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So I just learned about Fourier transforms and I know that, if $f(x)$ is integrable and satisfies $\lim\limits_{|x|\rightarrow\infty}f(x)=0$ we have $\hat{\frac{\partial f}{\partial x_j}}(k)=2\pi ik_j\hat{f}(k)$. This can be used to obtain the so called fundamental solution to the heat equation $$\frac{\partial u}{\partial t}=\Delta u$$

where $\Delta$ denotes the laplacian, by applying this rule twice to the RHS, solving for the Fourier transform and retrieving $u$ through an inverse Fourier transform. My question is: how does one know that the solution to heat equation must disappear at infinity? Is this even true in general or is it just an imposed boundary condition?

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  • $\begingroup$ There's no purely mathematical reason for the solution to go to zero at infinity. But a solution that does not do so would not be describing a physically possible distribution of heat. $\endgroup$ – bob.sacamento Jun 15 '17 at 18:03
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You should ask your question at https://physics.stackexchange.com/ to get a discussion of physical reasons for choosing that boundary condition.

As for mathematical reasons... the Fourier transform is best behaved as an operator on $L^2$ functions; that is, functions for which $\int_{-\infty}^{\infty} |f(x)|^2 \, \mathrm{d} x \neq \infty$.

So, to use an argument involving Fourier transforms, it's nice to first restrict the problem to $L^2$ functions.

I strongly expect actual condition your source wanted is not $\lim_{|x| \to \infty} f(x) = 0$ — instead, one of the three following conditions that often arise in Fourier theory was actually intended:

  • $f$ is an $L^2$ function as described above
  • $f$ is smooth and rapidly decreasing (i.e. a "schwartz function", also sometimes called a "test function"), which involves many more vanishing limits than just the one specified
  • $f$ has compact support: that is, there exists a $B$ such that $f(x) = 0$ for all $|x| > B$

and either you misinterpreted the source, the source was abusing notation, or the source is generally confused about what conditions it needs to apply.

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    $\begingroup$ integrating by parts the Fourier transform of the derivative with respect to $x$ the boundary terms cancel if $f$ goes to zero at infinity. Why is this wrong? $\endgroup$ – user438666 Jun 15 '17 at 18:42

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