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I'm trying to solve the following problem:

Assume that we are given $f(t,x) \in C^{\infty}(\mathbb R \times \mathbb R^n)$ such that $f(t, \cdot) \in \mathcal S(\mathbb R^n)$ for all $t>0$ (where $\mathcal S(\mathbb R^n)$ is Schwartz space). Assume furthermore that $f$ is a solution to the heat equation on $\mathbb R^n$ with $ \begin{align} \partial_tf(t,x) &= \Delta f(t,x) \qquad t>0 \\ \lim_{t\to 0} f(t,x) &= g(x) \end{align} $ for some $g\in \mathcal S(\mathbb R^n)$.

I want to show that under these conditions we must have $f(t,x) = (K_t\ast g)(x)$, where $\ast$ denotes convolution and $ K_t(x) = \frac{1}{(4\pi t)^{n/2}}e^{-\langle x, x\rangle/4t}$ is the heat kernel.

What I'm having trouble with is the following: The idea is to consider the Fourier transform $\hat f$ of $f$ and to derive the ordinary differential equation

$\partial_t \hat f(t,k) = - k^2 \hat f(t,k)$

But how do I even know that $\hat f(t,k)$ is differentiable with respect to $t$? What I would like to do is differentiate under the integral sign here:

$\partial_t \hat f(t,k) =\partial_t \int_{\mathbb R^n} f(t,x)e^{-ikx}\, dx$

But I don't know how to justify it. Could anyone help me out?

Thanks a lot! =)

1 Answers 1

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You cannot, in general, interchange the differentiation and the integration here. I take the following example from this MO answer. Let $\phi(x)$, $x\in\mathbb{R}$, be an arbitrary bump function ($\phi \in C^\infty_c(\mathbb{R})$). Consider the function $f(x,t)$ defined to be identically 0 if $t \not\in (-\pi/2,\pi/2)$ and $f(x,t) = \phi(x - \tan t)$ otherwise. One checks that this function is indeed $C^\infty(\mathbb{R}^2)$: for each $|t| \geq \pi/2$, and for any $x$, one can find a small neighborhood of $(x,t)$ such that $f$ vanishes identically there. And $f(x,t)$ is a composition of smooth functions on $(-\pi/2,\pi/2)\times \mathbb{R}$.

Notice also that for any $k$, the function $\partial_t^kf(t,x)$, restricted to a fixed time slice $t = t_0$, is Schwartz (and in fact has compact support).

Now, taking the spatial Fourier transform you get that $\hat{f}(t,\xi) = e^{i \tan(t) \xi} \hat{\phi}(\xi)$ for $t\in (-\pi/2, \pi/2)$. Computing the time derivative you get $ |\frac{d}{dt} \hat{f}(t,\xi)| = |\xi| \sec^2(t) |\hat{\phi}(\xi)| $ which is not continuous at $t = \pm \pi/2$.


As a side remark, in Stein and Shakarchi's Fourier Analysis, the analogous statement to the one you want to prove is given with the additional assumption that the restrictions $f(t,x)$ to $t$ are uniformly Schwartz, which then imply that one can interchange the differentiation with integral.

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    I guess, if the claim isn't to be found (in this strength) in any standard textbook on Fourier analysis, then it is either simply wrong, unproven or at least quite hard to prove. So I will accept your answer. Thank you again for thinking about it and providing me with a counterexample to my question about the differentiation under the integral sign! :)2012-01-19