Given: self-adjoint, monotonic increasing sequence in $L(H)$ such that $\|T_n\|

Question

Let $H$ be a Hilbert space, $(T_n)\subseteq L(H)$ a sequence such that $T_n^\ast=T_n$ and $T_n\le T_{n+1}$ for all $n\in \mathbb{N}$. There exists a constant $C>0$ such that $\|T_n\|<C$ for all $n\in\mathbb{N}$. The claim is: then there exists a $T\in L(H)$ such that $T=T^*$ and such that $T_n\to T$ strong, $n\to\infty$.

I stuck a little bit. My try: Let $n\in\mathbb{N}$. $T_n$ self-adjoint implies, that $\langle T_nx,x\rangle\in\mathbb{R}$ for all $n\in\mathbb{R}$, $x\in H$. So, the sequence $(\langle T_nx,x\rangle )$ is a real sequence which is monotonic increasing, because $\langle T_nx,x\rangle\le \langle T_{n+1}x,x\rangle$ for all $n\in\mathbb{N}$, and the sequence is bounded, because $|\langle T_nx,x\rangle |\le \|T_nx\|\|x\|\le C\|x\|^2$ for all $x\in H$. Therefore $(\langle T_nx,x\rangle )$ is convergent in $\mathbb{R}$.

Now, how to continue? The next step I made is to prove that $(\langle T_nx,y\rangle )$ is convergent for all $x,y\in H$ with the polarization identity, but I did it wrong.. Could anybody help me to prove this, that $(\langle T_nx,y\rangle )$ is convergent for all $x,y\in H$ in $\mathbb{R}$?

What I did next, if we know that $\lim\limits_{n\to\infty}\langle T_nx,y\rangle =:F(x,y)\in \mathbb{R}$ exists: It is $|F(x,y)|=\lim\limits_{n\to\infty}|\langle T_nx,y\rangle|\le C\|x\|\|y\|$ for all $x,y\in H$, therefore $$F_x:H\to\mathbb{C},\; y\mapsto \overline{F(x,y)}$$is a linear, bounded functional. The Riesz-Representationtheorem gives us, that there exists $Tx\in H$ such that $F_x(y)=\langle y,Tx\rangle=\overline{F(x,y)}$. If you conjugate both sides, we obtain $F(x,y)=\langle Tx,y\rangle$ for all $x,y\in H$. Then I proved that $T$ is linear and bounded and that $T$ is self-adjoint, there was no problem.

But how to prove $\|T_nx-Tx\|\to 0$ for all $x\in H$, $n\to\infty$?

But if you know an other proof, let me know it. I hope I did no mistakes. Regards

Edit: My try with the polarization identity $\langle x,y\rangle =\frac{1}{4}(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2)$ for $x,y$ of a complex Hilbert space $H$: $\forall n\in\mathbb{N}$, $x,y\in H$, it is : $\langle T_nx,y\rangle =\frac{1}{4}(\|T_nx+y\|^2-\|T_nx-y\|^2)+\frac{i}{4}(\|T_nx+iy\|^2-\|T_nx-iy\|^2)=\frac{1}{4}(\langle T_nx+y,T_nx+y\rangle-\langle T_nx-y,T_nx-y\rangle)+\frac{i}{4}(\langle T_nx+iy,T_nx+iy\rangle-\langle T_nx-iy,T_nx-iy\rangle)$. Can I argue now, that all the summands converge, because $(\langle T_nx,x\rangle )$ converges for all $x\in H$?

Answer 1

The polarisation identity is more general than only for the inner product. If $s$ is any sesquilinear form on a complex vector space, then we have

\begin{align} s(x+y,x+y) &= s(x,x) + s(x,y) + s(y,x) + s(y,y),\\ s(x-y,x-y) &= s(x,y) - s(x,y) - s(y,x) + s(y,y),\\ s(x+iy,x+iy) &= s(x,x) - i s(x,y) + is(y,x) + s(y,y),\\ s(x-iy,x-iy) &= s(x,x) + is(x,y) - is(y,x) + s(y,y), \end{align}

and so

$$s(x+y,x+y) - s(x-y,x-y) + is(x+iy,x+iy) - is(x-iy,x-iy) = 4s(x,y).$$

For bilinear forms on a real vector space, you need symmetry to have a polarisation identity.

In our situation, we take $s(x,y) = \langle T_n x,y\rangle$, and find

$$4\langle T_n x,y\rangle = \langle T_n(x+y),x+y\rangle - \langle T_n(x-y),x-y\rangle + i\langle T_n(x+iy),x+iy\rangle - i\langle T_n(x-iy),x-iy\rangle,$$

which shows the convergence of $\langle T_n x,y\rangle$ since we know that $\langle T_n z,z\rangle$ converges for all $z\in H$.

From that, you have (correctly, it seems) argued that there is a self-adjoint $T\in B(H)$ such that $T_n x$ converges weakly to $T x$ for all $x\in H$.

It remains to see that the convergence is in fact norm convergence. The sequence $S_n := T - T_n$ is a uniformly bounded sequence of (non-strictly) positive self-adjoint operators on $H$. The sequence of (positive self-adjoint) square roots $R_n$ of $S_n$ is therefore also uniformly bounded, hence equicontinuous. For every $x\in H$, we have

$$\lVert R_n x\rVert^2 = \langle R_n x, R_n x\rangle = \langle R_n^2 x, x\rangle = \langle S_n x,x\rangle \to 0.$$

By the equiconitnuity of the $R_n$, it follows that $\lVert S_n x\rVert \to 0$, i.e. $\lVert Tx - T_n x\rVert \to 0$.

Answer 2

I think you've done almost all the work, the missing piece is the following (modified) polarization identity:

$\left<T_nx,y\right> = \frac{1}{4} \sum_{k=0}^3 i^k \left<T_n(x+i^ky),x+i^ky\right>$ (check this!)

Now it should be clear that every summand converges because $(<T_n z,z>)$ converges for all $z \in H$