Is $L^1(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$ dense in $L^2(\mathbb{R}^n)$?

163 Views Asked by At

I'm reading Partial Differential Equations, by L. Evans, and in section $4.3$ is a definition of the Fourier transform for functions in $L^2(\mathbb{R}^n)$. He starts considering a function $u\in L^2(\mathbb{R}^n)$, and a sequence $\{u_k\}_{k=1}^\infty \subset L^1(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$, with $u_k\rightarrow u$ in $L^2(\mathbb{R}^n)$.

So my question is: why there exists such sequence $\{u_k\}_{k=1}^\infty$ for every function in $L^2(\mathbb{R}^n)$? Is $L^1(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$ dense in $L^2(\mathbb{R}^n)$? Thank you!