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This is Ex $2.3.6$ in Dym and Mckean's Fourier Series and Integrals.

One can define the operator $\hat{f}(\gamma)$ for $f \in L^2(\mathbb{R})$ as $\lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_b^a f(x) e^{-2 \pi i x \gamma} dx$ where the limit takes place in $L^2$.

It turns out this is a length preserving isomorphism in $L^2$, with inverse operator $\check{f}(x) = \lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_b^a f(\gamma) e^{2 \pi i x \gamma} d\gamma$.

I am now asked to show that for $f \in L^2$, if $\gamma^n \hat{f} (\gamma) \in L^2$ then $f$ is $C^\infty$ and all derivatives $D^n f \in L^2$.

Here is my proof which doesn't work yet: Let $g = \check{[(2 \pi i \gamma) \hat{f}]}$ and the idea is to show that $g = f'$ via $f(x) = f(0) + \int_0^x g(s) ds$. First you can see that $\int_0^x g(s) dx \lt \infty $ a.e. in $x$ since $g \in L^2$ by assumption and the integral can be expressed as the inner product $(g, \mathbb{1}(0, x))$ .

Then you can substitute $g$ to get $f(0) + \int_0^x \lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_b^a 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} d\gamma\, dt$.

Now the bad part is that the inner integral is an $L^2$ limit whereas I want to be able to say $f(0) + \int_0^x \int_{-\infty}^{\infty} 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} d\gamma\, dt = f(0) + \int_{-\infty}^{\infty} \int_{0}^{x} 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} dt \,d\gamma = f(0) + \int_{-\infty}^{\infty} \hat{f}(\gamma)(e^{2 \pi i x \gamma} - 1) \,d\gamma = \check{\hat{f}}(x) - \check{\hat{f}}(0) + f(0) = f(x)$ which would complete the proof.

I don't know how to justify these operations though.

Edit: Here is something I overlooked. Inner product with $\mathbb{1}(0,x)$ is a continuous operator on $L^2$ and therefore we can pull the inner limits out to get the simpler expression $f(0) + \lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_0^x \int_{a}^{b} 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} d\gamma\, dt $. Interchange of integrals is justified because of Fubini and I think this completes the proof.

Mark
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    You should be able to write $f'$ as a limit of a difference quotient and use dominated convergence. (Write the difference quotient with Fourier inversion). – Eric Thoma Jan 03 '16 at 02:01
  • Using which dominating function? – Mark Jan 03 '16 at 02:01
  • it should not be very different of the https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma – reuns Jan 03 '16 at 02:08
  • Defining a function using the value $f(0)$ is not technically correct but can be fixed by using a Lebesgue point $p$ of $f$ instead of $0$. See http://math.stackexchange.com/questions/1832691/how-to-define-f0-when-f-is-a-function-in-l2/1832709#1832709 – Mark Jun 20 '16 at 03:22

1 Answers1

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Let $y, \xi \in \mathbb{R}$ be fixed. Note $$ |e^{2\pi i\xi y} \left( \frac{e^{2\pi i \xi h} - 1}{h} \right) \hat{f}(\xi)| \leq |L \xi \hat{f}(\xi)| $$ where $L$ is a Lipschitz constant for $\exp$. We can prove that the RHS is integrable in $\xi$.

We have $$ \int | \xi \hat{f}(\xi) | d\xi = \int_{[-1,1]} |\xi \hat{f}(\xi)|d\xi + \int_{[-1,1]^c}|\xi \hat{f}(\xi)|d\xi. $$ Using Cauchy-Schwarz we have $$ \int_{[-1,1]} |\xi \hat{f}(\xi)|d\xi \leq ||\xi \hat{f}||_{L^2} || 1 ||_{L^2(-1,1)} = \sqrt{2}||\xi \hat{f}||_{L^2} $$ which is finite by assumption. We have by Cauchy-Schwarz $$ \int_{[-1,1]^c} \xi \hat{f}(\xi)d\xi = \int_{[-1,1]^c} \frac{1}{\xi} \cdot \xi^2\hat{f}(\xi)d\xi \leq || \xi^2 \hat{f}||_{L^2} || \xi^{-1} ||_{L^2([-1,1]^c)}. $$ This quantity is also finite since $\xi^2 \hat{f} \in L^2$ by assumption.

Consider the difference quotient $$ \frac{f(y+h) - f(y)}{h} = \frac{1}{h} \int e^{2\pi i \xi y} \left( e^{2\pi i \xi h} - 1 \right) \hat{f}(\xi) d\xi. $$ Now use dominated convergence theorem to pass a limit in $h$ inside the integral. Make sure you explain why Fourier inversion works in this case.

Eric Thoma
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  • How do you show that $L \xi \hat{f}(\xi)$ is integrable? – Mark Jan 03 '16 at 02:27
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    Also, the integral is not a standard integral, but it is an $L^2$ limit of functions of $y$ – Mark Jan 03 '16 at 02:33
  • You can prove $\xi \hat{f} \in L^1((-\infty,-1) \cup (1, \infty))$ by upping $n$ in the condition $\xi^n \hat{f} \in L^2$. And $L^2(-1,1) \subset L^1(-1,1)$ using Cauchy Schwarz integrating against the constant function. – Eric Thoma Jan 03 '16 at 02:34
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    @Mark In this case the Fourier inversion integral is well-defined I believe. The function $\hat{f}$ itself is only defined via an L^2 limit of Schwartz functions though. – Eric Thoma Jan 03 '16 at 02:35
  • @Mark You are correct in general these integrals are $L^2$ limits. If your function $f$ is in $L^1 \cap L^2$ though, then the integral for $\hat{f}$ may be interpreted literally, and it agrees with the $L^2$ limits. An analogous statement holds for the inverse transform. – Eric Thoma Jan 03 '16 at 02:40
  • Incidentally, integrability lets me complete my method as well via Fubini theorem. Thanks! – Mark Jan 03 '16 at 04:02
  • actually can you elaborate on the proof that $\xi \hat{f} \in L^1$? – Mark Jan 03 '16 at 04:12
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    @Mark I have edited my answer. A similar proof shows $\hat{f} \in L^1$, or more generally $\xi^n \hat{f} \in L^1$. – Eric Thoma Jan 03 '16 at 04:26