# Why does $\lVert L(x) \rVert \leq \lVert L \rVert\,\lVert x \rVert$?

Why does $\lVert L(x) \rVert \leq \lVert L \rVert\,\lVert x \rVert$?

If $L$ is a linear map between Banach spaces $V$ and $W$, why is this true? Also, is this true for $L$ not a linear map?

Thanks!

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No, it's no true for arbitrary linear maps. For linear maps between normed spaces this property is equivalent to continuity. See en.wikipedia.org/wiki/… – Martin Sleziak Nov 25 '11 at 23:05
Thanks for clarifying! – badatmath Nov 25 '11 at 23:10
Now after seeing Glougloubarbaki's answer I see that I commented too fast without thinking. What I meant was that boundedness (i.e. the existence of a constant $C$ such that $\lVert L(x) \rVert \le C \lVert x \rVert$) is equivalent to continuity, if we work with linear map $L$. – Martin Sleziak Nov 26 '11 at 12:06

It is true even in the case of Banach spaces.

Indeed, recall the definition of $\|L\|$ : $$\|L\| = \sup_{\|x\|_V =1} \|L x\|_W$$ so that if $\|L\|$ is finite (which however needs not be the case in infinite dimension) then for all non-zero $x \in V$, if we let $t=\|x\|_V$ and $u = x / t$ (of norm 1) then : $$\|L x\|_W = t \|L u\|_W \leq t \|L\|= \|L\| \|x\|_V$$ by definition of $\|L\|$.

This is completely false for $L$ non linear even in finite dimension as we crucially use $L$ linearity for $L(tx) = t L(x)$.

In the case where $\|L\|$ is infinite then the inequality is technically true but not very useful. $L$ is continuous if and only if $\|L\|$ is finite (and in this case $L$ is called bounded which should not be confused with actual boundedness on all of $V$ (obviously for linear maps only the null map is bounded).

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Thank you, that was helpful! – badatmath Nov 25 '11 at 23:23

Since $\|L\|$=inf {$k:\|L(x)\|\le k\|x\|, \text{for all } x \in V$}, it follows that whatever value of $k$, we have $\|L\|\le k$.

Thus, $\|L(x)\|\le ||L\|\|x\|$ for all $x\in V$.

This is not true in general if $L$ is not a linear map.

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