Uniform convergence for heat kernel on unit circle

In summary, the conversation revolves around using the Weierstrass M-test to show that a family of functions/kernels, represented by Ht(x), is uniformly convergent for a seminar. The solution involves finding a sequence of positive numbers that will always be greater than or equal to Ht(x), with the use of the M-test. The question asks if the sequence can include the variable t, and if this would allow for the use of the M-test to show uniform convergence. The argument presented in the conversation points towards the possibility of using the M-test, but the correctness of the argument is unsure.
  • #1
stripes
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Homework Statement



I would like to use the Weierstrass M-test to show that this family of functions/kernels is uniformly convergent for a seminar I must give tomorrow.

[itex]

H_{t} (x) = \sum ^{-\infty}_{\infty} e^{-4 \pi ^{2} n^{2} t} e^{2 \pi i n x} .

[/itex]

Homework Equations


The Attempt at a Solution



I just need to find a sequence of positive numbers that will always be greater than the heat kernel Ht(x) for all x of course. But must it be greater than or equal to Ht(x) for all t as well? That being said, it might prove difficult to find an appropriate sequence...

Can I include t in my sequence of positive numbers? It might make it easier. At first I was just thinking of something as simple as (15/16)^n...if someone could guide me in the right direction, I would appreciate it, thanks!
 
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  • #2
Does this work:

Using the Weierstrass M-Test, we'll consider the sequence of terms

[itex]| e^{-4 \pi ^{2} n^{2} t} e^{2 \pi i n x} | = \frac{| e^{2 pi i n x} |}{e^{-4 \pi ^{2} n^{2} t}}[/itex]

Notice that on a ring of radius 1, for all time greater than or equal to zero, this sequence will be largest when x = 1 and t = 0. Obviously when t = 0, the sequence does not converge. But we are not concerned at time t=0 since we are given initial condition u(x, 0) = f(x).

We wish to find a sequence of terms, [itex]M_{n}[/itex], so that [itex]| e^{-4 \pi ^{2} n^{2} t} e^{2 \pi i n x} | \leq M_{n}[/itex] for all n.

So we have

[itex]| e^{-4 \pi ^{2} n^{2} t} e^{2 \pi i n x} | = | \frac{ e^{2 \pi i n x} }{e^{4 \pi ^{2} n^{2} t}} | = \frac{| e^{2 \pi i n x} |}{e^{4 \pi ^{2} n^{2} t}} \leq \frac{| e^{2 \pi i n} |}{e^{4 \pi ^{2} n^{2} t}} = \frac{1}{e^{4 \pi ^{2} n^{2} t}} = M_{n}.[/itex]

My question is, since the heat kernel on the circle is defined as a variable of x, [itex]H_{t}(x),[/itex], can my [itex]M_{n}[/itex] include the "variable" t? Will this allow me to use the M-test to show uniform convergence? I ask because uniform convergence cannot depend on the variable x. But can it depend on t?

Is my argument above correct? If someone could just have a quick read over what I wrote and tell me where I have gone wrong, I would really appreciate it.
 
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