# Weak convergence in Hilbert space under a continuous linear transformation

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!

• Oct 13 '20 at 5:29

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 $$G$$.
"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.
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 $$T(x)$$.