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By the closed graph theorem an operator $T$ is continuous (equivalently bounded) if and only if it its graph is closed. An operator with a closed graph is called a closed operator.

So we have

$$ T \ \text{bounded} \Longleftrightarrow T \ \text{continuous} \Longleftrightarrow T \ \text{has closed graph} \Longleftrightarrow T \ \text{closed}. \quad (*) $$

But I often see closed operators mentioned in the context of unbounded operators. That is unbounded operators can be closed. But in $(*)$ above it seems that a closed operator is equivalent to a bounded operator?

csss
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    The operator in the closed graph theorem is defined on a Banach. Unbounded closed operators are only defined on a dense subset of a Banach. You can always extend them linearly to the whole space. Then, you can deduce from the closed graph theorem, that there is no way to extend a closed unbounded operator to the whole space while keeping its graph closed. –  Jul 13 '18 at 13:14
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    @scentofthetrees You should post that as an Answer, for the benefit of future readers. (Heh, if you do please say "a Banach space" instead of "a Banach"...) – David C. Ullrich Jul 13 '18 at 13:38
  • @DavidC.Ullrich Since in conferences I say "a Banach" and everyone understands, I will keep using it. –  Jul 13 '18 at 13:46
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    @scentofthetrees Have you ever had an editor let you get away with "a Banach" in a publication? Just curious... – David C. Ullrich Jul 13 '18 at 14:05
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    @DavidC.Ullrich This is not a publication, just an internet forum, and you are certainly not my editor. –  Jul 13 '18 at 14:13
  • That said, in vast swathes of functional analysis, you absolutely have to mind your Banach spaces and Banach algebras… – Branimir Ćaćić Jul 13 '18 at 15:57
  • @scentofthetrees : It helps the site to show that questions get answered. I suggest that you paste your answer. I'll +1 it if you do. It's a good, clean answer. – Disintegrating By Parts Jul 13 '18 at 17:55
  • @BranimirĆaćić : Just not in this problem. – Disintegrating By Parts Jul 13 '18 at 17:58

1 Answers1

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Let $T : X\rightarrow Y$ be a linear operator from a Banach space $X$ to a Banach space $Y$. Then the closed graph theorem states that the following are equivalent

  • $T$ is continuous;

  • $T$ is bounded;

  • $T$ is closed.

A linear operator $T : \mathcal{D}(T)\subsetneq X\rightarrow Y$ on a linear, dense domain $\mathcal{D}(T)$ does not satisfy the hypotheses of the closed graph theorem. So $T$ can be closed without being continuous. An example is the differentiation operator $T : C^1[0,1]\subset C[0,1]\rightarrow C[0,1]$. This operator is linear, densely-defined and closed, but is not bounded.

Disintegrating By Parts
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