Why $L^2\cap L^p $ is not dense in $L^{\infty}$?

Question

Ein Euclidean space $\mathbb{R}^n$. Why $L^2\cap L^p$ is not dense in $L^{\infty}$? I have that $L^2\cap L^p$ is dense in $L^p$ with $1\leq p<\infty.$ Indeed, for $g\in L^p$ with $1\leq p<\infty$, let $(g_j)_{j}$ with $g_j=\rho_j*g$ where $\rho_j$ is a mollifier function. By definition of molllifier function, for all $j$, $\rho_j\in \mathcal{C}_{c}^{\infty}$. Therefore, $\rho_j\in L^q$ for any $q\in [1,\infty[$. In particulary, $\rho_j\in L^1\cap L^{2p/(3p-2)}$. Because $\rho_j\in L^1$, by Young's inequality, $|g|_{p}=|\rho_j*g|_{p}\leq |\rho_j|_{1}|g|_{p}$. Similary, $|g_j|_{2}\leq |\rho_j|_{2p/(3p-2)}|g|_{p}$. Therefore $g_j\in L^2\cap L^p.$ Now, $g_j\to g \in L^p$. Conclusion, $L^2\cap L^p$ is dense in $L^p$.

Why $L^2\cap L^p$ is not dense in $L^{\infty}$?

Answer 1

Let $f(x)=1$ for all $x$. Suppose there exists $g \in L^{2}$ such that $\|f-g\|_{\infty} <\frac 1 2$. Then $|f(x)-g(x)| <\frac 1 2$ for almost all $x$ so $|g(x)| > \frac 1 2$ for almost all $x$. Hence $\int |g|^{2} =\infty$ which is a contradiction.