$\overline{L^2(\mathbb R)\cap L^1(\mathbb R)}^{L^2(\mathbb R)}=L^2(\mathbb R)$

Question

While reading a proof in a book they used the following result: $$ \overline{L^2(\mathbb R)\cap L^1(\mathbb R)}^{L^2(\mathbb R)}=L^2(\mathbb R) $$ saying that it's well known !! But all I can see is only one inclusion $\overline{L^2(\mathbb R)\cap L^1(\mathbb R)}^{L^2(\mathbb R)}\subset L^2(\mathbb R)$ which is trivial. Am I missing something because I can't see where the other inclusion comes from ? any help will be great thank you for your time.

Answer 1

The set $\mathbb L^1(\mathbf R)\cap\mathbb L^2(\mathbf R)$ contains the collection of linear combination of indicator functions of sets of finite measure (denoted $S$), which is dense in $\mathbb L^2(\mathbf R)$. Indeed, assume that $f$ belongs to $\mathbb L^2(\mathbf R)$ and $0\leqslant f(x)\leqslant M$. Then we can find a sequence $(s_n)\subset S$ such that $s_n\uparrow f$, $s_n\geqslant 0$ and $\sup_{x\in\mathbf R}|f(x)-s_n(x)|\to 0$. Therefore, for each $R$, $$\int_{\mathbf R}|f(x)-s_n(x)|^2\mathrm dx\leqslant \int_{\mathbf R\setminus [-R,R] }|f(x)|^2\mathrm dx +2R\left(\sup_{x\in\mathbf R}|f(x)-s_n(x)|\right)^2,$$ from which we infer that $$\limsup_{ n\to\infty}\int_{\mathbf R}|f(x)-s_n(x)|^2\mathrm dx\leqslant \int_{\mathbf R\setminus [-R,R] }|f(x)|^2\mathrm dx, $$ and since $R$ is arbitrary, we derive by monotone convergence that $f$ belongs to the closure in $\mathbb L^2(\mathbf R)$ of $S$.

Using again monotone convergence, one can show that each non-negative element of $\mathbb L^2(\mathbf R)$ belongs to the closure in $\mathbb L^2(\mathbf R)$ of $S$. To conclude, split an element of $\mathbb L^2(\mathbf R)$ into its negative and positive part.

Answer 2

If $f \in L^{2}$ is supported in $[-R,R]$, then it is also in $L^{1}$ because of Cauchy-Schwarz: $$ \int_{-R}^{R}|f|dx \le \left(\int_{-R}^{R}|f|^{2}dx\right)^{1/2} \left(\int_{-R}^{R}1^{2}dx\right)^{1/2} $$ Therefore, $f_{R} = \chi_{[-R,R]}f \in L^{1}\cap L^{2}$ for all $0 < R < \infty$. And $f_{R}\rightarrow f$ in $L^{2}$ as $R\rightarrow\infty$.