Closed Graph Theorem in Royden-Fitzpatrick

Question

The Closed Graph Theorem Let $T : X \to Y$ be a linear operator between spaces $X$ and $Y$. Then $T$ is continuous if and only if it is closed

Here is the proof:

It is clear that $T$ is closed if it continuous. To prove the converse, assume $T$ is closed. Introduce a new norm $||\cdot||_*$.on $X$ by $$||x||_* = ||x|| + ||T(x)||$$ for all $x \in X$. The closedness of the operator $T$ is equivalent to the completeness of the normed linear space $(X,||\cdot||_*)$. On the other hand, we clearly have $$||\cdot|| \le ||\cdot||_*$$ on $X$. Since $(X,||\cdot||_*)$ and $(X,||\cdot||)$ are Banach spaces it follows from the preceding corollary that there is a $c \ge 0$ for which $$||\cdot||_* \le c ||\cdot ||$$ on $X$. Thus for all $x \in X$, $$||T(x))|| \le ||x|| + ||T(x)|| \le c||x||.$$ Therefore $T$ is bounded and hence is continuous.

Why is it true that if $(X,||\cdot||_*)$ is complete, then $T$ is closed? I was able to prove the converse, but I don't see why this implication is true. Also, why is $(X,||\cdot||_*)$ Banach, i.e., why is it complete?

Answer 1

For $x_{n}\rightarrow x$ and $Tx_{n}\rightarrow y$, we are to show that $y=Tx$, this means $T$ is closed. It is now assumed that $(X,\|\cdot\|_{\ast})$ being complete, and we also see that $\|x_{n}-x_{m}\|_{\ast}=\|x_{n}-x_{m}\|+\|Tx_{n}-Tx_{m}\|\rightarrow 0$ as $n,m\rightarrow\infty$, the completeness of $(X,\|\cdot\|_{\ast})$ gives some $x'\in X$ such that $\|x_{n}-x'\|\rightarrow 0$, but then $\|x_{n}-x'\|\rightarrow 0$, so $x=x'$. As a result, $\|Tx_{n}-x\|=\|Tx_{n}-x'\|\rightarrow 0$, so $y=Tx$.

Answer 2

You have to assume that $X$ and $Y$ are Banach spaces. Without completeness of these spaces the result is false.

Completeness of $(X,\|.\|_{*})$: If $(x_n)$ is Cauchy in this space then $\|x_n-x_m\| \to 0$ as $n, m \to \infty$ and $\|Tx_n-Tx_m\| \to 0$ as $n, m \to \infty$. By completeness of $X$ and $Y$ we see that there exists $x$ and $y$ with $x_n \to x$ and $Tx_n \to y$. Now $(x_n,Tx_n)$ is a sequence in the graph of $T$ converging to $(x,y)$. Since the graphs is closed, $(x,y)$ belongs to the graph which means $y=Tx$. Now you can see from the definition of $\|.\|_{*}$ that $\|x_n-x\|_{*} \to 0$.

The other question has been answered by user284331