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Let $U, V$ be separable Banach spaces.

Suppose we have a bounded, linear operator $C : U\to V$.

Questions are the following

  • Shall $C$ be continuous since $V$ is a Banach space?
  • In general, is a bounded linear operator necessarily continuous (I guess the answer is no, but what would be a counter example?)
  • Are things in Banach spaces always continuous?
HelloWorld
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user106357
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    Bounded linear operators are continuous. (Think about how Lipschitz condition implies uniform continuity for functions on real line). Things in Banach spaces aren't always continuous though. –  Nov 08 '13 at 08:26
  • actually why don't we directly learn: "bounded linear operators" equivalent to "Lipschitz continuous one", but only equivalent to continuous ones? – Noix07 Jan 05 '16 at 12:31
  • @user27126: Is every continuous operator bounded? I know the answer is yes in finite-dimensional spaces, but how about in infinite-dimensional ones? I am thinking maybe, e.g., the differentiation operator which I know is unbounded is also continuous? – MSIS Feb 01 '21 at 05:53
  • Differentiation is not necessarily continuous, consider the space of smooth 2 $\pi$ periodic functions and the sequence sin(nx)/x and the sup-norm. – Taleofwoe Jan 15 '22 at 23:13

2 Answers2

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An operator $C$ is bounded iff the set {$\|Cx\|:\|x\|\leq 1$} is bounded $\Leftrightarrow$ there is a $M<\infty:\|Cx\|\leq M\|x\|$ for every $x\in U$.

Let $ε>0$. If $x,y\in U:\|x-y\|<ε/M$, then $$\|Cx-Cy\|=\|C(x-y)\|\leq M\|x-y\|<\varepsilon.$$ Thus $C$ is not only continuous but uniformly continuous also.

So, a bounded operator is always continuous on norm-spaces. Banach space is a norm-space which is complete, thus things are not different there.

jII
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Haha
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This property is unrelated to the completeness of the domain or range, but instead only to the linear nature of the operator. Yes, a linear operator (between normed spaces) is bounded if and only if it is continuous.

Added @Dimitris's answer prompted me to mention, beyond the fact that the implication on normed spaces indeed is an equivalence, that it's the converse which holds in the wider context of topological vector spaces, while the proposition mentioned here fails: there are bounded discontinuous linear operators, yet every continuous operator remains bounded.

Community
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Jonathan Y.
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  • Thanks Jonathan and Sanchez.In this case, so instead of saying "a bounded linear operator $C: U\to V$", one can say $C\in\mathcal{L}(U,V)$, where $\mathcal{L}(U,V)={C: U\to V,,|,, C,,\text{is linear and continuous}}$? – user106357 Nov 08 '13 at 08:39
  • You may, only I've only seen this notation used as a synonym of $\mathrm{Hom}(U,V)$, the space of all linear maps between them. Perhaps another notation would serve you better. – Jonathan Y. Nov 08 '13 at 08:42
  • $\text{Hom}(X,Y)={C: U\to V,,|,C,,\text{is linear}}$, which means $C$ is not necessarily continuous. In finite dimensions, the two notations are the same. But this is not true in infinite dimensions. Correct me if I'm wrong – user106357 Nov 08 '13 at 08:50
  • You're correct. I was saying I've only seen $\mathcal{L}(U,V)$ being used as an alternate notation to $\mathrm{Hom}(U,V)$. – Jonathan Y. Nov 08 '13 at 08:59
  • The standard notation that I have encountered is $B(U,V)$, meaning the space of bounded linear operators that map the entirety of the linear space $U$ into some/all of the linear space $V$. This is identical to the space of continuous linear operators $U\rightarrow V$. When $U=V$ then it is simply notated $B(V)$. – Stromael Nov 09 '13 at 21:21