I have a question about Fourier transforms:

What does the following question means?

" Does the Fourier transform of the function $f(x)= \frac{1}{\sqrt{1 + x^2}}$ belongs to $L_2(\mathbb{R})$?"

We know that $L_2(\mathbb{R})= \{ f: \mathbb{R} \rightarrow \mathbb{C} \mid \int_{- \infty}^{+ \infty} |f(t)|^2 dt < \infty \} $ and we know that $\hat{f}(t) = \int_{- \infty}^{+ \infty} e^{-itx} f(x) dx$.

Now for to show that $\hat{f}(t) \in L_2(\mathbb{R})$, what do we need to prove?

$\textbf{(1)}$ $~$ Is it enough to show that $f(x) \in L_1(\mathbb{R}) \cap L_2(\mathbb{R})$, then we have some results that shows $\hat{f}(t) \in L_2(\mathbb{R})!$

$\textbf{(2)}$ $~$ Or we have to prove $\int_{- \infty}^{+ \infty} |\hat{f}(t)|^2 dt < \infty$?

If $\textbf{(2)}$ is true? Can you please help me to show that?

Here linked-to result someone has found its Fourier transform, but I cannot understand it! I mostly prefer to find it by usual integration ways such as variable changes or substitution or even by inverse transform theorem!

Please let me know if I am wrong about using $\textbf{(1)}$ for to prove that $\hat{f}(t) \in L_2(\mathbb{R})$?


  • 2
    $\begingroup$ In fact $f\notin L^1$, so the Fourier transform is not given by that integral. Do you know the Plancherel Theorem? The point to this problem is to see if you do (in fact you need to know the Plancherel Theorem before you can even define $\hat f$ for this $f$!) $\endgroup$ Dec 4, 2017 at 22:58
  • $\begingroup$ @DavidC.Ullrich Thanks! Yes now I see that by setting $x = tan(u)$ the integral diverges in $L_1(\mathbb{R})$. And no I did not know about the Plancherel Theorem. But I looked at it, which makes our job easier. That means that for to prove $\hat{f}(t) \in L_2(\mathbb{R})$ we just need to prove that $f(t) \in L_2(\mathbb{R})$. Am I right? Which actually is. $\endgroup$
    – Nikita
    Dec 4, 2017 at 23:05
  • $\begingroup$ @DavidC.Ullrich But as I looked on details, I think the Plancherel Theorem also says that if $f \in L_1(\mathbb{R}) \cap L_2(\mathbb{R})$ then $\hat{f}(t) \in L_2(\mathbb{R})$! So can you please let me know how will it work? Thanks! $\endgroup$
    – Nikita
    Dec 4, 2017 at 23:36

2 Answers 2


As already stated it depends upon how you really define the Fourier transform. If you define it as an improper integral the question becomes non-trivial (otherwise it is just Plancherel). For $n\geq 1$ let $$ f_n(x) = f(x)\; {\rm \bf 1}_{[-n,n]}(x) = \frac{1}{\sqrt{1+x^2}} {\rm \bf 1}_{[-n,n]}(x) $$ where ${\rm \bf 1}_{[-n,n]}$ is the indicator function for the interval $[-n,n]$. Then $f_n$ is in both $L^1\cap L^2$ and the Fourier transform $$ \hat{f}_n(t) = \int_{-n}^n \frac{e^{itx}}{\sqrt{1+x^2}} dx $$ converges for every $t\neq 0$. This follows from partial integration: $$ \hat{f}_n(t) = \left[ \frac{e^{itx}}{it\sqrt{1+x^2}} \right]_{-n}^n + \int_{-n}^n \frac{\sin(tx)}{t} \frac{x}{\sqrt{1+x^2}^3} dx . $$ As $n\rightarrow \infty$, the first term goes to zero and the latter converges absolutely. Thus, you may define $\hat{f}(t)=\lim_n \hat{f}_n(t)$ for every $t\neq 0$ and the question is then if $\hat{f}\in L^2$?

The answer is yes: First note, that $f_n \rightarrow f$ in $L^2$ so $(f_n)_{n\geq 1}$ is Cauchy in $L^2$.

By Plancherel $\|\hat{f}_n-\hat{f}_m\|_2 = \|f_n-f_m\|_2$, so $(\hat{f}_n)_{n\geq 1}$ is also Cauchy in $L^2$, whence converges in $L^2$ to some function $g$.

Any convergent sequence in $L^2$ has, however, a subsequence that converges a.e. and we already know that for $t\neq 0$ (i.e. a.e.) the limit is $\hat{f}(t)$, so $\hat{f}=g$ (mod 0) is in $L^2$

  • $\begingroup$ H.Rugh Many thanks for your knowledgeable answer! $\endgroup$
    – Nikita
    Dec 16, 2017 at 2:30

Answering confusion in the comments about the Plancherel theorem:

A rough version of the theorem is

$\hat f\in L^2(\mathbb R)$ if and only if $f\in L^2(\mathbb R)$.

That's true, but it really needs to be stated more carefully. Because for example it's not clear what $\hat f$ even means given just $f\in L^2$. A more careful version starts here:

If $f\in L^1\cap L^2$ then $\hat f\in L^2$ and $\lVert\hat f\rVert_2=\lVert f\rVert_2$.

There there's no problem with the definition of $\hat f$, since $f\in L^1$. $\newcommand{\ft}{\mathcal F}$ At this point it's convenient to introduce the notation

$$\ft f=\hat f.$$

Now $\ft$ is defined on a dense subspace of $L^2$ and $\ft$ is an isometry (in the $L^2$ norm) on that dense subspace. Hence $\ft$ extends to an isometry $\ft:L^2\to L^2$. When people talk about $\hat f$ for $f\in L^2$ they're referring to that extension. (And now we're done with the problem in the question: Since $f\in L^2$ it follows that $\hat f\in L^2$.)

It follows easily that $\ft$ is its own inverse, almost:

If $f\in L^2$ then $\ft^2f(t)=f(-t)$.

Proof: The $L^1$ inversion theorem shows that this holds for, say, $f$ in the Schwartz space. Since the Schwartz space is dense and $\ft$ is continuous on $L^2$ (as is the operator $f(t)\mapsto f(-t)$) it holds for all $f\in L^2$.

In particular $\ft$ maps $L^2$ onto $L^2$: that is, $\ft$ is an isometric isomorphism, which is to say $\ft$ is unitary. (And in fact $\ft^4=I$, which leads easily to a decomposition of $L^2$ into four eigenspaces...)


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