{X(\Omega)}=\Vert g\Vert_{X(\Omega)}$. There is an essentially
unique r.i. space $\bar{X}(0,1)$ of functions on the interval $(0,1)$
consisting of all $g:(0,1)\rightarrow R$ such that $g^{\ast}(t)=f^{\ast}(t)$
for some function $f\in X(\Omega)$.} $X(\Omega)$, it follows readily from
(\ref{rea}) that for all Lip functions, we have
\begin{equation}
\left\|  f\right\|  _{LS(X)}:=\left\|  \left(  f^{\ast\ast}(t)-f^{\ast
}(t)\right)  \frac{I(t)}{t}\right\|  _{\bar{X}}\leq\left\|  \nabla f\right\|
_{X}.\label{corres}
\end{equation}

\begin{remark}
For $L^{1}$ norms these Poincar\'{e} inequalities are a simple variant of
Ledoux's inequality $(i)$. Indeed, let $m_{f}$ be a median\footnote{i.e.
$\mu\left(  f\geq m\right)  \geq1/2$ and $\mu\left(  f\leq m\right)  \geq
1/2$.} of $f$, then it is easy to see that
\begin{equation}
\int_{{\Omega}}\left|  f-m_{f}\right|  d\mu\leq\frac{1}{2I(1/2)}\int_{{\Omega
}}\left|  \nabla f(x)\right|  d\mu(x).\label{pr1}
\end{equation}
\end{remark}

The novelty of our inequalities, and the corresponding associated spaces
$LS(X),$ is that they incorporate the isoperimetric profiles associated with
the geometry in question. These spaces are not necessarily normed, although
often they are equivalent to normed spaces (cf. \cite{PU}), and, in the
classical cases, lead to optimal Sobolev-Poincar\'{e} inequalities (cf.
\cite{MMP}, \cite{mami0}, \cite{MM1} and the references therein).

We now investigate the optimality of the Poincar\'{e} type inequality
(\ref{corres}). The following result is new in the context of r.i. spaces.

\begin{theorem}
\label{optimal}Let $(\Omega,\mu)=(R^{n},\mu_{r}^{\otimes n}),$ with $\mu
_{r}(x)=$ $\varphi(x)dx,I_{\mu_{r}^{\otimes n}}(t)\approx\varphi
(F^{-1}(t)),\;t\in\left[  0,1\right]  $, where $F^{-1}$ is the inverse of the
distribution function associated to the density $\varphi(x)dx$\footnote{This
choice of $I$ is motivated by the results in \cite{Bo}, \cite{Bob} and
\cite{BFR}.}. Let $X(\Omega)$, $Y(\Omega)$ be r.i. spaces. Then, the following
statements are equivalent
\begin{equation}
(i):\text{ }\forall\text{ }f\in Lip(\Omega),\,\left\|  f-\int f\right\|
_{Y}\preceq\left\|  \nabla f\right\|  _{X}.\label{poin}
\end{equation}
\[
(ii):\text{ }\left\|  \int_{t}^{1}f(s)\frac{ds}{I(s)}\right\|  _{\bar{Y}
}\preceq\left\|  f\right\|  _{\bar{X}},\text{ }\forall\text{ }0\leq f\in
\bar{X},\text{with }supp(f)\subset(0,1/2).
\]
Moreover,

\begin{enumerate}
\item [(a)]If the operator $Q_{I\text{ }}f(t)=\frac{I(t)}{t}\int_{t}
^{1}f(s)\frac{ds}{I(s)}$ is bounded from $\bar{X}$ to $\bar{X},$ then the next
inequality can be added to the list of equivalences
\begin{equation}
(iii):\text{ }\left\|  f\right\|  _{\bar{Y}}\preceq\left\|  f^{\ast}
(t)\frac{I(t)}{t}\right\|  _{\bar{X}}.\label{perdida}
\end{equation}

\item[(b)] On the other hand if $Q_{I\text{ }}$ is not bounded from $\bar{X} $
to $\bar{X},\ $but $\left\|  f\right\|  _{X}\simeq\left\|  f^{\ast\ast
}\right\|  _{\bar{X}}$, then the next inequality can be added to the list of
equivalences
\begin{equation}
\left\|  f\right\|  _{\bar{Y}}\preceq\left\|  f\right\|  _{LS(X)}+\left\|
f\right\|  _{L^{1}.}\label{harhar}
\end{equation}
\end{enumerate}
\end{theorem}

