Extend the Fourier transform over $L^2(\mathbb R^n)$

Question

Using Plancherel theorem, we have that the Fourier transform is an isometry over $L^2(\mathbb R^n)$. But anyway. In my course it's written that Plancherel theorem is extremely important since it allow us to prolonge the fourier transform from $S(\mathbb R^n)$ (i.e. the schwarz space) to $L^2(\mathbb R^n)$. Indeed, $$\int_{\mathbb R^n}f(x)e^{2\pi i x\cdot \xi}dx$$ doesn't converge in the absolute sense if $f\in L^2(\mathbb R^n)$. However, given such $f$, we simply pick a sequence $\{f_n\}\subset S(\mathbb R^n)$ (that exist by density of $S(\mathbb R^n)$ in $L^p(\mathbb R^n)$) with $$f_n\underset{n\to \infty }{\longrightarrow } f$$ in the $L^2$ sense. Then, using Plancherel's theorem, $\{\hat f_n\}_{n\in \mathbb N}$ is a cauchy sequence in $L^2(\mathbb R^n)$, and hence we can defined $$\hat f=\lim_{n\to\infty }\hat f_n$$ using the fact that $L^2(\mathbb R^n)$ is complete.

My questions are the following:

1) Finally, when $f\in L^2(\mathbb R^n)$ does $\int_{\mathbb R^n}f(x)e^{2\pi ix\cdot \xi}dx$ exist or not ?

2) And does $\hat f$ that is the limit of the $\{\hat f_n\}$ is given by $$\hat f(\xi)=\int_{\mathbb R^n}f(x)e^{2\pi ix\cdot \xi}dx.$$

I'm a little bit lost in all that.

Answer 1

1. In the sense of Lebesgue integration, the integral defining the Fourier transform of $f$ makes sense if and only if $f \in L^1$. That's easy to see: the integral has to converge absolutely, and the modulus of the integrand is just $|f|$. Since $L^2(\mathbb{R}^n) \setminus L^1(\mathbb{R}^n)$ is nonempty, there are $L^2$ functions for which the integral does not converge. One example would be $x^{-2/3} 1_{[1,\infty)}(x)$ on $\mathbb{R}$.
2. I don't really understand what you mean here. If you mean "if I Schwarz-approximate the FT of an $L^1$ function, do I get the FT in the sense of Lebesgue integration?", the answer is yes. (If it weren't, this Schwarz trick would be awful.) If you mean "is the Schwarz limit of the FT of an $L^2$ function the FT of it in the sense of distribution theory?", the answer is again yes. (This means that $\langle \hat{f},\hat{g} \rangle = C \langle f,g \rangle$ for any Schwarz function $g$, where $C$ is a constant depending only on the FT normalization you are using.) It is in this sense that we identify this Schwarz-approximation limit as "the FT" of a function for which the FT integral doesn't converge. I think in pretty much any other sense the answer to this question is no.

Answer 2

To clarify the situation

Using your definition, the Fourier transform is defined as following \begin{align*} \mathfrak F: \mathcal S(\mathbb R^n)&\longrightarrow \mathscr F(\mathbb R^n):=\{f:\mathbb R^n\longrightarrow \mathbb R\mid f\text{ is a function}\}\\ f&\longmapsto \hat f \end{align*} where \begin{eqnarray}\mathfrak F(f)(\xi)=\hat f(\xi)=\int_{-\infty }^\infty f(x)e^{2\pi ix\cdot \xi}\mathrm d x.\tag{1}\end{eqnarray}

You can also define \begin{align*} \mathcal F: L^1(\mathbb R^n)&\longrightarrow \mathscr F(\mathbb R^n)\\ f&\longmapsto \hat f \end{align*} but this definition is less interesting than the previous one since $\mathcal F$ will not be extendable to $L^2$ as easily (the only way to extend it to $L^2$ is to consider the restriction to function with compact support, but I'm not even sure if it's really enough). Anyway. Now, what you said (and your arguments are true) is that you can prolonge $\mathfrak F$ on $L^2(\mathbb R^n)$ using density of $\mathcal S(\mathbb R^n)$ in $L^2(\mathbb R^n)$ and completness of $L^2(\mathbb R^n)$.

This doesn't mean that for $f\in L^2$ unspecified it's Fourier transform is going to be given by $(1)$. It just mean that you can defined \begin{align*} \tilde{\mathfrak F}:L^2(\mathbb R^n)&\longrightarrow \mathscr F(\mathbb R^n)\\ f&\longmapsto \tilde{\mathfrak F}(f):=\lim_{n\to \infty }\mathfrak F(f_n),\quad f_n\in\mathcal S(\mathbb R^n) \end{align*} such that $\left.\tilde{\mathfrak F}\right|_{\mathcal S(\mathbb R^n)}=\mathfrak F$. Notice that $$f=\lim_{n\to \infty }f_n,$$ in the $L^2-$sense. Then $\tilde{\mathfrak F}$ is called the prolongement of $\mathfrak F$ and is well defined by the argument you said.

Therefore, for $f\in L^2$ unspecified, if $f$ is in $\mathcal S(\mathbb R^n)$, then $\tilde{\mathfrak F}(f):=\hat f$ in the sense of $(1)$, and if $f\in L^2(\mathbb R^n)\backslash \mathcal S(\mathbb R^n)$, then $$\tilde{\mathfrak F}(f):=\lim_{n\to \infty }\hat f_n,$$ where $$\hat f_n(\xi)=\int_{|x|\leq n}f(x)e^{2\pi i x\cdot \xi}\mathrm d x$$ and the limit is of course taken in the $L^2-$sense.

Maybe you can notice that it's the same as $g:]0,1]\longrightarrow \mathbb R$ defined by $g(x)=\frac{\sin(x)}{x}$. You can prolonge $g$ on $0$ mean that there is $\hat g:[0,1]\longrightarrow \mathbb R$ s.t. $\hat g|_{]0,1]}=g$ and $\hat g(0)=\lim_{n\to \infty }g(x)$, but it doesn't mean that $g(0)=0$ (since $g$ is not defined at $0$).