Sequence of continuous linear functionals over a sequence of Hilbert spaces Let $H_n$ be a sequence of complex Hilbert spaces such that $H_{n+1}\subsetneq H_n$ and $\bigcap_{n=1}^\infty H_n=\{v_0\}$
Let $T_1:H_1\to \mathbb C$ be a continuous linear functional such that $T_1(v_0)=0$
Let, for each $n \in \mathbb N$, $T_n:H_n\to\mathbb C$ the continuous linear functionals obtained by restriction of $T_1$ to $H_n$
I would like to know if $\lim_{n\to\infty}\lVert T_n \rVert = 0$
Thanks for any suggestion.
 A: Yes, your statement is true even for reflexive1 Banach spaces instead of Hilbert spaces.
Let $x_n\in H_n$ with $\|x_n\| = 1$ and $|T_n(x_n)| = \|T_n\|$. Such an element exists because $H_n$ is reflexive2.
Let us first assume that $H_1$ is separable.
Now, because the closed unit ball is weakly compact and $H_1$ is separable there is a subsequence $(x_{n_k})_k$ of $(x_n)_n$ which converges weakly to some $x\in H_1$. Because each $H_n$ is closed and convex in $H_1$, it is also weakly closed. For each $n\in\mathbb N$, almost all $x_{n_k}$ are an element of $H_n$, therefore $x\in H_n$ for all $n\in\mathbb N$, which yields $x=0$.
But now, $x_{n_k}\to 0$ weakly which yields $\|T_{n_k}\| = |T_{n_k}(x_{n_k})| = |T_1(x_{n_k})|\to 0$. Since the sequence of $\|T_n\|$ is decreasing, this yields the claim.
If $H_1$ is not separable then replace $H_n$ with $X_n := H_n\cap \overline{\operatorname{span}}(\{x_i: i\in\mathbb N\})$. The restriction of $T_n$ from $H_n$ to $X_n$ preserves the norm and $X_1$ is separable.
1I don't know if reflexivity is necessary but my proof relies on it. It is definitely not true for each Banach space as pointed out in the comments. Thanks for the example. 
2 We don't need reflexivity at this point it just makes the proof more simple. It would suffice to have $|T_n(x_n)|\geq \frac 12\|T_n\|$ which we always find.
