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Let $H,G$ be Hilbert spaces. Assume that $T: H \rightarrow G$ is a continuous linear transformation. I want to show that if $(x_n)$ is a weakly convergent sequence in $H$, then $(Tx_n)$ is a weakly convergent sequence in G.

Here is what I've done so far:

By definition, we have $\langle x_n, v\rangle \rightarrow \langle x,v \rangle$, for all $v \in H$, and some $x \in H$.

Since $G$ is continuous, we have that $T(x_n) \rightarrow T(x)$ in G.

Hence, $\langle T(x_n),v \rangle \rightarrow \langle T(x),v \rangle$.

I'm not sure if this is correct, as I the only things I've used are definition of weak convergence, and continuity of $T$.

I'm new to weak convergence. If the proof is wrong, could you please let me know which part is wrong, and how should I fix it?

Thank you!

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Since $T$ is a continuous linear operator, $$\langle h, T^*g \rangle_H = \langle Th, g \rangle_G$$ where $T^*$ is the adjoint of $T$. Since $x_n$ converges weakly to $x$ in $H$, it follows that for all $g \in G$, \begin{align} \lvert \langle T(x_n),g \rangle_G - \langle T(x), g \rangle_G \rvert &= \lvert \langle T(x_n) - T(x), g \rangle_G \rvert \\ &= \lvert \langle T(x_n - x), g \rangle_G \lvert \\ &= \lvert \langle x_n - x, T^*(g) \rangle_H \rvert \\ &= \lvert \langle x_n, T^*(g) \rangle_H - \langle x, T^*(g) \rangle_H \rvert \\ &\rightarrow 0 \end{align} and, hence, $T(x_n)$ converges weakly to $T(x)$ in $G$.

When you say,

"Since $G$ is continuous, we have $T(x_n) \rightarrow T(x)$ in $G$. Hence, $\langle T(x_n), v \rangle \rightarrow \langle T(x), v \rangle$."

you may have been using the definition of norm convergence along with the continuity of $T$ and the inner product.

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Actually this holds in much generality for all Banach spaces $X$. Indeed if $f\in X^*$, where $X^*$ is the continuous dual of $X$—i.e., the Banach space of continuous linear functionals on $X$—then observe that $f\circ T$ is a functional in $X^*$ since $T$ is continuous and linear. It immediately follows that $f\circ T(x_n)\to f\circ T(x)$ for all $f\in X^*$ whenever $x_n$ converges weakly to $x$, and hence $T(x_n)$ converges weakly to $T(x)$.

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