Why is $ \|T(x) \|\le \| T\| \|x \|$?

Question

Let $X$ and $Y$ be normed linear spaces and let $T$ be a bounded linear operator from $X$ to $Y$. The norm of $T$ is defined as $$\|T \|=\sup\{\|T(x) \|\;:\;\|x \|\le 1\}. $$

From the definition of the norm, we can say that $\|T(x) \|\le \|T \|\;\forall \;x,$ and that $\|T \|\|x \|\le \|T \|$ as $\|x\|\le 1$.

However I don't understand why we write $\|T(x)\|\le \|T\|\|x\|$?

You can write |T| instead of \Vert T \Vert. – Ragib Zaman May 27 '13 at 08:55 — Ragib Zaman, May 27 '13 at 08:55

score 7 · Accepted Answer · answered May 27 '13 at 08:46

Clearly $\Vert T(x) \Vert \le \Vert T \Vert$ if $\Vert x \Vert = 1$.

Now, just remember that we're dealing with linear operators.

So if $\Vert x \Vert = c$, then $\Vert \frac{x}{c} \Vert = 1$.

So $$\left\Vert T(x) \right\Vert = c \left\Vert T\left(\frac{x}{c} \right) \right\Vert \le c \Vert T \Vert$$

Why is $ \|T(x) \|\le \| T\| \|x \|$?

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