Suppose $L,R$ are not necessarily bounded operators on a hilbert space $H$. Show that, if $L,R$ satisfy $$ \langle Lx,y \rangle = \langle x,Ry\rangle $$ for all $x,y \in H$, then $L$ is bounded.
I tried looking at the uniform boundedness theorem, but I cannot go much further than noting that $L,R$ resemble the property of an adjoint operator (they will be each others adjoint once you prove they are bounded).
Any thoughts?
On
Note that for any fixed $y \in H$, $\sup_{||x|| \leq 1} |\langle Lx, y \rangle| = \sup_{||x|| \leq 1} |\langle x, Ry \rangle| = ||Ry||$, so by uniform boundedness principle, $\sup_{||x|| \leq 1, ||y|| \leq 1} |\langle Lx, y \rangle| < +\infty$. But then $||L|| = \sup_{||x|| \leq 1} ||Lx|| = \sup_{||x|| \leq 1, ||y|| \leq 1} |\langle Lx, y \rangle| < +\infty$, so $L$ is bounded.
On
The question can be reduced to the Hellinger-Toeplitz theorem. Namely $\langle Lx,y\rangle =\langle x,Ry\rangle $ implies $$\langle (L\pm R)x,y\rangle =\langle x,(L\pm R)y\rangle$$ Thus the operators $L+R$ and $L-R$ are bounded, so are $L$ and $R.$
The Hellinger-Toeplitz theorem can be proved by the closed graph theorem (the most popular method) or by the uniform boundedness principle see.
This is a generalisation of the Hellinger-Toeplitz theorem, which can incidentally be proved the same way as said theorem:
Consider $(x_n)_n \subset H$ and $(x,y) \in H^2$ such that $\begin{cases} x_n \to x\\ L(x_n) \to y\end{cases}$ when $n \to \infty$.
Then, for all $\varphi \in H$: $$\begin{cases} \langle L(x_n), \varphi\rangle \xrightarrow[n \to \infty]{} \langle y, \varphi\rangle\\ \langle L(x_n), \varphi \rangle = \langle x_n, R(\varphi)\rangle \xrightarrow[n \to \infty]{} \langle x, R(\varphi)\rangle = \langle L(x), \varphi \rangle\end{cases}$$ Thus we obtain, by uniqueness of the limit and the fact that this is true for all $\varphi \in H$: $L(x) = y$, thus the graph of $L$ is closed, hence $L$ is bounded by the closed graph theorem.