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Let $G(t,x)$ be the heat kernel $$ G(t,x)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}, \quad t>0, \:x\in\mathbb{R}. $$

Here is one approximation to $G(t,x)$:

$$ G_\epsilon(t,x)=e^{-t/\epsilon} \sum_{k=1}^\infty \left(\frac{t}{\epsilon}\right)^k \frac{1}{k!} G(k\epsilon,x). $$

The question is: can one show that for some $a>0$ and $C>0$

$$ \int_{\mathbb{R}}\left|G(t,x)-G_\epsilon(t,x)\right| d x \le e^{-t/\epsilon}+C \left(\frac{\epsilon}{t}\right)^{1/3},\quad \text{for $0<\epsilon/t\le a$}? $$

Thanks a lot for any hints!

-----EDIT------

Thanks Professor Lucia for his nice solution. It turns out that it is not sufficient for what we actually need. Here is the revised question:

Fix $\epsilon>0$. Is there a constant $C>0$ and $0<\beta<1/2$, such that

$$ \int_{\mathbb{R}}\left|G(t,x)-G_\epsilon(t,x)\right| d x \le e^{-t/\epsilon}+C \left(\frac{\epsilon}{t}\right)^{\beta},\quad \text{for all $t>0$}? $$

The exponent $1/3$ is mysterious. But any order $\beta<1/2$ will be fine. Or probably this will never happen?

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    $\begingroup$ As I pointed out in a comment below, this does follow from what I wrote. For $t/\epsilon >1$ use the earlier solution, which provides a stronger bound. For $t/\epsilon \le 1$, just use display (1) from my answer which then produces a bound of $e^{-t/\epsilon} + C (t/\epsilon) (1+|\log (t/\epsilon)|)$ which again is stronger than what you want. $\endgroup$ – Lucia Apr 22 '14 at 21:19
  • $\begingroup$ Thanks professor Lucia for your help. We would like to give some credits to your help in our research paper. We could refer your account at mathoverflow, or we may refer to your real name. In the later case, you may send me an email at anandmathoverflow@gmail.com. Thanks a lot! $\endgroup$ – Anand Apr 22 '14 at 22:40
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One can prove a stronger estimate in fact. Suppose $u$ and $v$ are positive with $v>u$ say. Note that $$ |G(u,x)-G(v,x)| \le \int_{u}^v \Big| \frac{d}{dt} G(t,x)\Big| dt =\int_{u}^{v} \frac{e^{-x^2/2t}}{\sqrt{2\pi t}} \Big|\frac{x^2}{2t^2}-\frac{1}{2t}\Big| dt. $$ Integrating this over $x\in {\Bbb R}$ we obtain $$ \int_{-\infty}^{\infty} |G(u,x)-G(v,x)| dx \le \int_u^v \frac{1}{2t} \Big(\frac{1}{\sqrt{2\pi t}} \int_{-\infty}^{\infty} e^{-x^2/(2t)} \Big(\frac{x^2}{t}+1\Big) dx \Big) dt \le C \int_{u}^{v} \frac{dt}{t} = C \log \Big(\frac{v}{u}\Big), $$ for some constant $C$.

Now we use this in the problem; assume throughout that $t/\epsilon$ is bounded away from zero, say it is at least $1$. We have (the first term accounts for the missing $k=0$ term) $$ |G(t,x) - G_{\epsilon}(t,x)| \le e^{-t/\epsilon} G(t,x) + e^{-t/\epsilon}\sum_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} |G(t,x)-G(k\epsilon, x)|. $$ Integrating both sides over $x \in {\Bbb R}$ and using our first estimate we get $$ \int_{{\Bbb R}} |G(t,x)-G_{\epsilon}(t,x)| dx \le e^{-t/\epsilon} + Ce^{-t/\epsilon}\sum_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} |\log (k\epsilon/t)|. \tag{1} $$

It remains lastly to estimate the sum over $k$ above. Note that $|\log (v/u)| \le |u-v|/(\min(u,v)) \le |u-v|(1/u+1/v)$. So the sum over $k$ is $$ \le \sum_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} |k - t/\epsilon|\Big( \frac{1}{k} +\frac{\epsilon}{t}\Big). $$ Using Cauchy-Schwarz $$ \sum_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} \frac{1}{k}|k-t/\epsilon| \le \Big(\sum_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^{k} \frac{1}{k!} \frac{1}{k^2}\Big)^{\frac 12} \Big( \sum_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} (k-t/\epsilon)^2\Big)^{\frac 12}, $$
and it is easy to see that the first factor above is $O((\epsilon/t)e^{t/(2\epsilon)})$ and the second factor is $O((t/\epsilon)^{\frac 12} e^{t/(2\epsilon)})$ so that our quantity above is $O((\epsilon/t)^{\frac 12} e^{t/\epsilon})$. Similarly $$ \sum_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} \frac{\epsilon}{t} |k-t/\epsilon| = O\Big( \Big(\frac{\epsilon}{t}\Big)^{\frac 12} e^{t/\epsilon}\Big). $$ Using these estimates in (1) we get $$ |G(t,x)-G_{\epsilon}(t,x)| \le e^{-t/\epsilon} + C_1 \Big(\frac{\epsilon}{t}\Big)^{\frac 12}, $$ which is stronger than the bound you wanted.

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    $\begingroup$ Dear Professor Lucia, thank you so much for your constant help! Your solution is amazing. This question is from an old research paper without any proof. Since we want to make some extension, we should first know how to prove the Gaussian case. By the way, your constant $C$ can be chosen as $1$. Thanks a lot! $\endgroup$ – Anand Apr 19 '14 at 19:27
  • $\begingroup$ Dear Professor Lucia, actually the exponent $1/2$ instead of $1/3$ is too bad. We do need exponent less than $1/2$ because we need to integrate the square of the upper bound over $t$ around zero with $\epsilon$ fixed. Do you think this is possible? Thank you very much! $\endgroup$ – Anand Apr 22 '14 at 16:37
  • $\begingroup$ But $1/2$ is better than $1/3$! You are assuming that $\epsilon/t$ is small, and so $(\epsilon/t)^{1/2} \le C (\epsilon/t)^{1/3}$ for some constant $C$. $\endgroup$ – Lucia Apr 22 '14 at 16:42
  • $\begingroup$ Thanks Professor Lucia, you are right. But the property that we actually need is that for fixed $\epsilon$, the square of this upper bound is integrable at $t=0$. That's why in the old research paper, the exponent is less than $1/2$. Another comment is that one need to get rid of the requirement that $\epsilon/t \le a$ in order to integrate the square of the upper bound at $t=0$. This is doable when the exponent is $1/2$. Thanks a lot for your help! $\endgroup$ – Anand Apr 22 '14 at 19:34
  • $\begingroup$ If you are interested in $t/\epsilon$ small, then just use (1) from above, and note that only the term $k=1$ there is important. This will give a bound of $O(1)$ in the range $t/\epsilon$ small, which again is better than what you want. $\endgroup$ – Lucia Apr 22 '14 at 19:38

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