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The following theorem characterizes closable operators. Let $X$ and $Y$ be Banach spaces, let $\text{dom}(A)\subset X$ be a linear subspace, and let $A:\text{dom}(A)\to Y$ be a linear operator. Then the following are equivalent.

  1. $A$ is closable.
  2. The projection $\pi_X:\overline{\text{graph}(A)} \to X$ onto the first factor is injective.
  3. If $(x_n)_{n\in\mathbb{N}}$ is a sequence in $\text{dom}(A)$ and $y\in Y$ is a vector such that $\lim_{n\to\infty} x_n = 0$, and $\lim_{n\to\infty} Ax_n = y$, then $y=0$.

I have read a proof of this theorem where it shows 1. $\implies$ 3. $\implies$ 2 $\implies$ 1.

But I am looking to show that 1. $\implies$ 2. directly? Can this be done?

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$\newcommand{\dom}{\operatorname{dom}} \newcommand{\graph}{\operatorname{graph}}$ Suppose $A$ is closable. This mean that $A$ has a closed extension $B:\dom{B}\to Y$. Thus $$\graph(A)\subseteq \overline{\graph(A)}\subseteq\graph(B).$$ Consider the projection $\pi_X':\graph(B)\to X$. This projection is injective because $B$ a well-defined function. In other words $$\pi_X'(x,Bx)=\pi_X'(x',Bx') \implies x=x' \implies (x,Bx)=(x',Bx').$$ Therefore $\pi_X$ is injective as well. Indeed, the fact that that $\overline{\graph(A)}\subseteq \graph(B)$ implies that $$ \pi_X(x,y)=\pi_X(x',y') \implies \pi_X'(x,y)=\pi_X'(x',y') \implies (x,y)=(x',y').$$

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  • $\begingroup$ Great! Just one thing, how do we know for sure that $B$ is a well defined function? What statement can we make that formalizes this? $\endgroup$ – eurocoder Aug 19 '17 at 15:08
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    $\begingroup$ @eurocoder By the definition of $A$ being closable, there exists a closed extension $B$. This $B$ is a linear operator with closed graph that extends $A$. In particular, it is a function. Thus $x=x'$ implies $Bx=Bx'$. $\endgroup$ – John Griffin Aug 19 '17 at 15:11
  • $\begingroup$ Hmmm, so if we extend any linear operator we automatically have a well-defined function? $\endgroup$ – eurocoder Aug 19 '17 at 15:18
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    $\begingroup$ @eurocoder Every linear operator is a function. So if a linear operator $A$ extends to a linear operator $B$, then in particular $B$ is a function. $\endgroup$ – John Griffin Aug 19 '17 at 15:22
  • $\begingroup$ Ok thanks that clears everything up! $\endgroup$ – eurocoder Aug 19 '17 at 15:24

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