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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.
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.
Quote:
Originally Posted by Rebesques
scratch that.
We are asked to produce constants $\displaystyle a,\beta$ such that $\displaystyle \sup_{x\in\mathbb{R}^d,y\in B(0,a)}\frac{G(x+y,t)}{G(x,\beta t)}<\infty$.
This will be true if $\displaystyle \int_{\mathbb{R}^d} \frac{G(x+y,t)}{G(x,\beta t)} dx$ converges.
Bound the integral as (forgive me for not bothering to write down the constant $\displaystyle D$ exactly)
$\displaystyle D\int_{\mathbb{R}^d} \exp\left(\frac{-\beta+1}{4t\beta} \left\{|x+y|^2-|x|^2\right\}\right) dx \\ \leq \int_{\mathbb{R}^d} \exp\left(\frac{-\beta+1}{4t\beta} \left\{|y|^2-2x\cdot y\right\}\right) dx=J$
and use the fact that $\displaystyle |y|\leq a$ to obtain
$\displaystyle J\leq I(a)=\int_{\mathbb{R}^d} \exp\left(\frac{-\beta+1}{4t\beta} \left\{|a|^2-2x\cdot y\right\}\right) dx$.
The Moser-Trundinger inequality states that $\displaystyle I(a)<\infty$ for values $\displaystyle a^2\frac{-\beta+1}{4t\beta} \leq 4\pi$, which readily gives $\displaystyle a\leq C\sqrt{t}$.
Now, the second part is not trivial.
And actually, by demanding $\displaystyle \beta\geq1$, we can replace the finishing calculations by
$\displaystyle \frac{a^2}{4\beta t}\leq 4\pi\Rightarrow\ldots$.
Now, the latter part is messy - and quite interesting at that. I'll post it in smaller parts in the future.
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
$\displaystyle \sup_{x\in\mathbb{R}^{d},t>0,y\in B(0,\alpha\sqrt{t}).$
The inclusion of the $\displaystyle t>0$ is supposed to indicate that $\displaystyle t $ is allowed to freely vary just as $\displaystyle x$ is. I'm assuming your proof probably already makes this so, even if wasn't explicitly written into the $\displaystyle \sup$. Also, I find it interesting that you supposed more generally that $\displaystyle y$ is restricted to $\displaystyle y\in B(0,a)$, and then proceded to prove necessarily $\displaystyle a=\alpha\sqrt{t}$ (you used $\displaystyle C$ instead of $\displaystyle \alpha$ and got rid of the original $\displaystyle C$ in the problem statement by replacing the problem with a $\displaystyle \sup$ 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!
I think there's something wrong here. Since $\displaystyle t$ can vary with $\displaystyle x$, I don't think any conclusion can be made by performing an integral over $\displaystyle \mathbb{R}^{d}$ since this excludes the role of $\displaystyle t$. We're trying to produce the constant $\displaystyle C$ that you eliminated from consideration by replacing the original problem with a $\displaystyle \sup$. In particular, we are trying to quantify the sense in which the quotient is finite. The constant $\displaystyle a$ that you computed is given a priori. In other words, given $\displaystyle \alpha>0$, compute constants $\displaystyle \beta=\beta(\alpha),C=C(\alpha)$ such that
$\displaystyle \frac{G(x+y,t)}{G(x,\beta t)}\leq C$
holds for all $\displaystyle x\in\mathbb{R}^{d},t>0,y\in B(0,\alpha\sqrt{t})$. 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 $\displaystyle L^{1}(\mathbb{R}^{d})$ is not a subset of $\displaystyle L^{\infty}(\mathbb{R}^{d})$.
In particular, $\displaystyle \int_{\mathbb{R}^{d}}f(x)\;d\mu<\infty$ doesn't imply $\displaystyle \sup_{x\in\mathbb{R}^{d}}f(x)<\infty,$).
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 $\displaystyle a,\beta$ 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 $\displaystyle 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.
Thanks Rebesques -- especially for pointing out the bit about $\displaystyle \int f$ implying $\displaystyle \sup f<\infty$ when $\displaystyle f$ 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)
$\displaystyle \sup_{\epsilon>0}|\phi_{\epsilon}*f|(x)<A(Mf)(x)$ where $\displaystyle A=\int_{\mathbb{R}^{d}}\psi(x)\;dx$ and $\displaystyle \psi(x)=\sup_{y:|y|>|x|}|\phi(y)|$.
Since (I think) $\displaystyle G$ is equal to its least decreasing radial majorant (i.e. $\displaystyle \psi$ in the definition above), and $\displaystyle G(\cdot,t)$ plays the role of $\displaystyle \phi_{\epsilon}(x)=\phi(x/\epsilon)\epsilon^{-d}$, we might be able to transfer the proof onto $\displaystyle A(Mf)(x),$ which may be easier to work with than $\displaystyle G(\cdot,t)*f$. This just came to mind as I was responding to the above, so I haven't really looked into it yet.
