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Let $X,Y$ be inner-product spaces. Let $T\in L\left(X,Y\right)$ be a linear operator with adjoint operator $S\in L\left(Y,X\right)$ such that $$\langle Tx,y\rangle_Y=\langle x,Sy\rangle_X\quad\forall (x,y)\in X\times Y.$$

I need to show that the graph $\mathscr{G}(T)=\{(x,Tx)\in X\times Y\;|\;x\in X\}$ is closed. Since inner-product spaces always have an extension to a normed space, there is a norm (only, I don't know if this information will be of any use here). We also know that continuous and bounded operators are the same in those "nice" linear spaces.

Now, what is left to show is that $\mathscr{G}(T)=\overline{\mathscr{G}(T)}$; i.e. all limit points in the graph of $T$ are again in the graph of $T$. Let $(x_n,Tx_n)_{n\in\mathbb{N}}$ be a sequence in $\mathscr{G}(T)$ which goes to the point $(x,y)$; i.e. showing that $y\in \text{Im}(T)$. In other words, we need to show that $T$ is continuous in order to conclude $\lim_{n\to\infty}Tx_n=T(\lim_{n\to\infty}x_n)=Tx$. Now, maybe we can do the following: $$ \langle\lim_{n\to\infty}Tx_n,y\rangle_Y=\lim_{n\to\infty}\langle Tx_n,y\rangle_Y=\lim_{n\to\infty}\langle x_n,Sy\rangle_X=\langle\lim_{n\to\infty}x_n,Sy\rangle_X=\langle x,Sy\rangle_X=\langle Tx,y\rangle_Y=\langle T(\lim_{n\to\infty}x_n),y\rangle_Y\quad\forall y\in Y. $$ Now we have that $$ \langle\lim_{n\to\infty}Tx_n,y\rangle_Y-\langle T(\lim_{n\to\infty}x_n),y\rangle_Y=\langle\lim_{n\to\infty}Tx_n-T(\lim_{n\to\infty}x_n),y\rangle_Y=0\quad\forall y\in Y. $$ Hence, we choose $y=\lim_{n\to\infty}Tx_n-T(\lim_{n\to\infty}x_n)$ and therefore we can conclude that, using the definition of an inner-product: $\langle x,x\rangle=0\iff x=0$, we now have $\lim_{n\to\infty}Tx_n=T(\lim_{n\to\infty}x_n)$. Q.E.D.

Is this the right way to use the adjointness of $T$ and $S$ in order to show that $(x,y)$ is an element of $T$'s graph? Thanks for the time!

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    $\begingroup$ As stated, it is not true. If it were true, then by taking $X=Y$, we would get that every linear operator defined on $X$ is closed, hence by the Closed Graph Theorem, is bounded. But this is of course not true. Something is missing. $\endgroup$ – uniquesolution Mar 13 at 14:06
  • $\begingroup$ @uniquesolution This is exactly how the problem is proposed in the exercise class. $\endgroup$ – Algebear Mar 13 at 14:08
  • $\begingroup$ @uniquesolution What do you think of my attempt? Does that make it more clear on what I want to try to prove? $\endgroup$ – Algebear Mar 14 at 15:47
  • $\begingroup$ @uniquesolution No, you assume that every linear operator has an everywhere defined adjoint. That is where you slipped in the assumption of boundedness. $\endgroup$ – MaoWao Mar 14 at 16:04
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    $\begingroup$ @Algebear Yes, my first comment was directed at uniquesolution, who said the statement couldn't be true. You did show that $T$ has a closed graph. What you didn't show is that $T$ is continuous as you announced in the second paragraph (and you don't need to prove this for the original statement). $\endgroup$ – MaoWao Mar 14 at 20:46

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