Let $H$ be a separable Hilbert space, and $T:{\frak Dom}(T) \to H$ a densely defined self-adjoint operator. As is well-known, for $\mathrm{spec}(T)$ the spectrum of $T$, and $f:\mathrm{spec}(T) \to \mathbb{C}$ a bounded continuous function, we can associate via the function calculus a well-defined bounded operator $f(T):H \to H$.
Let us now assume that $T$ is diagonalisable, which is to say, assume that $H$ admits an orthonormal basis $\{b_k\}_{k \in \mathbb{N}_0}$ such that $T(b_k) = c_k b_k$, for all $k$, where $c_k \in \mathbb{R}$. Is it true that $f(T)$ is the unique bounded operator on $H$ satisfying $f(T)(e_k) = f(c_k)e_k$, for all $k$?
Yes, because $A(e_k) = f(c_k) e_k$ determines $A$ on the linear span $V$ of the orthonormal basis, and if $A$ is bounded that determines it on the closure of $V$, which is $H$.