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!
Best Answer
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 $C$.