Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
No, it's no true for arbitrary linear maps. For linear maps between normed spaces this property is equivalent to continuity. See… – 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
up vote 6 down vote accepted

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).

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.