Is $L^1(X) \cap L^2(X)$ a closed subspace of $L^2(X)$ and $L^1(X)$? Suppose that $X$ be a locally compact Hausdorff space. Could we say that $L^1(X)\cap L^2(X)$ is closed subspace of $L^1(X)$ and $L^2(X)$?
 A: It's not. Consider $f_{n}(x)=\frac{1}{x^{1+1/n}}\in L^1([1,\infty))\cap L^2([1,\infty))\subset L^2([1,\infty))$. Then:
$$\|f_n(x)-\frac1x\|_2=\int_1^{\infty}\left(\frac{1}{x^{1+1/n}}-\frac{1}{x}\right)^2dx=\frac{1}{2/n+1}-\frac{2}{1/n+1}+1\to 0$$ as $n\to\infty$. However $\frac1x\not\in L^1([1,\infty))$.
A: Actually, for any locally compact Hausdorff $X$, and Radon measure $\mu$ on $X$,  the set of  all continuous compactly supported function $C_c(X)$ is dense in $L^p(X)$ for all $1\leq p<\infty$ (theorem 3.14 of "Real and complex analysis" by Walter Rudin). Hence $C_c(X)\subset L^1(X) \cap L^p(X)$, and so $L^1(X) \cap L^p(X)$ is dense in $L^p(X)$, for all $1\leq p<\infty$.( when $X=N$ , $\{\frac{1}{n}\}_{n\in N}\not\in\ell^1(N)$ but $\{\frac{1}{n}\}_{n\in N}\in\ell^2(N)$. Hence $\ell^1(N)\cap\ell^2(N)\neq \ell^2(N)$, consequently $\ell^1(X)\cap\ell^2(X)$ is not closed in $\ell^2(X)$, becuase $\{\frac{1}{n}\}_{n\in N}\in\overline{\ell^1(X)\cap\ell^2(X)}^{\|.\|_2}$ and $\{\frac{1}{n}\}_{n\in N}\not\in\ell^1(X)\cap\ell^2(X)$. Similarly $\overline{\ell^1(X)\cap\ell^2(X)}^{\|.\|_1}=\ell^1(X)\neq \ell^1(X)\cap\ell^2(X)$)
