Let $\{\psi_{n}\}_{n\in \mathbb{N}}$ a sequence of unit elements of a given Hilbert space $\mathscr{H}$. These are not necessarily orthogonal. For each of these vectors, define the projection operator $P_{n}$ which sends $\mathscr{H} \ni \varphi \mapsto \langle \psi_{n},\varphi\rangle \psi_{n}$. Define a new operator: $$T = \sum_{n\in \mathbb{N}}\lambda_{n}P_{n}$$ with $\lambda_{n} \ge 0$ and $\sum_{n\in \mathbb{N}}\lambda_{n} = 1$. I am trying to prove the following: for every continuous function $f$ on $\sigma(T)$ $$f(T) = \sum_{n\in \mathbb{N}}f(\lambda_{n})P_{n}$$
My attempt is the following: first, note that $T$ is linear, bounded and self-adjoint. By the spectral theorem, since $\sum_{n=1}^{N}\lambda_{n}P_{n} \to T$ then: $$f(T) = \sum_{n\in \mathbb{N}}f(\lambda_{n}P_{n}) = \sum_{n\in \mathbb{N}}f(\lambda_{n})f(P_{n})$$
The result then would follow if $f(P_{n}) = P_{n}$. But I can't give a reasonable justification why this is true. Any help?
As noted in the comments, this statement does not hold. As a counterexample, consider the following vectors and values over $\Bbb C^2$: $$ \psi_1 = \pmatrix{1\\0}, \quad \psi_2 = \frac 1{\sqrt{2}}\pmatrix{1\\1}, \quad \lambda_1 = \lambda_2 = 1. $$ If you like, you can extend this to an infinite sequence by selecting $\psi_n$ arbitrarily and taking $\lambda_n = 0$ for $n \geq 3$. If we take $f:x \to x^2$, then we have $$ f(\lambda_1 P_1 + \lambda_2 P_2) = \pmatrix{3/2 & 1/2\\1/2 & 1/2}^2 = \frac 12 \pmatrix{5 & 2\\2 & 1},\\ f(\lambda_1)P_1 + f(\lambda_2) P_2 = \lambda_1 P_1 + \lambda_2 P_2 = \frac 12 \pmatrix{3 & 1\\1& 1} . $$