If both $f$ and $\hat{f}$ are in $L^1(\mathbb{R})$, then they are both in $L^2(\mathbb{R})$.

Question

I want to prove this statement, and my very naive idea is to use Schwarz inequality in some way. However it seems $\lVert \cdot \rVert_{L^2}$ is likely on the RHS of $\leq$.

Answer 1

Since $$\lVert \hat{f} \rVert_{L^2}\leqslant \lVert \hat{f} \rVert_{L^1}^{\frac{1}{2}} \lVert \hat{f} \rVert_{L^\infty}^{\frac{1}{2}}\leqslant \lVert \hat{f} \rVert_{L^1}^{\frac{1}{2}} \lVert f \rVert_{L^1}^{\frac{1}{2}},$$ we have $\hat{f}\in L^2(\mathbb{R}).$

A similar proof gives $f \in L^2(\mathbb{R}).$