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?

  • $\begingroup$ Actually, I think it is all wrong cause I am assuming it is linear $\endgroup$ – Euler_Salter Dec 3 '18 at 23:12
  • 1
    $\begingroup$ The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine. $\endgroup$ – Kavi Rama Murthy Dec 3 '18 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.