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


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.


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!

  • 1
    $\begingroup$ Your formula is not true in $X = \{0\}$. $\endgroup$ – 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 $\epsilon \to 0$.


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