Are there Hilbert space $H$ and adjoint operator $T$ s.t., $||T||=1$, $(Tx,x) (\forall x \in H)$ and $T^n$ not convergent uniformly?
I proved $T^n$ convergent strongly by using spectrum theory. But I can't prove it convergent uniformly, so I guess counterexample exists.
On
What about $H=l^2$ and $T$ defined by $$ (Tx)_n = \frac{n}{n+1}x_n. $$ Then for the unit vector $e_k$ we have $$ (T^m-T^n)e_k = \left(\left(\frac{k}{k+1}\right)^m-\left(\frac{k}{k+1}\right)^n\right)e_k, $$ and $\|(T^m-T^n)e_k\|$ cannot be made small for large $m,n$ uniformly with respect to $k$.
Consider a separable Hilbert space $H$ with an orthonormal basis $(e_n)_n$ and define $T : H \to H$ as $T = \operatorname{diag}(\alpha_n)_n$, that is $Tx = \sum_{n=1}^\infty \alpha_n \langle x, e_n\rangle e_n$ where $(\alpha_n)_n$ is a bounded sequence of scalars.
We have $\|T\| = \sup_{n\in\mathbb{N}} |\alpha_n|$ and $T$ is a positive self-adjoint operator if and only if $\alpha_n \ge 0, \forall n \in \mathbb{N}$.
Notice that $T^k = \operatorname{diag}(\alpha^k_n)_n$ so $$T^ke_n = \alpha_n^k e_n$$
and this converges to $0$ when $k\to\infty$ if and only if $|\alpha_n| < 1$.
Thus if $|\alpha_n| < 1, \forall n \in \mathbb{N}$ the only candidate for the strong limit (and hence the uniform limit as well) of $(T^k)_k$ is $0$.
Therefore $(T^k)_k$ converges uniformly if and only if $T^k \to 0$ uniformly if and only if $$\|T\|^k \to 0$$ if and only if $\|T\| < 1$.
So just take a sequence with $\sup_{n\in\mathbb{N}} |\alpha_n| \ge 1$, such as the one @daw suggests: $\alpha_n = \frac{n}{n+1}$.
The only other option for the existence of the strong limit is that for some $n\in\mathbb{N}$ we have $\alpha_n = 1$.
In that case the candidate for the limit is $\operatorname{diag}(\beta_n)_n$ where $\beta_n =\begin{cases} 1,\text{if } \alpha_n = 1\\ 0, \text{if } |\alpha_n| < 1\end{cases}$, and $T^k \to \operatorname{diag}(\beta_n)_n$ uniformly if and only if $$\sup_{n\in\mathbb{N}, |\alpha_n| < 1} |\alpha_n| < 1$$