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!
Here is a straightforward way to prove the result. Let $(x,z)\in \overline{G(T)}$ be any element then there exist a sequence $(x_n,z_n)=(x_n,Tx_n)$ in $G(T)$ such that $(x_n,Tx_n)\rightarrow (x,z)$ in $X\times Y$. Then, $x_n \rightarrow x$ in $X$ and $T(x_n)\rightarrow z$ in $Y$. By using the continuity of inner product we have, for any $y\in Y$, $ \langle x,Sy\rangle=\lim_{n\rightarrow \infty}\langle x_{n},Sy\rangle$ and $\lim_{n\rightarrow \infty}\langle Tx_n,y\rangle=\langle z,y\rangle$. Now, for any $y\in Y$, we have $$\ \langle Tx, y\rangle =\langle x,Sy\rangle,\ (\text{by using the properties of adjoint operator } S )\\ =\lim_{n\rightarrow \infty}\langle x_n,Sy\rangle, (\text{by using the observation above }) \\= \lim_{n\rightarrow \infty}\langle Tx_n,y\rangle ,(\text{by using properties of adjoint})\\ = \langle z,y\rangle, (\text{by using the aboveobservation above }) .$$ So, we have $\langle Tx, y\rangle = \langle z,y\rangle, \forall y\in Y.$ This implies that $z=Tx$. The, $(x,z)=(x,Tx)\in G(T)$ and since $(x,z)\in \overline{G(T)}$ is arbitrary point, we get $\overline{G(T)}\subseteq G(T).$ So we get, $G(T)=\overline{G(T)}$, i.e $G(T)$ is closed in $X\times Y.$ Hence, the operator $T: X\rightarrow Y$ is closed. In similar way one can prove the adjoint operator $S:Y\rightarrow X$ is closed operator.