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I'm having a problem applying the closed graph theorem, which I think stems from distributions still being very new to me.

I am reading a proof in Stein and Weiss, Introduction to Fourier Analysis on Euclidean Spaces, 4.13 in Chapter 1, in the Further Results section, which begins:

Suppose for the sake of contradiction that the Fourier transform of every function $f\in L^p (\mathbb{R})$, as a tempered distribution, is a function. The closed graph theorem easily shows, for all $f\in L^p, p>2$, there is a constant $A$ so that the following holds:

$$\int_{|x|\le 1} | \ \hat f(x)| \ dx \le A ||f||_p. $$

I don't see how this follows from the closed graph theorem. Any help would be greatly appreciated.

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up vote 2 down vote accepted

Note that "function" here should be taken to mean "locally integrable function". (This is the natural interpretation in the context of distributions: the question asked is whether the Fourier transform of an $L^p$ function, which exists as a distribution, can be represented by a proper function; in this context the function as a distribution must be locally integrable.)

Suppose the contrary, that $\hat{f} \in L^1_{\text{loc}}$ for every $f\in L^p$, this necessarily implies that $\hat{f} \in L^1_{B(0,1)}$. Closed graph theorem states, in this context,

We have that $\mathcal{F}: L^p \to L^1_{B(0,1)}$ is a linear operator between Banach spaces. If the graph is closed (i.e. if $(f_n,\mathcal{F}f_n)$ converges in $L^p \times L^1_{B(0,1)}$ to $(f,g)$, then $g = \mathcal{F}f$), then the map is continuous (i.e. $\|\mathcal{F}f\|_{L^1_{B(0,1)}} \leq A\|f\|_p$).

To verify the hypothesis of the closed graph theorem, in other words show the statement in the first parenthesis is true, you should use the fact that $\mathcal{F}$ is a continuous mapping of tempered distributions, and note that a sequence of functions converging in $L^1_{\text{loc}}$ also converges as distributions.

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