I am wondering is it next true:
Suppose that $f(t)$ is non-negative and non-decreasing function on $[0,\infty)$ and let $A$ be a positive operator on some infinite-dimensional separable Hilbert space. Do we then have $\lambda_i(f(A))=f(\lambda_i(A))$, where $\{\lambda_i\}$ is a sequence of eigenvalues of $A$.
Also, I am little confused, because I am not sure that $f(A)$ is well defined. As far I know we need that $f$ is bounded Borel function on spectrum of $A$ to define $f(A)$. If we add that $f$ is continuous in our statement then everything is okay, but do we really need continuity (it looks for me that is too much to require).
First for the eigenvalues: By the spectral theorem (and since $A$ is a positive (hence self-adjoint) operator), there is a measure space $(X, M, \mu)$ and some (measurable) function $g : X \to [0,\infty)$, such that $A$ is unitarily equivalent to the multiplication operator
$$ M_g : L^2(\mu) \to L^2 (\mu), h \mapsto g\cdot h. $$
Thus, it suffices to prove the claim for $M_g$ instead of $A$.
If $\lambda$ is an eigenvalue of $M_g$, then there is some $h \in L^2 (\mu)$ with $h \not 0$ and $g \cdot h = M_g h = \lambda h$ almost everywhere.
This implies $g \equiv \lambda$ (almost everywhere) on the set $V := \{x \mid h(x) \neq 0\}$. Note that $V$ has positive measure, since $h \not 0$. Now, let $$V_n := \{x \mid |h(x)| \geq 1/n\}.$$ Then $\mu(V_n) <\infty$ since $h \in L^2(\mu)$. Furthermore, since $V = \bigcup_n V_n$ is not a null-set, we get $0 < \mu(V_n) < \infty$ for some $n \in \Bbb{N}$.
Now
$$ f(M_g) \chi_{V_n} = M_{f\circ g} \chi_{V_n} = \chi_{V_n} \cdot (f\circ g) \equiv \chi_{V_n} f(\lambda), $$ because of $g \equiv \lambda$ on $V_n \subset V$. Thus, the indicator function $\chi_{V_n} \in L^2(\mu)$ is an eigenvalue of $f(M_g)$ (which is unitarily equivalent to $f(A)$) for the eigenvalue $f(\lambda)$.
Note that we did not use monotonicity of $f$ at all. Measurability would have been sufficient.
For the claim that in general $f(\sigma(A)) \neq \sigma(f(A))$, simply take $A : L^2([0,1]) \to L^2([0,1]), f \mapsto x \cdot f$. Then $\sigma(A) = [0,1]$. Now, let
$$ f\left(x\right)=\begin{cases} x, & 0\leq x<1,\\ 5, & x\geq1. \end{cases} $$ Then $f(\sigma(A)) = [0,1) \cup \{5\}$, which is not closed. In particular, $f(\sigma(A)) \neq \sigma(f(A))$, since $\sigma(f(A))$ is compact.