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In Real and Complex Analysis pg. 187, (d) property of Theorem 9.13 says: If $f\in L^2(\mathbb R)$ then: \begin{equation}\int_{-n}^{n}f(t)e^{-it\xi}\, dt\to \hat f\text{ and }\int_{-n}^{n}\hat f(\xi)e^{it\xi}\, d\xi\to f\text{ w.r.t. }\left\|\cdot\right\|_2\end{equation} For the first convergence, as $f\in L^2$, $f_n=f\chi_{[-n,n]}\in L^1\cap L^2$ and $f_n\to f$ in $\left\|\cdot\right\|_2$. Plancherel's Theorem implies \begin{equation}\left\|\hat f-\int_{-n}^{n}f(t)e^{-it\xi}\, dt\right\|_2= \left\|\hat f-\hat f_n\right\|_2=\left\|f-f_n\right\|_2\to 0\end{equation} All goof so far. Then Rudin goes on to write that the other convergence result is proven similarly. My question is how is it proven (similarly)?

Going by the previous result I defined $g_n=\hat f\chi_{[-n,n]}\in L^1\cap L^2$ and $g_n\to \hat f$ in $\left\|\cdot\right\|_2$. But in that case \begin{equation}\int_{-n}^{n}\hat f(\xi)e^{it\xi}\, d\xi=\mathcal F^*(g_n)\end{equation} where $\mathcal F^*$ is the Fourier cotransformation defined for functions in $L^1$ by \begin{equation}\mathcal F^*f=\int_{-\infty}^{\infty}f(\xi)e^{it\xi}\, d\xi\end{equation} and by a density argument for functions in $L^2$. Now I can prove that $\mathcal F^*$ has similar properties to the Fourier Transform, for example if $f\in L^2$ then $\mathcal F^*f\in L^2$ as well. My approach was to use Plancherel's Theorem as before: \begin{equation}\left\|f-\int_{-n}^{n}\hat f(\xi)e^{it\xi}\, d\xi\right\|_2= \left\|f-\mathcal F^*g_n\right\|_2=\left\|\mathcal Ff-\mathcal F\mathcal F^*g_n\right\|_2\end{equation} I would like to say $\mathcal F\mathcal F^*g_n=g_n$ a.e. but this is true if $\mathcal F^*g_n\in L^1$ which I don't know. Any suggestions on how I should proceed (that are preferably similar to Rudin's proof)?

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  • $\begingroup$ Oh, sorry. But $\mathcal F^* f_n$ is definitely not correct in light of your previous edit. $\endgroup$
    – Potato
    Aug 19, 2013 at 16:53
  • $\begingroup$ @Potato This result is the next theorem in Rudin's book, that uses the Theorem I'm asking for in its proof. I don't follow as to why $\mathcal F^*f_n$ is incorrect. $\endgroup$
    – Optional
    Aug 19, 2013 at 16:54
  • $\begingroup$ Because the integrand in the previous expression contains $\hat f$, not $f$. $\endgroup$
    – Potato
    Aug 19, 2013 at 16:58
  • $\begingroup$ By the way, why in the equality of the last line do you apply $\mathcal{F}$ to both for Plancherel? Why not go backwards? If you already accept that $f = \mathcal{F}^\ast \mathcal{F} f$, and you prove Plancherel for $\mathcal{F}^\ast$ instead (which should be trivial as it differs by a conjugate). $\endgroup$
    – Evan
    Aug 19, 2013 at 16:58
  • $\begingroup$ @Evan Oh, there are in fact two $f_n$s in the question, and I was looking at the earlier one. $\endgroup$
    – Potato
    Aug 19, 2013 at 17:00

2 Answers 2

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I think what you want to prove is a little trickier than Rudin lets on. At least, it's not as direct as before. Here's the proof I know. It appears in Stein and Weiss's Introduction to Fourier Analysis on Euclidean Spaces.

Using Fubini's theorem, for any $g\in L^2 \cap L^2 $ you have

$$\langle g , \mathcal F^* \mathcal F f \rangle = \langle \mathcal F g , \mathcal F f\rangle.$$

Then, using Plancherel's theorem, this is equal to $\langle g, f\rangle$. Since $F^* \mathcal F f$ and $f$ give the same inner product against any function in $L^2$ (we actually verified it on a dense subset, but same thing), they are equal (as members of $L^2$).

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  • $\begingroup$ Yeah I just checked it (for $f_n$ in the place $f$). I don't think that Rudin had this in mind when he said "similarly" $\endgroup$
    – Optional
    Aug 19, 2013 at 17:24
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    $\begingroup$ @Optional I'm not sure. If there's a better answer, I would be very interested to see it. $\endgroup$
    – Potato
    Aug 19, 2013 at 17:26
  • $\begingroup$ What troubles me is that Rudin had this in the 2nd edition of his book, but removed it for the 3rd... $\endgroup$
    – Optional
    Aug 19, 2013 at 17:27
  • $\begingroup$ Wait, there seems to be a problem here. We have $f_n\in L^2$ and so $\mathcal Ff_n\in L^2$ and $\mathcal F^*\mathcal Ff_n\in L^2$. But the formula of $\mathcal F^*$ is only true for functions in $L^1$ (like $f_n$) and not necessary for $\mathcal Ff_n$. We must thus prove the first part of the Theorem for $\mathcal F^*$ and then use the continuity of the inner product. $\endgroup$
    – Optional
    Aug 19, 2013 at 18:39
  • $\begingroup$ @Optional Yes, sorry, you do need a density argument. I was a little sloppy writing this up. $\endgroup$
    – Potato
    Aug 19, 2013 at 18:43
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Here's an excerpt from my edition of Rudin. It's an international student edition that predates the third edition, but I'm not sure if it's the second or the first. In any case, it has what you want. He uses a convolution argument to get the function you want to invert into $L^1$, then applies the $L^1$ inversion formula.

enter image description here

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  • $\begingroup$ Thanks for that. I wonder if he meant that this is similar to the first part. Anyway, the other proof you gave me is much better and closer to what I had in mind. $\endgroup$
    – Optional
    Aug 19, 2013 at 17:37

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