The Heat Kernel $H_t(x)$ is a good kernel for $t \to 0$, understanding proof from Stein's Fourier Analysis

Question

I am covering Stein's Fourier Analysis, and I've stumbled on this proof that I have trouble understanding.

We are trying to show that $H_t(x) = \sum_{n\in Z}e^{-4\pi^2n^2t}e^{2\pi inx}$ is a good kernel, so we have to show:

1. $\int_{|x| \leq 1/2} H_t(x) dx = 1 $,
2. $\int_{|x| \leq 1/2} |H_t(x)|dx \leq M $
3. For every $\delta > 0 $, $\int_{\delta < |x| < 1/2} |H_t(x)|dx \rightarrow 0$ as $t \rightarrow 0 $.

Now I understand that the heat kernel on the circle is the periodization of the heat kernel on the real line, and that heat kernel on the real line is a good kernel.

I am stuck on showing that the condition i) holds. I've tried integrating $H_t(x) = \sum_{n\in Z}e^{-4\pi^2n^2t}e^{2\pi inx}$, but that didn't lead me anywhere.

In the textbook, the proof says that we can write $H_t(x) = H_{t,R}(x) + \epsilon_t(x)$, where $|\epsilon_t(x)| \leq c_1e^{-c_2/t}$ with $c_1, c_2 > 0$ and $t < 0 \leq 1$.

Now I understand that we obtain the first part by the periodization from which it follows that $\epsilon_t(x) = \frac{1}{\sqrt{4\pi t}}\sum_{|n| \geq 1}e^{-(x+n)^2/4t}$, however I do not understand how we get the estimate in the next line $\leq Ct^{-1/2}\sum_{n \geq 1} e^{-cn^2/t}$.

Then, in the line of the proof, it concludes that $\int_{\delta < |x| < 1/2} |H_t(x)|dx \rightarrow 0$ as $t \rightarrow 0 $, because $H_{t,R}$ does. How can we conclude this?

Any help would be appreciated, thank you

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Proof in the textbook

Answer 1

• Condition (1) is trivial, since $$\int_{|x|\leq \frac 1 2}H_t(x)dx = \sum_{n\in Z}e^{-4\pi^2n^2t}\int_{|x|\leq \frac 1 2}e^{2\pi inx}dx=\sum_{n\in Z}e^{-4\pi^2n^2t}\delta_{n=0}$$

• Condition (2): As you noticed, $H_t$ is obtained by periodizing the Gaussian kernel $$\mathcal H_t(x) = \frac{e^{-u^2/(4t)}}{\sqrt{4\pi t}}$$ Indeed, the periodized Gaussian kernel is a periodic function (duh!) whose $n$-th Fourier coefficient is $$\begin{split} \int_{|x|<\frac 1 2} \left ( \sum_{k\in\mathbb Z} \mathcal H_t(x+k)\right)e^{-2i\pi nx}dx &= \sum_{k\in\mathbb Z}\int_{|x|<\frac 1 2} \mathcal H_t(x+k)e^{-2i\pi nx}dx\\ &\stackrel{u=x+k}{=}\sum_{k\in\mathbb Z}\int_{k}^{k+1}\mathcal H_t(u)e^{-2i\pi nu}du\\ &= \int_{-\infty}^{+\infty}\mathcal H_t(u)e^{-2i\pi nu}du\\ &= \widehat{\mathcal H}_t(n) \end{split}$$ Consequently, writing the periodized heat kernel as a Fourier series, we get $$\sum_{k\in\mathbb Z} \mathcal H_t(x+k) = \sum_{n\in\mathbb Z} \widehat{\mathcal H}_t(n)e^{2i\pi nx}$$ Note that this is also known as the Poisson summation formula. Substituting the expression of the Fourier transform of $\mathcal H_t$, i.e. $\widehat{\mathcal H}_t(\xi)= e^{-4\pi^2\xi^2 t}$, we obtain $$\sum_{k\in\mathbb Z} \mathcal H_t(x+k)=H_t(x)\tag{Poisson}$$ It immediately follows from $\mathcal H_t(x+k)>0$ that $H_t(x)>0$, and condition (2) is verified with $M=1$.

• Condition (3): Rewrite the Poisson formula as $$H_t(x) = \mathcal H_t(x) + \sum_{|k|\geq 1}\mathcal H_t(x+k)$$ and notice that if $|x|<\frac 1 2$ and $|k|\geq 1$, $$\left|\mathcal H_t(x+k)\right|=\frac{e^{-(x+k)^2/(4t)}}{\sqrt{4\pi t}} \leq \frac{e^{-(|k|-\frac 1 2)^2/(4t)}}{\sqrt{4\pi t}}$$ Now, and this is where you were puzzled: Notice that $|k|-\frac 12 \geq \frac {|k|} 2$ and therefore $$\left|\mathcal H_t(x+k)\right|\leq \frac{e^{-|k|^2/(16t)}}{\sqrt{4\pi t}}=Ct^{-\frac 1 2} e^{-ck^2/t}$$ Now, the rest of the proof is identical: for $0<t<1$, we have $\frac {k^2} t \geq \frac 1 2 \left (k^2 + \frac 1 t\right)$, thus $$\left|\mathcal H_t(x+k)\right| \leq Ct^{-\frac 1 2} e^{-ck^2/2}e^{-c/(2t)}$$ and $$0\leq \sum_{|k|\geq 1}\mathcal H_t(x+k) \leq Ct^{-\frac 1 2}e^{-c/(2t)}\sum_{|k|\geq 1}e^{-ck^2/2} \leq c_1 e^{-c_2/t}$$ and it follows that for any $\delta>0$, as $t\rightarrow 0$ $$\int_{\delta<x<\frac 1 2}|H_t(x)-\mathcal H_t(x)|dx \rightarrow 0$$ Condition (3) is then clear since it is easy to verify that it also applies to $\mathcal H_t$ itself.