So, I'm proving that, given a spectral family $(E_t)_{t \in \mathbb{R}}$ in a Hilbert space $\mathscr{H}$ and a continuous function $f \in C(\mathbb{R}, \mathbb{C})$, the conditions below are equivalent:
$\int_{-\infty}^{+\infty} f dE_t = \lim_{a \rightarrow -\infty, b\rightarrow +\infty} \int_a^b fdE_t$ exists;
$\int_{-\infty}^{\infty} |f(t)|^2 d\langle x, E_t x \rangle < \infty$;
$l_x(y) = \int_{-\infty}^{+\infty} \overline{f(t)} d\langle x, E_t y \rangle$ is a bounded linear functional in $\mathscr{H}$.
I'm not sure about an argument to show that 3 $\implies$ 2:
Given $a, b \in \mathbb{R}$ with $a < b$, let $y_{a, b} = \int_a^b fdE_t x$ for some $x \in \mathscr{H}$. Then $\|y_{a, b} \|^2 = \int_a^b |f|^2 d\langle x, E_t x \rangle$ and then, because $l_x$ is bounded we would have
$$\|l_x\| \|y_{a, b} \| \geq \langle x, \int_{-\infty}^{+\infty} \overline{f(t)}dE_t y_{a, b}\rangle = \langle \int_{-\infty}^{+\infty} f dE_t x, \int_a^b fdE_t x \rangle = \lim_{\substack{a\rightarrow -\infty \\ b \rightarrow +\infty}} \langle \int_a^b |f|^2 dE_t x, x \rangle $$
$$= \int_{-\infty}^{+\infty} |f|^2 d\langle x, E_t x \rangle \ .$$ And that would imply 2. But it looks like I'm using 1 in the last step (I don't know if $\int_{-\infty}^{+\infty} fdE_t$ exists...). What is wrong with this argument and how could I show this implication?
Using the same idea, the same objects as introdced in OP, we write $$ \begin{aligned} \|l_x\| \|y_{a, b} \| &\geq \left\langle x, \int_{-\infty}^{+\infty} \overline{f(t)}\;dE_t y_{a, b} \right\rangle \\ &= \left\langle \int_{-\infty}^{+\infty} f \;dE_t x\ ,\ \int_a^b f\; dE_t x \right\rangle \\ &= \left\langle \int_a^b f \;dE_t x\ ,\ \int_a^b f\; dE_t x \right\rangle \\ &= \int_a^b |f|^2 \; d\langle E_t x, x \rangle \\ &= \|y_{a,b}\|^2\ . \end{aligned} $$ So the family $(y_{a,b})$ is bounded, now we take a increasing (w.r.t. inclusion) sequence of intervals $[a,b]$ that cover in union $\Bbb R$.
(Note: I had to go some 30 years back in the memory, hope there is no big gap...)