# $(T_n)_{n\in\mathbb{N}}\subseteq L(H)$, $T_n\to T$ weak, why does there exist $C>0$ such that $\|T_n\|\le C$ for all $n\in\mathbb{N}$?

Let $H$ be a Hilbert space, $(T_n)_{n\in\mathbb{N}}\subseteq L(H)$ a sequence such that $T_n^*=T_n$ and $T_n\le T_{n+1}$ for all $n\in \mathbb{N}$. There exists a map $T\in L(H)$ such that $T^*=T$ and $T_n\to T$ weak, $n\to \infty$. Then there exists a constant $C>0$ such that $\|T_n\|\le C$ for all $n\in\mathbb{N}$.

I'm stuck again. I want to prove it with the uniform boundedness principle. My try: Consider the canonical embedding $j:H\to H^{**}, h\mapsto j(h),\;j(h)(h^*)=h^*(h)$. The sequence $(\langle T_n,x\rangle)_n$ converges and therefore $\langle T_n,x\rangle$ is bounded for all $y\in H$.

Fix $x\in H$, why is $\{j(T_nx):n\in\mathbb{N} \}\subset H^{**}$ pointwise bouded? And why is $\sup_{n\in\mathbb{N}}\|j(T_nx)\|<\infty$?

The rest is clear: $j$ is an isometry, therefore $$\sup_{n\in\mathbb{N}}\|j(T_nx)\|=\sup_{n\in\mathbb{N}}\|T_nx\|<\infty$$ for all $x\in H$. With the uniform boundedness principle it follows, that $\sup_{n\in\mathbb{N}}\|T_n\|<\infty$ and therefore there exists a $C>0$ such that $\|T_n\|\le C$ for all $n\in\mathbb{N}$.

Note: maybe there are unnecessary requirements, but this is only a part of the whole homework.

• For each $x$, the sequence $(y_n)$ defined by $y_n = T_n(x)$ is weakly convergent. Hence it is weakly bounded. Aug 22, 2015 at 14:44
• ok, $y_n\in H$. But I still don't understand why this implies that the set $\{ j(T_nx):n\in\mathbb{N} \}$ is pointwise bounded in $H^{**}$. Could you elaborate it? Aug 23, 2015 at 13:10

Let's state something more general first. The uniform boundedness principle implies that for any Banach space $X$, every weakly convergent sequence $\{x_n\}\subset X$ is bounded. Indeed,
1. We have the canonical isometric embedding $j:X\to\ X^{**}$.
2. The sequence $\{ j(x_n)y\}$ has a limit for every $y\in X^*$.
3. Thus, the family of operators $\{j(x_n)\}$ is pointwise bounded on $X^*$
5. Since $j$ is an isometry, $\{x_n\}$ is norm-bounded.
Back to your question. For every pair $x,y\in H$, the sequence $\langle T_n x,y\rangle$ converges. This means $\{T_n x\}$ converges weakly. By the above, $\{T_nx \}$ is bounded. Apply UBP again to conclude $\{\|T_n\|\}$ is bounded.