# Thread: Scaling Estimate on the Heat Kernel

1. ## Scaling Estimate on the Heat Kernel

I posted this question here differential equations - Some Scaling Estimate for Heat Kernel - Mathematics Stack Exchange, but still haven't received a response. It contains the description and some of my attempted progress.

Much appreciation to anyone that can help me prove this! I need the result for some further work I'm doing.

2. ## Re: Scaling Estimate on the Heat Kernel

Off the top of my head, I'd say this is an application of the Moser-Trundinger inequality. It's going to take some work.

3. ## Re: Scaling Estimate on the Heat Kernel

Originally Posted by Rebesques
Off the top of my head, I'd say this is an application of the Moser-Trundinger inequality. It's going to take some work.

scratch that.

We are asked to produce constants $a,\beta$ such that $\sup_{x\in\mathbb{R}^d,y\in B(0,a)}\frac{G(x+y,t)}{G(x,\beta t)}<\infty$.

This will be true if $\int_{\mathbb{R}^d} \frac{G(x+y,t)}{G(x,\beta t)} dx$ converges.

Bound the integral as (forgive me for not bothering to write down the constant $D$ exactly)

$D\int_{\mathbb{R}^d} \exp\left(\frac{-\beta+1}{4t\beta} \left\{|x+y|^2-|x|^2\right\}\right) dx \\ \leq \int_{\mathbb{R}^d} \exp\left(\frac{-\beta+1}{4t\beta} \left\{|y|^2-2x\cdot y\right\}\right) dx=J$

and use the fact that $|y|\leq a$ to obtain

$J\leq I(a)=\int_{\mathbb{R}^d} \exp\left(\frac{-\beta+1}{4t\beta} \left\{|a|^2-2x\cdot y\right\}\right) dx$.

The Moser-Trundinger inequality states that $I(a)<\infty$ for values $a^2\frac{-\beta+1}{4t\beta} \leq 4\pi$, which readily gives $a\leq C\sqrt{t}$.

Now, the second part is not trivial.

4. ## Re: Scaling Estimate on the Heat Kernel

And actually, by demanding $\beta\geq1$, we can replace the finishing calculations by

$\frac{a^2}{4\beta t}\leq 4\pi\Rightarrow\ldots$.

Now, the latter part is messy - and quite interesting at that. I'll post it in smaller parts in the future.

5. ## Re: Scaling Estimate on the Heat Kernel

Rebesques,

Thanks for the resposne! I was starting to give up hope that anyone would produce a solution and put a bounty up on MSE. Do you have an account there where you could put this answer? If not, I'll do it on your behalf.

Anyway, I haven't had a chance to go through your proof yet as I just noticed your response, but I wanted to respond with my thanks ASAP and I look forward to what you come up with for the second part (I've been making some progress myself, and I will try to update this if it gets to a significant point).

Some preliminary observations though. It seems the sup should be written as

$\sup_{x\in\mathbb{R}^{d},t>0,y\in B(0,\alpha\sqrt{t}).$

The inclusion of the $t>0$ is supposed to indicate that $t$ is allowed to freely vary just as $x$ is. I'm assuming your proof probably already makes this so, even if wasn't explicitly written into the $\sup$. Also, I find it interesting that you supposed more generally that $y$ is restricted to $y\in B(0,a)$, and then proceded to prove necessarily $a=\alpha\sqrt{t}$ (you used $C$ instead of $\alpha$ and got rid of the original $C$ in the problem statement by replacing the problem with a $\sup$ of the indicated quotient). Like I said, I haven't gone through it yet, so I'll reserve any additional comments until then. Thanks again!

6. ## Re: Scaling Estimate on the Heat Kernel

I think there's something wrong here. Since $t$ can vary with $x$, I don't think any conclusion can be made by performing an integral over $\mathbb{R}^{d}$ since this excludes the role of $t$. We're trying to produce the constant $C$ that you eliminated from consideration by replacing the original problem with a $\sup$. In particular, we are trying to quantify the sense in which the quotient is finite. The constant $a$ that you computed is given a priori. In other words, given $\alpha>0$, compute constants $\beta=\beta(\alpha),C=C(\alpha)$ such that

$\frac{G(x+y,t)}{G(x,\beta t)}\leq C$

holds for all $x\in\mathbb{R}^{d},t>0,y\in B(0,\alpha\sqrt{t})$. The quantifiers of all the parameters for this entire problem (especially part (b)) are quite a headache if you ask me. Considering how confused I am, for all I know your solution actually does prove the estimate with the proper quantifiers.

One more thing, all other concerns above aside, I don't know how you get from convergence of the integral to the indicated pointwise estimate since $L^{1}(\mathbb{R}^{d})$ is not a subset of $L^{\infty}(\mathbb{R}^{d})$.
In particular, $\int_{\mathbb{R}^{d}}f(x)\;d\mu<\infty$ doesn't imply $\sup_{x\in\mathbb{R}^{d}}f(x)<\infty,$).

7. ## Re: Scaling Estimate on the Heat Kernel

One point at a time:

(1) t is taken as positive and constant throughout the calculations (the problem's requirement only comes as a pointwise inequality on t, and in fact that cannot change)

(2) The constants $a,\beta$ are taken as parameters and are treated algebraically all the way up to the last step - where they must be suitably chosen, so that the M-T inequality can be applied.

(3) Actually $G(\cdot,t)/G(\cdot,\beta t)\in C^{\infty}(\mathbb{R}^d)$, so we have the assertion at hand.

ps. I think i do have an account on MSE, but had no idea you can put up a trophy for a question. Oh well.

8. ## Re: Scaling Estimate on the Heat Kernel

Thanks Rebesques -- especially for pointing out the bit about $\int f$ implying $\sup f<\infty$ when $f$ is continuous everywhere!

Look forward to your solution to part (b). It seems to me that that one could apply the estimate (incidentally proved in a previous problem on the present assignment I am working on)

$\sup_{\epsilon>0}|\phi_{\epsilon}*f|(x) where $A=\int_{\mathbb{R}^{d}}\psi(x)\;dx$ and $\psi(x)=\sup_{y:|y|>|x|}|\phi(y)|$.

Since (I think) $G$ is equal to its least decreasing radial majorant (i.e. $\psi$ in the definition above), and $G(\cdot,t)$ plays the role of $\phi_{\epsilon}(x)=\phi(x/\epsilon)\epsilon^{-d}$, we might be able to transfer the proof onto $A(Mf)(x),$ which may be easier to work with than $G(\cdot,t)*f$. This just came to mind as I was responding to the above, so I haven't really looked into it yet.