Let $B(X,Y)$ be the family of all bounded maps from $X$ to $Y.$ Then, we have for arbitrary $T\in B(X,Y), $ \begin{align} \Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert= \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert=\sup\limits_{x\neq 0} \frac{\Vert T x \Vert}{\Vert x \Vert}.\end{align}


Since $T$ is bounded and linear, $\exists K\geq 0$ such that for all $x\in X,\;\Vert T x \Vert \leq K \Vert x \Vert.$ If $\Vert x \Vert\leq 1$, then $\Vert T x \Vert \leq K \Vert x \Vert\leq K.$

Thus, \begin{align}\tag{1}\label{1} \sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert\leq \inf\{ K\geq 0:\Vert T x \Vert \leq K \Vert x \Vert,\forall\;x\in K \}=\Vert T \Vert\end{align} By definition of $\inf,$ for every $\epsilon> 0,\exists \;x_{\epsilon}\in X,x_{\epsilon}\neq 0$ such that \begin{align} \Vert T x_{\epsilon} \Vert>\left(\Vert T \Vert -\epsilon\right)\Vert x_{\epsilon} \Vert.\end{align} Let $u_{\epsilon}=\frac{x_{\epsilon}}{\Vert x_{\epsilon} \Vert},$ then $u_{\epsilon}=1$ and $ \Vert T u_{\epsilon} \Vert>\Vert T \Vert -\epsilon.$ We obtain from $ \eqref{1} $

\begin{align} \tag{2}\label{2}\Vert T \Vert \geq\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert\stackrel{\text{how?}}{\geq} \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert\stackrel{\text{how?}}{\geq} \sup\limits_{\Vert x_{\epsilon} \Vert\neq 0} \Vert T\left(\frac{x_{\epsilon} }{\big \Vert x_{\epsilon} \Vert}\right)\big\Vert\stackrel{\text{how?}}{\geq}\Vert T \Vert -\epsilon.\end{align} Since $\epsilon>0$ was arbitrary, then \begin{align} \Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert= \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert=\sup\limits_{x\neq 0} \frac{\Vert T x \Vert}{\Vert x \Vert}.\end{align} Can you please explain the how's in $ \eqref{2} ?$


2 Answers 2


Answering your "how" questions regarding the inequalities:

The first one is because you're taking the sup over a smaller set. When $A \subset B$, and $f$ is a real valued map whose domain contains $A$ and $B$, we have $$\sup_{x \in A} f(x) \le \sup_{x \in B} f(x).$$ Here $B = \{x \, : \, \|x\| \le 1\}$ and $A = \{ x \, : \, \|x \| = 1\}$.

The second is again just restricting to a smaller set. For each $\epsilon$, the vector $x_\epsilon/\|x_\epsilon\|$ has norm $1$, so if we only consider the sup over those vectors, we are again considering a smaller set, and thus get the inequality for the same reason as above.

The third is by the definition of $x_\epsilon$. We've chosen $x_\epsilon$ such that $$\left \| T\left(\frac{x_\epsilon}{\| x_\epsilon\|} \right)\right \| \ge \|T \| - \epsilon.$$

  • $\begingroup$ Thanks, I am grateful! $\endgroup$ Dec 17, 2018 at 21:50
  1. Since $\bigl\{\lVert Tx\rVert\,|\,\lVert x\leqslant1\bigr\}\supset\bigl\{\lVert Tx\rVert\,|\,\lVert x=1\bigr\}$, $\sup\bigl\{\lVert Tx\rVert\,|\,\lVert x\leqslant1\bigr\}\leqslant\sup\bigl\{\lVert Tx\rVert\,|\,\lVert x=1\bigr\}$
  2. Because $\left\lVert\dfrac{x_\varepsilon}{\lVert x_\varepsilon\lVert}\right\rVert=1$.
  3. It was proved before that $\left\lVert T\left(\dfrac{x_\varepsilon}{\lVert x_\varepsilon\lVert}\right)\right\rVert\geqslant\lVert T\rVert-\varepsilon$.
  • $\begingroup$ Thanks, I am very grateful for the help! $\endgroup$ Dec 17, 2018 at 21:49
  • $\begingroup$ I'm glad I could help. $\endgroup$ Dec 17, 2018 at 21:51

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