$L^2$ Compactness with the Fourier Transform, Fréchet-Kolmogorov-Riesz theorem

Question

I'm having difficulties proving the following.

Let $G \subset L^2(\mathbb R^d)$ be bounded, meaning $\sup_{f \in G} \|f\|_2 <\infty$. Then the following two are equivalent:

a) $lim_{R\to \infty}\sup_{f \in G} \|\mathcal F (f)\|_{L^2(\mathbb R^d\setminus B_R(0))}=0$;

b) $lim_{y\to 0}\sup_{f \in G} \|f(\cdot +y)-f(\cdot)\|_2=0$.

where $\mathcal F (f)(y)=\frac 1 {(2 \pi)^{d/2}} \int _{\mathbb R^d}f(x)e^{-iy\cdot x}dx$ is the fourier transform of $f$.

I can show that a) implies b) using the identity $$\|f(\cdot +y)-f(\cdot)\|_2=\|(e^{ix\cdot y}-1)\mathcal F(f)(x)\|_2$$ but I'm stuck trying to prove the converse. Any help would be greatly appreciated. Thanks!

Answer 1

While I searched the internet, I found a paper Pego (1985). Your statement is the same as Theorem 1 in that paper.

The proof uses a Gaussian function to connect (b). The proof is readable.