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In Grafakos's book, in order to the define Fourier transform on $L^2$, he claims that the Fourier transform is an isometry on $L^1\cap L^2$. Since Fourier transform is an isometry, thus $\hat{f}\in L^2$. However, I couldn't figure out why $\hat{f}\in L^1$.

89085731
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    Actually some of the nicest functions you can think of don't have their Fourier transform in $L^1(\mathbb{R})$. A good exercise is to show that there is a function $f\in L^1(\mathbb{R})\cap L^{\infty}(\mathbb{R})\cap C^{\infty}(\mathbb{R})$ such that $\hat{f}\not\in L^1(\mathbb{R})$. One instance where $\hat{f}\in L^1(\mathbb{R})$ is when $f$ is such that $f\in L^1(\mathbb{R})\cap C^1(\mathbb{R})$ and $f'\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$. – NeedForHelp Jan 25 '17 at 02:33

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In general, it isn't. Here, "on" doesn't mean what it usually means, that is one space to itself, but rather that it is an $L^2$ isometry from $L^1\cap L^2$ to $L^2$. A good example is the Fourier transform of $\chi_{[-1,1]}$. Its Fourier transform has finite $L^2$ norm, but it is not Lebesgue integrable.

Cameron Williams
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Assuming the Fourier transform of $f\in L^1(\mathbb{R})$ is the function $\hat{f}:\mathbb{R}\to\mathbb{C}$ defined by $$ \hat{f}(\xi):=\int_{-\infty}^{\infty}f(x)e^{-ix\xi}\,dx $$

Here's a way to see that $f:=\chi_{[-1,1]}$ (which is in $L^1(\mathbb{R})\cap L^2(\mathbb{R}))$ is such that $\hat{f}\not\in L^1(\mathbb{R})$. Suppose the contrary. Then the Inversion Theorem would apply and we would have $$ \hat{\hat{f}}(-x)=2\pi f(x) $$ at every point where $f$ is continuous, so (since $f$ is pair) $$ \hat{\hat{f}}(x)=2\pi\chi_{[-1,1]}(x) $$ for all $x\neq1,-1$. So $\hat{\hat{f}}$ would be continuous (since it would be the FT of $\hat{f}\in L^1(\mathbb{R})$ by hypothesis) and such that $$ \hat{\hat{f}}=\begin{cases}2\pi&&|x|<1\\0&&|x|>1\end{cases} $$ This is impossible, so $\hat{f}\not\in L^1(\mathbb{R})$.

NeedForHelp
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  • This approach is really good for showing that the abstract Fourier transform of the characteristic function of an open set isn't in $L^1$. Great argument! – Cameron Williams Jan 25 '17 at 03:20
  • @Cameron Williams I have no merit! This argument was shown to me when I took a class dealing with Fourier analysis. – NeedForHelp Jan 25 '17 at 03:36