In Rudin's R&CA, the fourier inversion theorem states that if $f\in L^1$ AND $\hat f \in L^1$, and we define $g$ as $$g(x)=\int_{\mathbb R} \hat f(t) e^{itx}\text{d}t,$$ then $g=f$ a.e.

Now, if $g_n=\chi_{[-n,n]}$, how do I prove that $g_n*g_1$ is the Fourier transform of an $f$ in $L^1$?
It is clear that $g_n*g_1$ is in $L^1$, and an obvious next step would be to apply the inverse Fourier transform to it to get (up to a multiplicative constant) $n\text{sinc} (x)\text{sinc}(nx)$. It is easy to prove that this is in $L^1$, however, I am confused by what I just did, since the theorem quoted above requires that both $f$ and $\hat f$ be in $L^1$. Is it valid to first compute the inverse transform and then prove that it is in $L^1$? Once you do that, how do you know that this function really transforms into the original?
Perhaps the exercise had in mind a clearer route, since the first part of the question only requires that I prove the existence of an absolutely integrable function whose Fourier transform is $g_n*g_1$. Once this is done, the above computations would be rigorous. The problem now becomes, how do I prove said function exists?


I'm not sure what you're confused about. Anyway:

Let $f=g_n*g_1$. You easily calculate $\hat g_n$ and $\hat g_1$ and see that they're in $L^2$, hence $\hat f=\hat g_n\hat g_1\in L^1$.

So $f,\hat f\in L^1$. Since $f$ is continuous the inversion theorem shows that $$f(x)=\int\hat f(\xi)e^{ix\xi}\,d\xi=\int\hat f(-\xi)e^{-ix\xi}\,d\xi =2\pi\hat h(x),$$where $h(\xi)=\hat f(-\xi)$. (At first we know only that $f=\hat h$ almost everywhere, but since $f$ and $\hat h$ are both continuous it follows that $f=\hat h$ everywhere.)

  • $\begingroup$ I think that you miss a factor $2\pi$: $$f(x)=\int\hat f(\xi)e^{ix\xi}\,d\xi=\int\hat f(-\xi)e^{-ix\xi}\,d\xi = 2\pi \hat h(x).$$ The Fourier transform must be defined as $$\hat f(t) = \frac{1}{2\pi} \int_{\mathbb R} f(x) \, e^{-itx} \, dx$$ when the inverse transform has the given form. $\endgroup$ – md2perpe Dec 28 '17 at 19:42
  • $\begingroup$ NO doubt. I never worry about where the $2\pi$'s go unless I'm writing something for publication. Different authors put them in different places - the Littlewood convention says $2\pi=1$ just for this reason. $\endgroup$ – David C. Ullrich Dec 28 '17 at 21:24

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