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Let $A$ be a linear operator on $L$ with domain $D(A)$. To show that $A$ is closable, i.e., it has a closed linear extension, why does it suffice to show that if $\{f_n\}\subset D(A), f_n\to 0,$ and $Af_n\to g\in L,$ then $g=0$? I would greatly appreciate any help.

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So we say that an operator is closed if it has a closed graph. This means that given an unbounded operator $T \in \mathcal{B}(H,K)$ that $\{x,Tx)\}$ is closed in $H \times K$. This explicitly means that given a sequence $x_n \to x $ in H such that $Tx_n \to y$ in K that $y=Tx$. Now, if you can show that this holds at a point in H then you can extend to the whole domain of T in the typical way.

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Given any operator $T$ on $L$, denote by $D(T)$ its domain and $\Gamma(T)$ its graph, i.e.

$$\Gamma(T)=\{(x,Tx):x\in D(T)\}.$$

Note that $\Gamma(T)$ is a subspace of $L\times L$, and $T$ is closed if and only if $\Gamma(T)$ is a closed subspace. With that in mind, $T$ is closable if and only if $\overline{\Gamma(T)}=\Gamma(T')$ for some operator $T'$ on $L$ (and clearly $T'$ must be closed). I will prove this upon request, but it is a good exercise.

Back to the problem at hand: assume that $\{f_n\}\subset D(A)$, $f_n\to0$, $Af_n\to g$ implies $g=0$. We want to show $\overline{\Gamma(A)}=\Gamma(A')$ for some operator $A'$ on $L$. Let $(f,g),(f,h)\in\overline{\Gamma(A)}$; it remains to show $g=h$. There exist sequences $\{(f^1_n,g_n)\},\{(f^2_n,h_n)\}\subset\Gamma(A)$ such that $(f_n^1,g_n)\to(f,g)$ and $(f_n^2,h_n)\to(f,h)$. Hence $f_n^1-f_n^2\to0$ and $A(f_n^1-f_n^2)=g_n-h_n\to g-h$. By assumption, this implies $g-h=0$, and we are done.

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