PDE Fourier transformation method regularity of solution Let the heat equation be given
$$u_t -\Delta u=0,\quad u(\cdot, 0)=u_0 $$
With Fourier transform one obtains the IVP
$$\hat{u}_t(k)+|k|^2\hat{u}(k)=0,\quad \hat{u}(k,0)=\hat{u}_0(k) $$
with the solution
$$\hat{u}(k,t)=\hat{u}_0e^{-|k|^2t}. $$
Doing an inverse transformation one obtains 
$$u(x,t)=(-2\pi)^{-n} \int_{\mathbb{R}^n} e^{-|k|^2t}\hat{u}_0(k) e^{ik\cdot x}dk $$
With a convolution trick we can write this as
$$u(x,t)=(4\pi t)^{-n/2} \int_{\mathbb{R}^n} e^{-|x-y|^2/4t}u_0(y) dy $$
Now all of that is clear to me, however my question is, in what sense is this an solution to the heat equation? I assume it is an classical solution, but I'm not sure wheter it fulfills the original equation point wise since the Fourier transformation involves integrals, so I would find it plausible that is an solution in the $L^2$ norm, like for example 
$$ ||u_t -\Delta u||_{L^2}=0 $$
Would be glad if you could point me to results in that regard.
Are there some general observations regarding the regularity of solution obtained through Fourier transformation?
 A: For $t>0$ it is a classical solution.
If $u_0$ is nice ($L^1$ or compactly supported distribution to give examples), then $u(\cdot,t)$ is a mollified version of $u_0$.
Mollification produces smooth functions, as long as $u_0$ is good enough for the convolution with the Gaussian to be well defined.
This mollification point of view shows that $u(x,t)$ is smooth in $x$ when $t>0$.
It is also smooth in $t$; you can differentiate the formula directly or perhaps appeal to $\partial_tu=\Delta u$.
A general rule of thumb is that if a smooth function is a solution in a weak sense, then it is a solution in the classical or strong sense.
If you want smoothness down to $t=0$ or even for $t<0$, it's a different ball game and depends on the regularity of $u_0$.
In general, to see the regularity of a function produced by the Fourier transform, look at the decay of the Fourier transform.
You found that $u(\cdot,t)$ is the inverse Fourier transform of $e^{-|k|^2t}\hat u_0(k)$.
This means that the decay on the Fourier side is greatly improved as compared to $u_0$.
When the Fourier transform decays faster than any power of $|k|$, the function is smooth.
The Fourier method in itself does not assume that the solution is a classical one.
In principle, it is enough that $u$ solves the PDE in the sense of, say, Schwartz distributions.
The way differentiation turns into multiplication by polynomials works for these distributions even though differentiation is defined weakly.
Distribution-valued ODEs (like $\partial_t\hat u=-|k|^2\hat u$) require some care to set up and solve properly, but it can be done.
As long as your function class turns weak derivatives to polynomials upon Fourier transform, the method is fine for weak solutions.
Or if you can use test functions (in the definition of a weak solutions) that resemble $e^{ik\cdot x}$, you can use the definition directly.
And it may sometimes happen that the Fourier transform method is not always perfectly justified, but it leads to a good guess for a solution which can be proven correct and unique by other means.
