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Let $T$ be a linear operator on a hilbert space $H$ such that $$ \langle Tx,y\rangle = i\langle x,Ty\rangle \qquad \forall\, x,y \in H $$ Show that $T$ is bounded.

We can show that $T$ is bounded by showing that $T$ has a closed graph. Now I was told that in order to show that the graph of $T$ is closed, it's sufficient to show that $\lim_n \langle x_n, Tx_n\rangle = \langle 0,y\rangle$ then $y=0$. Why is this sufficient?

egreg
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Olba12
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1 Answers1

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You need to show that if $x_n \to x$, and $T(x_n) \to y$, then $T(x)=y$.

Let $z_n = x_n-x$ and $z=y-T(x)$. Then $z_n \to 0$ and $T(z_n) \to z$. You need to show that $z=0$.

Since $z_n \to 0$ and $T(z_n)\to z$, you get $$ \langle z_n, T(z_n) \rangle \to \langle 0,z \rangle.$$

If you can prove that this implies $z=0$, you are done.

Eman Yalpsid
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N. S.
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