I'm trying to find the derivative of $Tr(A^x)$ with respect to ( real) $x$, where $A$ is a bounded, linear, positive semi-definite, hermitian operator on a Hilbert space $X$. If $A$ is a matrix, we get $\frac{d}{dx} Tr(A^x) = Tr(A^x \log(A))$. I want to show that the result holds if $A$ is a bounded, linear, positive semi-definite, hermitian operator too.
I will state my considerations and would be happy if one or another can tell if I missed something or add some hints how to proceed correctly!
Due to the properties of $A$, we know, that there exists a basis $(b_n)$ such that every $b_n$ is an eigenvector of $A$, i.e. $A b_n = \lambda_n b_n$. In more detail, we even know that the number of eigenvectors is countably infinity and the same applies to the number of eigenvalues. Furthermore, since $A$ is bounded, we deal with the "natural generalization" of a matrix, so I hope to find a similar argumentation by using the spectral decomposition (as in the finite-dimensional case, where $A$ can be represented by a matrix).
Since the number of eigenvalues is countable it is (similar to the finite-dimensional case) possible to express $A = \sum_i \lambda_i (b_i,b_i)$ and $A^x = \sum_i \lambda_i^x (b_i,b_i)$ . I assumed here, that the (real) power of an operator can be defined via its eigenvalues like it is possible with matrices. Unfortunately, I haven't found a book or some other reliable resource with a detailed definition.
If this is possible, I would proceed in the follwing way: $\frac{d}{dx} A^x = \frac{d}{dx} \sum_i \lambda_i^x (b_i,b_i) = \sum_i \lambda_i^x \log(\lambda_i) (b_i,b_i) $. Since $\lambda_i \geq 0$ (due to positive semi-definiteness of $A$) all summands are positive and the she sum is absolutely convergent, thus I can change the sum and the derivative. $\log(\lambda_i)$ is well-defined if $\lambda_i \neq 0$ (we already have $\lambda_i \geq 0$ due to positive-definiteness of $A$).
Finally, since $Tr(A) := \sum_j (A b_j,b_j)$, I would get the result $\frac{d}{dx} Tr(A^x) = \frac{d}{dx} \sum_j(A^x b_j, b_j) = \sum_j \frac{d}{dx}(A^x b_j,b_j) = \sum_j (\frac{d}{dx} A^x b_j,b_j) = Tr(\frac{d}{dx}A^x) = Tr(A^x \log(A))$.
Although I dealt with the infinite-dimensional case, my approach was nearly the same as in the finite-dimensional case. I wonder if I missed some functional calculus arguments, that make it more difficult, or if I can argue this way? Is my "intuitive" definition of the real power of an operator correct and does it exist for all bounded, linear, hermitian, positive semi-definite operator?