# Prove $\|T\| = \sup_{\|x\| < 1} \|Tx\|$

Let $$X, Y$$ be Banach spaces. And $$T \in B(X\rightarrow Y)$$. Prove that $$\|T\| = \sup_{\|x\| < 1} \|Tx\|$$

## Discussion

Having trouble seeing how to handle some of these ideas below. Please let me know if there's a simpler way or if I'm on the right track.

## Attempt

Since $$X, Y$$ are Banach they are normed spaces. Since $$T$$ is a bounded linear operator (I think that's what that notation means) in a normed space, it is continuous. Recall that by definition of a norm on an operator, the norm of $$T$$ is

$$\|T\| = \sup_{x \in X} \frac{\|Tx\|}{\|x\|}, \quad x\neq 0 \tag{\star}$$

Now, let $$(x_n)$$ be Cauchy in $$X$$. Since $$X$$ is Banach, $$(x_n)$$ is convergent to some $$x \in X$$. Note that $$\|x_m - x_n\|< \epsilon\,$$ for all $$m,n > N(\epsilon)$$ implies

$$\left| \|x_m\| - \|x_n\| \right| < \epsilon$$

and therefore

\begin{align} \frac{\|x_n\|}{\|x_m\|} \longrightarrow1 \quad &\text{as} \quad n,m \longrightarrow +\infty \\ \text{and} \\ \bigg\|\frac{x_n}{\|x_m\|}\bigg\| = \|z_{nm}\| \longrightarrow 1 \quad &\text{as} \quad n,m \longrightarrow + \infty \tag{1} \end{align}

But to make our argument align with the proposition, recall that every convergent sequence of real numbers has a monotone subsequence that converges; $$(\|z_{mn}\|)$$ is such a sequence. So take $$(x_{n_k})$$ to be a subsequence of $$(x_n)$$, where $$n_k$$ correspond to the indices that generate a monotone subsequence of $$(\|z_{mn}\|)$$. Then for all $$n_k > N(\epsilon$$), we have that $$(1)$$ holds and we can choose either $$m = n_k, n = n_{k+1}$$ or vice versa depending on whether the sequence defined by $$\|x_{m =n_k}\|$$ is decreasing or increasing in order to keep $$\|z_{mn}\| < 1$$. Then by $$(\star)$$

\begin{align} \|T\| = \sup_{x \in X} \frac{\|Tx\|}{\|x\|} &= \sup_{x \in X} \frac{\|T\left(\lim_{n\to\infty}x_n\right)\|}{\|\lim_{m\to\infty}x_m\|} \\ \\ &= \sup_{x \in X} \frac{\lim_{n\to\infty}\|Tx_n\|}{\lim_{m\to\infty}\|x_m\|} \tag{2} \\ \\ &= \sup_{x \in X} \lim_{n,m\to\infty} \bigg\|T\left(\frac{x_n}{\|x_m\|}\right)\bigg\| \tag{3} \\ \\ &= \sup_{\|z_{mn}\|< 1} \lim_{n,m\to\infty} \|Tz_{mn}\| \end{align}

where $$(2)$$ is justified by continuity of the norm and continuity of $$T$$, and $$(3)$$ is justified by linearity of the norm.

But now I don't know how to finish it off (if it's even right). Also, the proof just feels bad to me. Thanks for the help!

• Your formula is not true in $X = \{0\}$. – gerw Mar 20 at 6:48

$$\|Tx\| \leq \|T\|\|x\| \leq \|T\|$$ if $$\|x\|<1$$. Thus RHS $$\leq$$ LHS. For the other way take any $$x \neq 0$$ and consider $$y=\frac x {(1+\epsilon) ||x\|}$$. Then $$\|y\| <1$$ so $$\|Ty\| \leq \,$$ RHS. Taking sup over $$x$$ we get $$\|T\| \leq (1+\epsilon) \times$$ RHS. Let $$\epsilon \to 0$$.