Let $H$ be a Hilbert space and $(T_\alpha)$ an increasing net of self-adjoint operators that converges (in some topology) to an operator $T$. Then $(T_\alpha)$ is not necessarily norm-bounded I think (is this correct?). Assuming that the index set is non-empty, for any $\alpha_0$ we can consider the net $(T_\alpha)_{\alpha\geq\alpha_0}$. I want to prove that this net is actually norm-bounded. I think that $T_{\alpha_0}\leq T_\alpha\leq T$ and that $$-\|T_{\alpha_0}\|\leq\|T_\alpha\|\leq\|T\|$$ for all $\alpha\geq\alpha_0$, but I don't know how to make this precise or mathematically rigorous. In particular, I just assumed $T_\alpha\leq T$, since this is natural for increasing convergent sequences of real numbers. I'm not really familiar with nets. Any help would be greatly appreciated!
Tail of increasing convergent net of self-adjoint operators is bounded
c-star-algebrasnetsoperator-algebrasself-adjoint-operatorsstrong-convergence
Related Solutions
The sequence $(T_{n}x,x)$ is convergent: $(T'x,x)\geq(T_{n+1}x,x)\geq(T_{n}x,x)$ for each $n=1,2,...$
Then sequence $(T_{n}x,y)$ is also convergent by means of polarization.
We can define $(Tx,y)=\lim_{n}(T_{n}x,y)$. It is easy to see that $T$ is self-adjoint because of those $T_{n}$.
On the other hand, by Uniform Boundedness Principle, one can show that $\|T_{n}\|\leq C$ and hence $T$ is also bounded.
Indeed, consider the sesquilinear form $U(x,y)=(T_{n}x,y)$, then we have \begin{align*} \|T_{n}x\|^{2}&=U(x,T_{n}x)\\ &\leq U(x,x)^{1/2}U(T_{n}x,T_{n}x)^{1/2}\\ &=(T_{n}x,x)^{1/2}(T_{n}^{2}x,T_{n}x)^{1/2}\\ &\leq(T_{n}x,x)^{1/2}(T'(T_{n}x),T_{n}x)^{1/2}\\ &\leq(T_{n}x,x)^{1/2}\|T'\|^{1/2}\|T_{n}x\|, \end{align*} so $\|T_{n}x\|\leq(T_{n}x,x)^{1/2}\|T'\|^{1/2}\leq(T'x,x)^{1/2}\|T'\|^{1/2}$, so for each $x$, $\|T_{n}x\|$ is bounded, now we apply Uniform Boundedness Principle.
Now consider the sesquilinear form $S(x,y)=((T-T_{n})(x),y)$.
We have \begin{align*} \|(T-T_{n})x\|^{2}&=S(x,(T-T_{n})x)\\ &\leq S(x,x)^{1/2}S((T-T_{n})x,(T-T_{n})(x))^{1/2}\\ &=((T-T_{n})(x),x)^{1/2}((T-T_{n}x)^{2},(T-T_{n})(x))\\ &\leq((T-T_{n})(x),x)^{1/2}\|(T-T_{n})(x)\|^{1/4}\|(T-T_{n})^{2}(x)\|^{1/4}\\ &\leq C'((T-T_{n})(x),x)^{1/2}\|x\|^{1/2}\\ &\rightarrow 0. \end{align*}
Edit:
The boundedness of $\|T_{n}\|$ can be proved without appealing to $T'$.
Indeed, the sequence $(T_{n}x,y)$ is bounded by applying polarization to the boundedness of $(T_{n}x,x)$.
Now we consider $S_{n}(y)=(y,T_{n}x)$, for each fixed $x$, Uniform Boundedness Theorem gives $\|S_{n}\|\leq M_{x}$, then $|(T_{n}x,y)|\leq M_{x}$ for any $y$ with $\|y\|\leq 1$. We set $y=T_{n}x/\|T_{n}x\|$ to get $\|T_{n}x\|\leq M_{x}$, once again Uniform Boundedness Theorem gives $\sup_{n}\|T_{n}\|<\infty$.
If $T$ is densely defined with domain $\mathcal{D}(T)$ then $T^*$ is closed and
$$y \perp \text{Range}(T)\iff \forall x\in \mathcal{D}(T):\langle Tx,y\rangle=0 \iff y\in\mathcal{D}(T^*) \land T^*(y) = 0$$
This proves
$$\text{Range}(T)^\perp = \text{Ker}(T^*)$$
and proves, inter alia that $\text{Ker}(T^*)$ is a closed subspace because the orthogonal complement to any set is a closed subspace.
If $T$ is densely defined and closable then $T^*$ is densely defined (and as before closed). Additionally, $T^{**} = \overline{T}$ the closure of $T$ obtained by taking the closure of the graph of $T$. From this we have
$$\text{Range}(T^*)^\perp = \text{Ker}(T^{**})=\text{Ker}(\overline{T})$$
Best Answer
"Some topology" is pretty vague, and for sure one can come up with pathologic topologies where $T$ is not even normal. But since the tag is "strong convergence", that's what I'll assume.
Indeed the net needs not be bounded. This is already true in dimension one: take $(T_\alpha)_{\alpha\in\mathbb Z}$, with $$ T_\alpha=\begin{cases} \alpha,&\ \alpha\leq 0\\[0.3cm] 1-\tfrac1\alpha,&\ \alpha>0\end{cases}. $$ Then $T_\alpha\to1$, but the net is not bounded.
Regarding the tail, indeed from $T_{\alpha_0}\leq T_\alpha\leq\|T\|$ you can deduce that the tail of the net is bounded. For this you may use that $$\tag1 \|T_\alpha\|=\sup\{\langle T_\alpha x,x\rangle:\ \|x\|=1\}. $$ From $(1)$ you see that $\|T_\alpha\|\leq\max\{\|T\|,\|T_{\alpha_0}\|\}$.