Prove $L^1(\mathbb{R})\cap L^2(\mathbb{R})$ has empty interior both in $L^1(\mathbb{R})$ and in $L^2(\mathbb{R})$

Question

I suppose it has interior, so we can find an open ball of radius epsilon in $L^1(\mathbb{R})\cap L^2(\mathbb{R})$ and so that ball is contained in both the sets, so we can write: $$B_\varepsilon=\{u(x)\in L^1(\mathbb{R}):||u(x)||_{L^1}<\varepsilon\}=\{u(x)\in L^2(\mathbb{R}):||u(x)||_{L^2}<\varepsilon\}$$ But now I can't find the contradiction. I tried to compute the norms but I can't end with a contradiction. I can't find it at all. I don't know if this is the right way.

Answer 1

Suppose $f$ is an interior point in $L^{2}$. Consider $f+\frac 1 {nx} I_{x>1}$. Show that this sequence converges to $f$ in $L^{2}$ and none of these functions are in $L^{1}$. This proves that interior in $L^{2}$ is empty. For interior in $L^{1}$ consider $f+\frac 1 {n\sqrt x} I_{0<x<1}$.