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This is a problem our teacher gave us and I have a feeling he forgot to mention some additional data.

Let $H$ be a Hilbert space and $T : H \to H$ a everywhere-defined linear operator such that $T$ is self-adjoint, i.e. $\forall x,y \in H : \langle Tx,y\rangle=\langle x,Ty\rangle$.

Show that $T$ is bounded!

Using Cauchy-Schwarz inequality it is clear that if $T$ is idempotent then it's bounded, but with just the information initially provided I don't feel like it's possible to prove the boundedness.

Can somebody shed some light on this? Perhaps with a counter-example?

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    $\begingroup$ My first instinct is to used the closed graph theorem, since your operator is everywhere defined (ie more than densely defined) $\endgroup$ – Joel Jan 12 '18 at 16:39
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This follows from the Uniform Boundedness Principle. $\newcommand{\ip}[2]{\langle #1,#2\rangle}$ Say $B$ is the closed unit ball of $H$. For $x\in H$ define $\Lambda_x:H\to\Bbb C$ by $$\Lambda_x y=\ip{Tx}y.$$Note that each $\Lambda_x$ is bounded, in fact with norm $\|Tx\|$. For every $y\in H$ we have $$\sup_{x\in B}|\Lambda_xy|=\sup_{x\in B}|\ip x{Ty}|=\|Ty\|<\infty.$$So Uniform Boundedness implies that $$\sup_{x\in B}\|Tx\|=\sup_{x\in B}||\Lambda_x||<\infty,$$hence $T$ is bounded.

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    $\begingroup$ Indeed. Also, by the way, this argument shows that it is not possible to "fill in" a partially-defined unbounded operator artificially, using the Axiom of Choice, etc., to achieve an everywhere-defined (linear) operator, no matter how hard we try. $\endgroup$ – paul garrett Jan 12 '18 at 18:15
  • $\begingroup$ @paulgarrett Surely you mean "without using AC"? I mean if $V$ is a subspace of $X$ and $T:V\to Y$ is linear then it's easy to show (assuming AC) that $T$ extends to a linear map from $X$ to $Y$. $\endgroup$ – David C. Ullrich Jan 12 '18 at 18:59
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    $\begingroup$ I rather expect the intention was to say that we can't extend an unbounded self-adjoint (or just symmetric) operator to obtain an everywhere defined symmetric operator. $\endgroup$ – Daniel Fischer Jan 12 '18 at 19:09
  • $\begingroup$ @DanielFischer Ah, you're probably right - in any case that's true and yes it follows from the above. $\endgroup$ – David C. Ullrich Jan 12 '18 at 19:30
  • $\begingroup$ @DavidC.Ullrich, oop, yes, thanks! :) $\endgroup$ – paul garrett Jan 12 '18 at 20:46
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The Closed Graph Theorem is a good way. If $x_n\rightarrow x$ and $Tx_n\rightarrow y$, then $$ \langle y,z\rangle = \lim_n \langle Tx_n,z\rangle=\lim_n \langle x_n,Tz\rangle=\langle x,Tz\rangle = \langle Tx,z\rangle,\;\;\; z\in H. $$ This implies that $Tx=y$. Therefore $T$ has a closed graph. So $T$ is bounded.

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The operator you has mentioned

$T : H \to H$ a linear operator such that $\forall x,y \in H \; \langle\,Tx,y\rangle=\langle\,x,Ty\rangle $

is an everywhere-defined symmetric operator on a Hilbert space so it is self-adjoint and by Hellinger–Toeplitz theorem it is a bounded operator, but this is not always the situation because there are many self-adjoint operators defined in a dense subspace of $H$ and they are not-bounded.

Note: The self-adjoint operators defined in a dense subspace of $H$, we cannot extend them in an everywhere-defined symmetric operator, we can get only extension for them which is not symmetric the following example is a self-adjoint operator but not bounded.

Let $\Omega$ be any open subset of $ \mathbb{R}^n$ and $(T, D)$ such that $H= L^2(\Omega)$ and $$T= -\Delta \quad and \quad D=\{u\in L^2(\Omega) , \Delta u \in L^2(\Omega)\}$$

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