I'm trying to prove that if $U : H \to H$ is an bounded operator in a Hilbert space for which functional calculus is well defined (say, $U$ self adjoint and compact) and $g_n : \sigma(A) \to \mathbb{R}$ are continuous, uniformly bounded and pointwise covergent to certain $g : \sigma(A) \to \mathbb{R}$, then
$$ g_n(U) \to g(U) $$
in the sense of strong convergence.
So far, I've done the following: for each $h \in H$ we have the spectral measure $d\mu_h$ for which satisfies $$Q_h(f) = \langle f(U)h,h \rangle = \int_{\sigma(U)}f_nd\mu_h$$ for any continuous $f : \sigma(U) \to \mathbb{R}$. By dominated convergence we get $Q_h(g_n) \to Q_h(g)$ for all $h \in H$ and via a polarization identity the former gives $$\langle g_n(U)u,v\rangle \to \langle g(U)u,v\rangle$$ for all $u,v \in H$.
I haven't been able to conclude anything from here. Any hints you could provide will be greatly appreciated.
What you have shown is that for every $u \in H$, $g_n(U)u \rightharpoonup g(U)u$ in $H$. Recall that $x_n \to x$ in $H$ if and only if $x_n \rightharpoonup x$ and $\|x_n\| \to \|x\|$. This means that it is now enough to show that $\|g_n(U)u\| \to \|g(U)u\|$.
For this, notice that $$\|g_n(U)u\|^2 = \langle g_n(U)u , g_n(U)u \rangle = \langle g_n(U)^* g_n(U) u, u \rangle = \langle \overline{g_n}(U) g_n(U)u, u \rangle$$ where $\overline{g_n}$ is the complex conjugate of $g$. To conclude, notice that $\overline{g_n}(U) g_n(U) = (\overline{g_n}g)(U)$ and $\overline{g_n}g$ is a sequence of continuous functions which is uniformly bounded and converges pointwise to $\overline{g}g$. Hence $(\overline{g_n}g)(U)u \rightharpoonup (\overline{g}g)(U)u$ so that $$\|g_n(U)u\|^2 = \langle \overline{g_n}(U) g_n(U)u, u \rangle \to \langle (\overline{g}g)(U)u, u \rangle = \langle g(U)^*g(U)u, u \rangle = \|g(U)u\|^2$$ which gives the desired convergence of norms.