$\lVert f \rVert _{H^{s}(\mathbb{R}^d )} \leq C \lVert f \rVert_{L^{1} (\mathbb{R}^d )}$ cannot hold at $ s= -d/2$.

Question

How do I show that $\lVert f \rVert _{H^{s}(\mathbb{R}^d )} \leq C \lVert f \rVert_{L^{1} (\mathbb{R}^d )}$ cannot hold at $ s= -d/2$ ?

The definition is,

$\lVert f \rVert_{H^{s}} = \lVert \langle \nabla \rangle^{s}f \rVert_{L^2}$ and $\langle \nabla \rangle^{s} = (1+|\nabla|^2)^{\frac{s}{2}}$.

I tried there is a $f$ which is not $H^{s}$ but is $L^1$. I am gussing it is $\langle D \rangle ^{-s} \frac{1}{|x|^{d/2}}$. But I can't.

Answer 1

Idea: If you take $f(x) = \delta(x)$, then you see that $\hat f(\xi) = \text{const}$, say $1$. Hence we have \begin{align} \|f\|_{H^{-d/2}(\mathbb{R}^d)} = \int_{\mathbb{R}^d} \frac{d\xi}{(1+|\xi|^2)^{d/2}} \sim \int^\infty_0\frac{r^{d-1}dr}{(1+r^d)} = \infty. \end{align} However, we see that $\|f\|_{L^1(\mathbb{R}^d)} = 1.$ Thus the inequality fails for the delta function.

Spoiler:

Let $d=1$. Take the sequence $f_N(x) = N(2\pi)^{-1/2}\exp(-N^2x^2/2)$