Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

What does it mean to say that , A bounded linear operator is not "generally" bounded function. Can anybody explain ?

share|cite|improve this question
up vote 4 down vote accepted

A linear operator $T:V\to W$ is said to be "bounded" if there exists some $M\in\mathbb R$ such that $\|Tx\|\leq M\|x\|$ for all $x\in V$. A function $f:V\to W$ is said to be "bounded" if there exists some $M\in\mathbb R$ such that $\|f(x)\|\leq M$ for all $x\in V$. Note that this is a stronger condition than the previous one, as the upper bound for $\|T(x)\|$ varies with $x$ while the upper bound for $\|f(x)\|$ does not. In fact, any linear operator which is "bounded" as a function is simply $0$.

share|cite|improve this answer
So "generally" means "unless it is trivial". – timur May 23 '12 at 7:42

Let's choose our favorite bounded linear operator. At the moment, mine happens to be the identity operator $I$ on $\mathbb{R}$, a very boring bounded linear operator. If it's too boring, you can pretend I'm talking about the identity operator on $W^{k,p}$.

Then $||I||_{\text{operator}} = 1$, as clearly $||I(x)|| \equiv 1\cdot||x||$. And so it's a bounded linear operator. But $||I(x)||$ can itself be arbitrarily large, as $||x||$ can be arbitrarily large. Thus it's not a bounded function.

share|cite|improve this answer

A nonzero linear operator on any vector space (over $\mathbb R$ or $\mathbb C$) is never a bounded function: $\|A(tx)\| = |t| \|Ax\| \to \infty$ as $t \to \infty$ whenever $Ax \ne 0$.

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.