Uniform convergence of functions with values in a Hilbert space

Question

Let $(f_n)$ be any family of functions with value in a Hilbert space $\mathcal H\,$. Suppose that, for all $t\in \mathbb R\,$, $f_n(t)$ converges weakly in $\mathcal H$ to a limit $f(t)$. Then, it is well--known that if $\|f_n(t)\|_{\mathcal H}\to \|f(t)\|_{\mathcal H}\,,$ for all $t\,,$ then we have $$ f_n(t)\to f(t) \text { in } \mathcal H\,\qquad \text{ as } n\to \infty\,\qquad \forall t. $$ Now, suppose that $f_n$ converges weakly to $f$ in $\mathcal C_b(\mathbb R;\mathcal H)\,,$ where $\mathcal C_b(\mathbb R;\mathcal H)$ denotes the space of bounded continuous functions on $\mathbb R$ with value in $\mathcal H$ . Is there any analog of the above theorem that ensures $$ \sup_{t\in\mathbb R}\|f_n(t)-f(t)\|_{\mathcal H}\to 0 \quad \text{ as } n\to \infty\,. $$

Answer 1

No, there is not. The main problem is that your assumptions do not have anything to do with the variable $t$.

For a counterexample consider $\mathcal H=\mathbb R$ and $f_n(t)$ be the function that is zero everywhere except for the interval $(1-1/n,1+1/n)$, where it has a bump such that $$\max_{(1-1/n,1+1/n)}f_n(t)=1.$$ You can prove by hand that for all $t\in\mathbb{R}$ we have that $f_n(t)\rightarrow 0$, so you have both weak and strong convergence of the sequence $\{f_n(t)\}$ to $0$, and $0\in C_b(\mathbb{R},\mathcal{H})$, however $\sup || f_n(t)-0 ||$ does not converge to zero.