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I want to prove or disprove that the Fourier transform $\mathcal F \colon (\mathcal S(\mathbb R^d), \lVert \cdot \rVert_1) \to L^1(\mathbb R^d)$ is unbounded, where $\lVert\cdot \rVert_1$ denotes the $L^1(\mathbb R^d)$-norm.

Having thought about this for a moment, I believe it is indeed unbounded. So I tried to find a sequence of Schwartz functions $(f_n)_{n\in \mathbb N} \subseteq \mathcal S(\mathbb R^d)$ with $\forall n: \lVert f_n \rVert_1 = 1$ and $$\lVert \mathcal F f_n \rVert \to +\infty.$$ Of course I first thought about Gaussians but couldn't quite find a suitable sequence. Any help appreciated!

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    $\begingroup$ Some uncertainty principle thing going on here... $\endgroup$
    – copper.hat
    Jan 23, 2019 at 20:11

2 Answers 2

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You're right to consider Gaussians!

If you define the Fourier transform of a Schwarz function $f$ to be $$\mathcal F f(\xi)=\int_{\mathbb R^d}f(t)e^{-2i\pi\langle \xi, x\rangle}dx$$ then consider the family of Gaussian functions parametrized by $\sigma >0$ $$f_\sigma(t)= \frac 1 {(2\pi)^{\frac d 2}\sigma^d }e^{-\frac {\|x\|^2}{2\sigma^2}}$$ The corresponding Fourier transforms are $$\mathcal F f_\sigma(\xi)=e^{-2\pi\sigma^2\|\xi\|^2}$$ Now $$\|f_\sigma\|_1=\mathcal F f_\sigma(0) =1$$ while $$\|\mathcal F f_\sigma\|_1=\frac {C} {\sigma^d}$$ for some constant $C$.

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  • $\begingroup$ And this blows up as $\sigma \to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much! $\endgroup$
    – lasik43
    Jan 23, 2019 at 19:51
  • $\begingroup$ You're welcome! $\endgroup$ Jan 23, 2019 at 19:53
  • $\begingroup$ Minor typo, you meant to write $f_\sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $\sigma$ in $f_\sigma$ if you intended $\mathcal F f_\sigma$ to have this formula. (Conclusion is of course right) $\endgroup$ Jan 23, 2019 at 19:54
  • $\begingroup$ Note that you changed your Fourier transform definition to $$e^{-2i\pi\langle \xi, x\rangle} \int_{\mathbb R^d}f$$ $\endgroup$ Jan 23, 2019 at 20:09
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    $\begingroup$ Arghhh, need coffee... $\endgroup$ Jan 23, 2019 at 21:33
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Begin with $f=\mathbb1 _{[-1/2,1/2]}\in L^1(\mathbb R)$. This satisfies $$ \mathcal F f(\xi) = \frac{\sin(\pi \xi)}{\pi \xi} \notin L^1(\mathbb R)$$ Now take any $f_n \in \mathcal S $ converging to $f$ in $L^1$ and almost everywhere. We have $\|f_n\|_{L^1} < 2$ eventually. By dominated convergence, we have the pointwise convergence

$$\mathcal F f_n(\xi) = \int_{\mathbb R} f_n(x)e^{-2\pi i x\xi} dx \to \mathcal F f(\xi) $$ By Fatou's lemma, $$ \infty = \|\mathcal Ff \|_{L^1} \le \liminf_{n\to\infty}\|\mathcal Ff_n \|_{L^1}$$

For dimensions $d>1$, one can use $f = \mathbb 1_{[-1/2,1/2]^d} $.

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  • $\begingroup$ Thanks for this alternative approach! :) $\endgroup$
    – lasik43
    Jan 23, 2019 at 19:51
  • $\begingroup$ @bavor42 You're welcome :) $\endgroup$ Jan 23, 2019 at 19:56
  • $\begingroup$ @bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $\mathcal F \delta = 1 \notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up $\endgroup$ Jan 23, 2019 at 19:59
  • $\begingroup$ Yeah that makes sense. Still, 2 nice ways to look at it ;) $\endgroup$
    – lasik43
    Jan 23, 2019 at 20:06

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