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Terminology/Definition: The set $$ \Gamma(A):=\{(x,Ax):x\in\text{Dom}A\}\subseteq X\times Y $$ is called the graph of $A$. We define the norm in $X\times Y$ by $||(x,y)||:=||x||_X+||y||_Y$. We say that $A$ is a closed graph operator if $\Gamma(A)$ is a closed set in $X\times Y$. This means that whenever $x_n\in\text{Dom}A$, $x_n$ converges to $x$ and $Ax_n$ converges to $y$, then $x\in\text{Dom}A$ and $y=Ax$.

Currently I'm self studying functional analysis, namely closed graph operators. In the text, the other gives the following comment:

Let $A\colon X\to Y$ be a bounded operator between Banach spaces. It is clear that $A$ is a closed graph operator.

To be straightforward: it is not clear to me. Therefore, I worked it out, but I'm unsure if all my arguments are correctly justified. Here is my attempt:

Proof of comment. Let $x_n\in\text{Dom}A=X$ with $x_n\to x$ in $X$. Since $X$ is closed, $x\in X=\text{Dom}A$. Consider now the sequence $Ax_n$. Since $A$ is linear together with the continuity of the norm and the operator $A$, $$ \lim||Ax_n-Ax||= \lim||A(x_n-x)||= ||\lim A(x_n-x)||= ||A(\lim(x_n-x))||= ||A(0)||= ||0||= 0. $$ Therefore we have that $Ax_n\to Ax$.

Arctic Char
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Rough_Manifolds
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2 Answers2

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To show that $\Gamma(A)$ is closed, we will assume a sequence in $\Gamma(A)$ which converges to some element $(x, y)\in X\times Y$ and then show that $(x, y)$ is in $\Gamma(A)$.

This is more or less the same as what you do: if $(x_n, Ax_n) \to (x, y)$, then $x_n \to x$. From what you did, $Ax_n \to Ax$. Since $Ax_n$ also converges to $y$, $Ax = y$ since limit is unique. Thus $(x, y) = (x, Ax) \in\Gamma(A)$.

Remark This is indeed more general: for any continuous mapping $f:X\to Y$ between two topological spaces $X, Y$, where $Y$ is Hausdorff, the graph $$ \Gamma(f) = \{ (x, f(x)) \in X\times Y : x\in X\}$$ is a closed subset in $X\times Y$. See here

Arctic Char
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This is correct, but overcomplicated. We are given $A$ is continuous, and $x_n\to x$, so immediately $\lim Ax_n=A\lim x_n= Ax$.

K.Power
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