the graph of any continuous map is closed on Banach spaces.

Question

$X$ and $Y$ are Banach spaces. Prove that the graph of any continuous map $T: X \to Y$ is closed.

Is below a correct solution?

We have to show that any sequence $(x_n,y_n)$ in $\mathcal{G}(T) = \{(x,y) : x\in X \ \ \& \ \ \ Tx=y\}$ converges to a point $(x,y)$ in $\mathcal{G}(T)$.

Let's take a sequence $(x_n,y_n)$ in $\mathcal{G}(T)$. Since $X\times Y$ is Banach, we have that $(x_n,y_n) \to (x,y)$ in $X\times Y$. It remains to show that $Tx=y$.

We have that by completeness of $X$, that $x_n \to x$ for an $x \in X$, and by the continuity of $T$, $Tx_n \to Tx$, also we have that $Tx_n \to y$. So choose $N$ s.t. for all $n\geq N$, $\|Tx_n - Tx\|<\frac{\epsilon}{2}$ and $\|Tx_n - y\|<\frac{\epsilon}{2}$. So

$$\|Tx-y\|\le\|Tx_n - Tx\|+\|Tx_n - y\|<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$

Answer 1

Your proof needs some refinement.

In this proof, you don't need that $X$ and $Y$ are Banach spaces, it suffices that they are normed spaces. Now you take a sequence $(x_n, Tx_n)$ in $\mathcal{G}(T)$ and assume that this sequences converges to $(x, y) \in X \times Y$. What we have to show is $(x, y) \in \mathcal{G}(T)$. By definition of the convergence, it is clear that $x_n \to x$ and $Tx_n \to y$. But since $T$ is assumed to be continuous, $Tx_n$ converges to $Tx$. Thus $y = Tx$ and the graph is closed.

Answer 2

Your proof is messy.

We want to prove that if $\mathcal G(T) \ni(x_n, y_n) \to (x,y)$, then $(x,y)\in\mathcal G(T)$.