Let the domain of the Laplacian $D(\Delta)\subset L^2(\mathbb{R}^n)$ be given by $$D(\Delta)=\{\psi\in L^2(\mathbb{R}^n):|k|^2\hat\psi(k)\in L^2(\mathbb{R}^n)\}$$ where $\hat\psi$ denotes the Fourier transform of $\psi$. Assume also that $n\le3$. Show that each $\psi\in D(\Delta)$ is continuous.
There is a hint that says to show that $\hat\psi$ is $L^1$ by writing it as a product of $L^2$ functions (i.e. use Holder's inequality). My first thought was to use the fact that $|k|^2\hat\psi(k)\in L^2(\mathbb{R}^n)$. However, writing $$\int_{\mathbb{R}^n}|\hat\psi(k)|dk=\int_{\mathbb{R}^n}\bigg|\frac{1}{|k|^2}|k|^2\hat\psi(k)\bigg|dk$$ we run into the problem that $\frac{1}{|k|^2}$ is not in $L^2(\mathbb{R}^n)$. So how does one show that $\hat\psi\in L^1(\mathbb{R}^n)$?
I think I have a solution. Since the fourier transform is unitary, we have $\|\hat\psi\|_2=\|\psi\|_2$. Thus, $\hat\psi\in L^2(\mathbb{R}^n)$. So $$\int_{\mathbb{R}^n}|\hat\psi(k)|dk=\int_{\mathbb{R}^n}\frac{1}{\sqrt{1+k^4}}\sqrt{1+k^4}|\hat\psi(k)|dk$$ $$\le\int_{\mathbb{R}^n}\frac{1}{1+k^4}dk\int_{\mathbb{R}^n}(1+k^4)|\hat\psi(k)|^2dk$$ $$=(\|\hat\psi\|_2^2+\|k^2\hat\psi(k)\|_2^2)\int_{\mathbb{R}^n}\frac{1}{1+k^4}dk<\infty$$