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Math Help - Scaling Estimate on the Heat Kernel

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    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.
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    Super Member Rebesques's Avatar
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    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.
    Last edited by Rebesques; December 7th 2013 at 06:02 AM.
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    Re: Scaling Estimate on the Heat Kernel

    Quote Originally Posted by Rebesques View Post
    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.
    Last edited by Rebesques; December 7th 2013 at 07:53 AM.
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    Super Member Rebesques's Avatar
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    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.
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    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!
    Last edited by TaylorM0192; December 7th 2013 at 04:41 PM.
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    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,).
    Last edited by TaylorM0192; December 7th 2013 at 05:04 PM.
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    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.
    Last edited by Rebesques; December 7th 2013 at 05:44 PM.
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    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)<A(Mf)(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.
    Last edited by TaylorM0192; December 10th 2013 at 02:53 PM.
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