As a concrete illustration\footnote{For further examples we refer to
\cite{BH}, \cite{MiE}, and the references therein.} consider the family of
probability measures on the real line given by $d\mu_{r}(t)=\alpha_{r}
^{-1}e^{-\left|  t\right|  ^{r}}dt=\varphi_{r}(t)dt$, $1<r\leq2,$ where
$\alpha_{r}^{-1}$ is chosen to ensure that $\mu_{r}(\mathbb{R)=}1.$ These
probabilities form a \ scale between exponential and Gaussian measure. The
associated isoperimetric profile is given by $I_{\mu_{r}}(t)=\varphi_{r}
(F_{r}^{-1}(t)),$ where $F_{r}^{-1}$ is the inverse of the distribution
function associated to the density $\varphi_{r}(t)$ (cf. \cite{Bov})$.$ The
isoperimetric profiles $I_{\mu_{r}^{\otimes n}},$ associated to the product
probability measures $\mu_{r}^{\otimes n},$ is dimension free (see
\cite{BFR}): there is a universal constant $c(r)$ such that $I_{\mu_{r}
}(t)\geq\inf_{n\geq1}I_{\mu_{r}^{\otimes n}}(t)\geq c(r)I_{\mu_{r}}(t)$. As an
application let $n\geq2$, and apply Theorem \ref{optimal} to $X=L^{p}
(\mathbb{R}^{n},d\mu_{r}^{\otimes n}),$ $1\leq p<\infty,$ then (cf. also
\cite[Theorem 3]{MM1}),
\[
\int_{0}^{1}\left(  \left(  f-\int f\right)  ^{\ast}(s)\frac{I_{\mu_{r}}
(s)}{s}\right)  ^{p}ds\preceq\int_{\mathbb{R}^{n}}\left|  \nabla f(x)\right|
^{p}d\mu_{r}^{\otimes n}(x),
\]
with dimension free constants. In particular, since (see \cite[Lemma
16.1]{Bov}) $\lim_{t\rightarrow0^{+}}\frac{I_{\mu_{r}}(t)}{t\left(  \log
\frac{1}{t}\right)  ^{1/q}}=r$ , $1/r+1/q=1$, it follows easily that
\[
\int_{0}^{1}f^{\ast}(s)^{p}(\log\frac{1}{s})^{p/q}ds\preceq\int_{\mathbb{R}
^{n}}\left|  \nabla f(x)\right|  ^{p}d\mu_{r}^{\otimes n}(x)+\int
_{\mathbb{R}^{n}}\left|  f(x)\right|  ^{p}d\mu_{r}^{\otimes n}(x).
\]
Moreover, for this class of measures, $L^{p}(LogL)^{p/q}$ is the best possible
choice among all r.i. spaces $Y$ for which the inequality $\Vert f-\int
f\Vert_{Y}\preceq\Vert\nabla f\Vert_{L^{p}}$ holds. If $p=\infty,$ we have
\begin{equation}
\Vert f-\int f\Vert_{LS(L^{\infty})}=\left\|  \left(  \left(  f-\int f\right)
^{\ast\ast}(t)-\left(  f-\int f\right)  ^{\ast}(t)\right)  \frac{I_{\mu_{r}
}(t)}{t}\right\|  _{L^{\infty}}\preceq\left\|  \nabla f\right\|  _{L^{\infty}
}.\label{concen}
\end{equation}
The relation to concentration inequalities follows directly from our main
inequality. Indeed, we have
\[
\sup_{t<1}\left\{  (f^{\ast\ast}(t)-f^{\ast}(t))\frac{I_{\mu_{r}}(t)}
{t}\right\}  \leq\sup_{t}\left|  \nabla f\right|  ^{\ast\ast}(t)=\left\|
f\right\|  _{Lip},
\]
which, by the asymptotic properties of $I_{\mu_{r}},$ implies that
$f^{\ast\ast}(t)-f^{\ast}(t)\preceq\frac{\left\|  f\right\|  _{Lip}}{\left(
\log\frac{1}{t}\right)  ^{1/q}}$ ($0<t<1/2$). We may now proceed as in
[\cite{mami0}, Section 7].

Let us finally consider Sobolev embeddings into $L^{\infty}.$ Notice that from
inequality (\ref{rea}) we get
\[
\left\|  f\right\|  _{\infty}-2\int_{0}^{1/2}f^{\ast}(t)=\int_{0}
^{1/2}(f^{\ast\ast}(t)-f^{\ast}(t))\frac{dt}{t}\leq\int_{0}^{1/2}\left(
\frac{1}{t}\int_{0}^{t}\left|  \nabla f\right|  ^{\ast}(s)ds\right)  \frac
{dt}{I_{\mu_{r}}(t)}=\int_{0}^{1/2}\left|  \nabla f\right|  ^{\ast}(s)\int
_{s}^{1/2}\frac{ds}{I_{\mu_{r}}(s)s}.
\]
Using the asymptotics of $I_{\mu_{r}}(s)$ combined with the Poincar\'{e}
inequality \ref{pr1} yields
\[
\left\|  f-m_{f}\right\|  _{\infty}\preceq\int_{0}^{1/2}\left|  \nabla
f\right|  ^{\ast}(s)\frac{ds}{s\left(  \log\frac{1}{s}\right)  ^{1/q}}.
\]

\section*{Acknowledgements}

We are very grateful to the referee for suggestions to improve the
presentation.The first named author has been partially supported by Grants
MTM2007-60500, MTM2008-05561-C02-02 and by 2005SGR00556.


