The graph of an injection from $L^2$ to $L^1$ is closed Assume $L^2\subset L^1$. Consider $T$ a linear injection from $L^2$ to $L^1$, how can I prove that the graph of $T$, $G(T)=\{(f,Tf):f\in L^2\}$ is closed? I know that the following are equivalent:


*

*G(T) is closed.

*$\forall$ sequence $(f_n)_n$ in $L^2$, if $f_n\rightarrow f$ and $Tf_n\rightarrow g$, then $Tf=g$.

*$\forall$ sequence $(f_n)_n$ in $L^2$, if $f_n\rightarrow 0$ and $Tf_n\rightarrow g$, then $g=0$.


But I don't know how to proceed then. Any hint or help is appreciated.
Initially I also assumed $\mu$ is $\sigma$-finite, which is redundant as pointed below by @p4sch.
 A: Note that by the closed graph theorem this is equivalent to the continuity of $T$. Moreover, the condition $L^2 \subset L^1$ already implies that $\mu$ have to be a finite measure.
Proof: Assume that $\mu$ is not finite. Let $(A_n)_{n \in \mathbb{N}}$ be sequence of disjoint measurable sets with $\Omega = \bigcup_{n=1}^\infty A_n$ and $\mu(A_n) >1$. (That's possible, since $\mu$ is not finite, but $\sigma$-finite.) Define
$$f= \sum_{n=1}^\infty \frac{1}{\mu(A_n) n} 1_{A_n},$$
then $f \in L^2$, because $\|f\|_2^2 = \sum_{n=1}^\infty \frac{1}{\mu(A_n) n^2} \leq \sum_{n=1}^\infty \frac{1}{n^2} <\infty$, but $\|f\|_1 = \sum_{n=1}^\infty \frac{1}{n} = \infty$.
Thus, we know that $\mu$ is finite and by Hölder-inequality $\|f\|_1 \leq \mu(\Omega)^{1/2} \|f\|_2$. 

However, there are injective linear maps $T \colon L^2 \rightarrow L^1$, which are not continuous, provided that $L^2$ is infinite dimensional.

Let $(e_i)_{i \in I}$ be a Hamel basis of $L^2$ with $\|e_i\|_2=1$. Define $T(e_i) = r_i \|e_i\|_1^{-1} e_i $, where $r_i \neq 0$ and $r_{i_n} = n$ for some countable subset $(i_n) \subset I$. Now extend $T$ by linearity to $L^2$. Since $(e_i)$ is a basis, we see that $T$ is injective. Now $\|T(e_{i_n})\|_1 = n$, but $\|e_{i_n}\|_2=1$. Thus $T$ is not continious.
