# Fourier transform $\mathcal F \colon (\mathcal S(\mathbb R^d), \lVert \cdot \rVert_1) \to L^1(\mathbb R^d)$ unbounded?

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!

• Some uncertainty principle thing going on here... Jan 23, 2019 at 20:11

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 $$C$$.

• 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! Jan 23, 2019 at 19:51
• You're welcome! Jan 23, 2019 at 19:53
• 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) Jan 23, 2019 at 19:54
• Note that you changed your Fourier transform definition to $$e^{-2i\pi\langle \xi, x\rangle} \int_{\mathbb R^d}f$$ Jan 23, 2019 at 20:09
• Arghhh, need coffee... Jan 23, 2019 at 21:33

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 $$f = \mathbb 1_{[-1/2,1/2]^d}$$.

• Thanks for this alternative approach! :) Jan 23, 2019 at 19:51
• @bavor42 You're welcome :) Jan 23, 2019 at 19:56
• @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 Jan 23, 2019 at 19:59
• Yeah that makes sense. Still, 2 nice ways to look at it ;) Jan 23, 2019 at 20:06