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For $1<p<2$, denote $p':=1/(1-1/p)$. Let $1<p<2$ and $f \in L^p(\mathbb R^n)$ and $\hat f \in L^1(\mathbb R^n)$,

then how to show that $f(x)=\int_{\mathbb R^n} \hat f (y) e^{-2\pi I x\cdot y} dy, \forall x \in \mathbb R^n$?

I can show it if it were $p=1,2$. For the given problem, I was thinking like this; there is a sequence $\{g_n\}$ in $C_c^\infty (\mathbb R^n)$ such that $g_n$ converges to $f$ in $L^p$ and pointwise a.e. also. Then $\hat g_n$ converges to $\hat f$ in $L^{ p'} $, $p'>2$ . Although, since $\hat g_n$ is Schwarz class and $\hat f \in L^1 \cap L^{p'}$, we can take their Fourier transforms again, but we wouldn't know whether $\widehat {\hat g_n}= h_n$ , where $h_n(x)=g_n(-x)$, converges to $\widehat {\hat f} $ in some $L^q$ or not. I was thinking if the result for $p=1,2$ can be applied to get the desired claim, but I don't know.

Further work: $\hat f \in L^{p'} \cap L^1$, so $\hat f \in L^r, \forall r \in [1,p']$, where $1/p+1/p'=1$ so $p'>2$. Now $\exists f_1\in L^1, f_2 \in L^2$ such that $f=f_1+f_2$ and $\hat f = \hat f_1 + \hat f_2$. Then $\hat f_1 \in C_0$ and $\hat f_2 \in L^2$. Then $\hat f_1=\hat f-\hat f_2 \in L^2$. I don't know whether this is helpful or not.

user
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1 Answers1

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For $\epsilon>0$ consider the function $$f_\epsilon(x) = e^{-\epsilon |x|^2}f(x)$$ By Hölder's inequality $f_\epsilon \in L^1$. Also, $\widehat{f_\epsilon} = \gamma_\epsilon * \hat f$ where $\gamma_\epsilon = \widehat{e^{-\epsilon |x|^2}}$ is a tightly localized (when $\epsilon$ is small) Gaussian with integral $1$. Hence $\widehat{f_\epsilon}\in L^1$ and $\widehat{f_\epsilon} \to \hat f$ in $L^1$ as $\epsilon>0$.

Since $f_\epsilon, \widehat{f_\epsilon} \in L^1$, the $L^1$ Fourier inversion formula applies. As $\epsilon\to 0$, we have $f_\epsilon\to f$ pointwise and $\widehat{f_\epsilon} \to \hat f$ in $L^1$, where the latter suffices to pass to the limit in the integral $$ \int_{\mathbb R^n} \widehat {f_\epsilon} (\xi) e^{-2\pi i \xi\cdot x} d\xi \to \int_{\mathbb R^n} \widehat {f} (\xi) e^{-2\pi i \xi\cdot x} d\xi $$

Note that this applies to $f\in L^p$ with any $p>1$, not only $1<p<2$.

  • Where did we use $\hat f\in L^1$ ? How do you know $\widehat f_{\epsilon}\to \hat f$ in $L^1$ ? – user Nov 26 '17 at 18:32
  • Your first question is answered by the second question. To answer the second question, see here for the proof that convolution with "thin" Gaussians converges to original function in $L^1$, provided the function is in $L^1$. –  Nov 26 '17 at 19:01