Let $\mathscr{H}=\{f:\mathbb{D}\to \mathbb{C}| f(z)=\sum_{n=0}^{\infty}c_{n}z^n, \Vert f\Vert =\sum_{n}\vert c_{n}\vert^2 < \infty\}$. We define $$(T(t)f)=\sum_{n=0}^{\infty}(n+1)^{-t}c_{n}z^n.$$
Why is the operator $T$ positive and self-adjoint?
We assume the inter-product is $$(\sum c_{n}z^n, \sum d_{n}z^n)=\sum_{i+j=n}c_{i}\overline{d_{j}}z^n.$$ I do not know what is the self-adjoint of $T$?
If $T$ is self-adjoint, and $$(T(t)f,f)=\sum_{i+j=n}\frac{c_{i}\bar{c_{j}}}{(i+1)^t}\geq 0,$$ which inply the positive.
As written the question makes little sense. The inner product should be a number, but as written it seems to be a function (and, in that case, the formula as written cannot be right).
I will assume that the inner product is $$ \langle \sum_n c_nz^n,\sum_nd_nz^n\rangle=\sum_n c_n\overline{d_n} $$ (this makes $\mathcal H$ a Hilbert space as it is actually $\ell^2(\mathbb N)$ in disguise). In that case, you have $$ \langle T(t)f,f\rangle=\sum_n\frac{|c_n|^2}{(n+1)^t}\geq0. $$ Since the Hilbert space is complex, this is enough to imply that $T(t)$ is selfadjoint (because $\langle T(t)f,f\rangle$ is real), and that $T(t)$ is positive.