# Bounded operator on weakly convergent sequence maps to weakly convergent sequence

Suppose that $$H, K$$ are Hilbert Spaces. We have a sequence $$v_n\in H$$ that converges weakly to some vector $$v\in H$$, that is: $$\langle v_n - v, w\rangle \to 0 \quad \text{as} \quad n\to\infty$$ Also, we know that the operator $$T:H \to K$$ is bounded. I need to show that the sequence $$Tv_n\in K$$ converges weakly to $$Tv\in K$$.

My Proof

I tried proving it like this: Let $$\epsilon>0$$. Then since $$v_n$$ converges weakly to $$v$$ we have that there exists $$n_0\in \mathbb{N}$$ such that: $$n>n_0 \Longrightarrow |\langle v_n-v, w\rangle | < \epsilon$$ Now, let $$z\in K$$: \begin{align} |\langle Tv_n-Tv, z\rangle| &= |\langle T(v_n-v), z\rangle| && \text{as T is linear}\\ &= |\langle v_n-v, T^*z\rangle| && \text{as every bounded operator has a unique adjoint T^*}\\ &< \epsilon && \text{as v_n converges weakly to v} \end{align}

Is this okay?

• Actually, I think it is all wrong cause I am assuming it is linear – Euler_Salter Dec 3 '18 at 23:12
• The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine. – Kavi Rama Murthy Dec 3 '18 at 23:32