# Showing that if $A$ is a closable linear operator, then $\pi_X:\overline{\text{graph}(A)} \to X$ is injective?

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?

• 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? – eurocoder Aug 19 '17 at 15:08
• @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 $Bx=Bx'$. – John Griffin Aug 19 '17 at 15:11
• @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. – John Griffin Aug 19 '17 at 15:22