Prove that $\Vert T \Vert =\sup\limits_{\Vert x \Vert< 1}\Vert T x \Vert,\;\forall\;T\in B(X,Y).$

Question

Prove that $\Vert T \Vert =\sup\limits_{\Vert x \Vert< 1}\Vert T x \Vert,\;\forall\;T\in B(X,Y).$

Trial

It suffices to show that $\sup\limits_{\Vert x \Vert< 1}\Vert T x \Vert=\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert,$ since we have shown that $\Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert.$

Since $\{\|Tx\|:\|x \|<1 \}\subseteq \{\|Tx\|:\|x \|<1 \}$, then
$$\sup\limits_{\Vert x \Vert< 1}\Vert T x \Vert\leq \sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert.$$ How then, can I show the other side

$$\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert\leq \sup\limits_{\Vert x \Vert< 1}\Vert T x \Vert ?$$ Thanks for your time and help.

Answer 1

Suppose the supremum of $||T(x)||$ when $||x||\leq 1$ is $M$. It is enough to show that for each $\epsilon>0$ there is $x\in X$ such that $||x||<1$ and $||T(x)||>M-\epsilon$. So let $\epsilon>0$. We know there is $y\in X$ such that $||y||\leq 1$ and $||T(y)||>M-\frac{\epsilon}{2}$. If $||y||<1$ then we are done. Else, $||y||=1$. In that case pick a sequence $y_n\in X$ such that $||y_n||<1$ for each $n\in\mathbb{N}$ and $y_n\to y$. Since $T$ is a bounded operator it is continuous, hence $T(y_n)\to T(y)$. Any norm is continuous so $||T(y_n)||\to ||T(y)||$. So pick $x$ from the sequence such that the distance between $||T(x)||$ and $||T(y)||$ is smaller than $\frac{\epsilon}{2}$, and that's it.

Answer 2

Let $\|x\| = 1$ and note $x_n := (1 - \frac{1}{n})x$ converges to $x$. I think you can then show $\|T x_n\| \to \|T x\|$.