TITLE: Help with a problem about Artinian rings from Christian Peskine's book
QUESTION [9 upvotes]: I am stuck with problem 3, chapter 4, from the book of Peskine, An Algebraic Introduction to Complex Projective Geometry.

Let $A$ be a Noetherian ring. Assume that if $a \in A$ is neither invertible nor nilpotent, then there exist $b \in A$ such that $b$ is not nilpotent and that $ab = 0$. Show that $A$ is Artinian.

I have tried to prove that every prime is also maximal since I already have the Noetherian property and I can't see how to prove directly the dcc.

REPLY [5 votes]: Here is my answer. Assume that $A$ is not artinian. Then there exists a maximal ideal not contained in the union of all minimal primes. Hence
there exists an element $a$ not invertible and not contained in any minimal prime.
Consequently if $ab=0$ then $b$ is contained in all minimal primes and is therefore nilpotent, contradicting the hypothesis.


TITLE: Finding general solution for a nonhomogeneous system of equations
QUESTION [1 upvotes]: I have a system of differential equations:
$\begin{cases} x_1'=x_2+2e^t \\
 x_2'=x_1+t^2 \end{cases}$
And I want to find the general solution for it.
I started by finding the general solution for the homogeneous equations:
$\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=C_1\begin{pmatrix}e^t+e^{-t} \\ e^t-e^{-t}\end{pmatrix}+C_2\begin{pmatrix}e^t-e^{-t} \\ e^t+e^{-t}\end{pmatrix}$.
Now I need to find a "specific" solution for the nonhomogeneous equations but I have problems applying the method in which I make constants $C_1$ and $C_2$ a variable. Could anyone help me with this part?

REPLY [0 votes]: \begin{align*}
x_1&=C_1\cosh t+C_2\sinh t+te^t+\frac{e^t}{2}-t^2-2\\
x_2&=C_1\sinh t+C_2\cosh t+te^t-\frac{e^t}{2}-2t
\end{align*}


TITLE: Locus of vertex
QUESTION [0 upvotes]: A variable parabola of latus rectum $l $, touches a fixed equal parabola , the axes of the two curves being parallel . Then locus of the vertex of  the moving curve is a parabola , then what is the latus rectum of this curve .
I took the fixed curve as $y^2 =4bx$
And the variable curve is $(y-k)^2 = 4a(x-h)$
Where $(h,k)$ is the vertex . But now how to proceed ?

REPLY [1 votes]: I like projective geometry, so I'll be explaining using terms from there. Your first parabola can be written as
$$(x,y,1)\cdot\begin{pmatrix}0&0&2b\\0&-1&0\\2b&0&0\end{pmatrix}
\cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$
Its axis is the $x$ axis, evidenced by the fact that $(1,0,0)$ is the only point at infinity (i.e. last coordinate zero) which lies on that conic as well. Now I'd like to describe the tangent for a given direction. A direction can be described by a point at infinity. I'll use the point $(t,1,0)$ with parameter $t$. For $t\in\mathbb R$ this can describe any direction except horizontal, which is just fine since that parabola can't have a horizontal tangent in any case.
So we need the tangent through a given point. One way to compute that is by computing the polar line of that point. It will intersect the conic in the two contact points. Usually that leads to two tangents, but in our case, with the point chosen at infinity, one of the tangents will be the line at infinity, so we can pick the remaining (finite) contact point.
To compute a polar line, multiply the point with the matrix:
$$\begin{pmatrix}0&0&2b\\0&-1&0\\2b&0&0\end{pmatrix}
\cdot\begin{pmatrix}t\\1\\0\end{pmatrix}=
\begin{pmatrix}0\\-1\\2bt\end{pmatrix}$$
This is the line $y=2bt$. It intersects the conic at $x=\frac{y^2}{4b}=bt^2$. So $(bt^2,2bt)$ is the point with the given tangent direction. Now do the same for the second curve with $a$ instead of $b$. Setting the point of contact equal, you get
\begin{align*}
bt^2 &= at^2+h \\
2bt &= 2at+k
\end{align*}
You can eliminate the variable $t$ from this, e.g. using the resultant computed using the Sylvester matrix
$$0=\begin{vmatrix}
a-b & 0 & h \\
2(a-b) & k & 0 \\
0 & 2(a-b) & k
\end{vmatrix}=(a-b)(k^2 + 4ah - 4bh)$$
So the equation of the locus of the vertex is
$$k^2 = 4(b-a)h$$