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.

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.

Re: Scaling Estimate on the Heat Kernel

Re: Scaling Estimate on the Heat Kernel

And actually, by demanding , we can replace the finishing calculations by

.

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

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

The inclusion of the is supposed to indicate that is allowed to freely vary just as is. I'm assuming your proof probably already makes this so, even if wasn't explicitly written into the . Also, I find it interesting that you supposed more generally that is restricted to , and then proceded to prove necessarily (you used instead of and got rid of the original in the problem statement by replacing the problem with a 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!

Re: Scaling Estimate on the Heat Kernel

I think there's something wrong here. Since can vary with , I don't think any conclusion can be made by performing an integral over since this excludes the role of . We're trying to produce the constant that you eliminated from consideration by replacing the original problem with a . In particular, we are trying to quantify the sense in which the quotient is finite. The constant that you computed is given a priori. In other words, given , compute constants such that

holds for all . 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 is not a subset of .

In particular, doesn't imply ).

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 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 , 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.

Re: Scaling Estimate on the Heat Kernel

Thanks Rebesques -- especially for pointing out the bit about implying when 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)

where and .

Since (I think) is equal to its least decreasing radial majorant (i.e. in the definition above), and plays the role of , we might be able to transfer the proof onto which may be easier to work with than . This just came to mind as I was responding to the above, so I haven't really looked into it yet.