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

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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 $0$.

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So "generally" means "unless it is trivial". –  timur May 23 '12 at 7:42
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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.

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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 $Ax \ne 0$.

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