The lemma is from chapter 4.5 in PDE by Evans, I don't really understand the proof and have questions about it.
Lemma(Asymptotics). Suppose that $k,l:\mathbb{R}\to\mathbb{R}$ are continuous functions, that $l$ grows at most linearly and that $k$ grows at least quadratically. Assume also there exists a unique point $y_0\in\mathbb{R}$ s.t. $$ k(y_0)=\min_{y\in\mathbb{R}}k(y).$$ Then $$\lim_{\epsilon\to 0}\frac{\int_{-\infty}^{\infty} l(y)e^{\frac{-k(y)}{\epsilon}}dy}{\int_{-\infty}^{\infty}e^{\frac{-k(y)}{\epsilon}}dy}=l(y_0).$$
Proof. Write $k_0=k(y_0)$. Then the function $$\mu_\epsilon(y):=\frac{e^{\frac{k_0-k(y)}{\epsilon}}}{\int_{-\infty}^{\infty} e^{\frac{k_0-k(z)}{\epsilon}}dz}\quad\quad (y\in\mathbb{R})$$ satisfies $$\mu_\epsilon\geq 0,\int_{-\infty}^{\infty}\mu_\epsilon(y)dy=1,$$
$$\mu_\epsilon\to 0 \text{ exponentially fast for } y\neq y_0,\text{ as }\epsilon\to 0.$$
Consequently, $$\lim_{\epsilon\to 0}\frac{\int_{-\infty}^{\infty} l(y)e^{\frac{-k(y)}{\epsilon}}dy}{\int_{-\infty}^{\infty}e^{\frac{-k(y)}{\epsilon}}dy}=\lim_{\epsilon\to 0}\int_{-\infty}^{\infty}l(y)\mu_\epsilon(y)dy=l(y_0).$$
Now I am wondering:
- How to estimate the decay speed of $\mu_\epsilon$ as $\epsilon\to 0$ or more generally, the decay speed of integral in this form $$\int_{-\infty}^{\infty} e^{(k_0-k(y))/\epsilon}.$$
- Why "consequently" we get the desired result. In my opinion, we should try to prove that $\mu_\epsilon$ is a good kernel, i.e. it's positive, its integral is 1, and for every $R>0$, $\lim_{\epsilon\to 0}\int_{|y|>R}\mu_\epsilon(y)dy=0$.
Can anyone help me with my problems? Thanks!
Best Answer
I will answer (1). For question 2, the positivity and the fact that the integral is 1 follows from the positivity of the exponential the definition you wrote down and the fact that the in the definition is finite: since $k$ grows at least quadratically, there exists a positive constant $C$ and some real number $D$ such that $$k(y) \geq k_0 + D + C(y-y_0)^2.$$
(Use a compactness argument if you want to make this rigorous.)
Note that this means $$\int_{-\infty}^\infty e^{(k_0 - k(z))/\epsilon}dz \leq \int_{-\infty}^\infty e^{-D -C(z-y_0)^2/\epsilon}dz < \infty.$$ Appropriately, this gives us positivity and the integral of $\mu_\epsilon$ being 1:
$$\int_{-\infty}^\infty \mu_\epsilon(y)dy := \int_{-\infty}^\infty \frac{e^{(k_0 - k(y))/\epsilon}}{\int_{-\infty}^\infty e^{(k_0 - k(z))/\epsilon}dz }dy = 1.$$
Now, we aim to prove the exponential decay statement that Evan gives, where we could use the quadratic growth of $k$ along with Laplace's method, but we will use the following coarse bound instead.
Fix $y \neq y_0$. Take $\epsilon > 0$. Then by the inequality we gave above, \begin{align*} \mu_{\epsilon}(y) &= \frac{e^{(k_0 - k(y))/\epsilon}}{\int_{-\infty}^\infty e^{(k_0 - k(z))/\epsilon}dz} \\ &\leq \frac{e^{(k_0 - k(y))/\epsilon}}{\int_{y_0 - \delta}^{y_0 + \delta} e^{(k_0 - k(z))/\epsilon}dz} \\ &= \frac{1}{\int_{y_0 - \delta}^{y_0 + \delta} e^{(k(y) - k(z))/\epsilon}dz} \end{align*}
Since $y_0$ is the unique minimum of $k$, $k$ is continuous, and $y \neq y_0$, we can pick $\delta$ small enough such that for all $z$ such that $|z - y_0| < \delta$, $$k(z) \leq (k(y) - k(y_0))/2 = \eta > 0.$$ This $\eta$ will depend on $y$ and $\delta$, but for our purposes this doesn't matter. We end up with $$\mu_\epsilon(y) \leq \frac{1}{\int_{y_0 - \delta}^{y_0 + \delta} e^{\eta/\epsilon}dz}$$
which exhibits the exponential decay that Evan's said it had. In particular, we have shown that outside of a neighborhood $(y_0-\delta, y_0 + \delta)$ of $y_0$, $$\mu_\epsilon(y) \leq \frac{e^{-\eta/\epsilon}}{2\delta}.$$
From this point, it isn't hard to show that $\mu_\epsilon$ satisfies the properties of a good kernel using a proof similar to what I did above (after fixing $R > 0$, make $\eta = \min_{|x| \geq R} \frac{k(x) - k(y_0)}{2}.$)