if I have a self-adjoint operator $T:D(T) \rightarrow L^2$, then I define its unitary exponential operator by $$e^{iT}(f) := \lim_{k \rightarrow \infty} e^{iT_{k}}(f),$$
where $T_k(f):=\frac{1}{2} k^2((T+i k)^{-1} + (T-ik)^{-1})(f) \rightarrow T(f),$ for $f \in D(T).$ Notice, that $T_k$ are bounded operators.
Now, assume that our operator has a purely discrete spectrum $\lambda_0,\lambda_1,\lambda_2 ,....$ and $f:=\sum_{n=0}^{\infty} a_n e_n$, where $e_n$ are eigenfunctions to our operator. How do I show that $$e^{iT}(f)= \sum_{n=0}^{\infty}e^{i \lambda_n} a_n e_n?$$
For $\lambda \in \mathbb{R}$ and $k > 0$, $$ \frac{k^{2}}{2}\left[\frac{1}{\lambda+ik}+\frac{1}{\lambda-ik}\right] =\lambda \frac{k^{2}}{\lambda^{2}+k^{2}} $$ So the above function converges to $\lambda$ as $k\rightarrow\infty$. You have $$ e^{iT_{k}}x = \sum_{n}\exp\left\{i\lambda_{n}\frac{k^{2}}{\lambda_{n}^{2}+k^{2}}\right\}a_{n}e_{n}. $$ Therefore, $$ \left\|e^{iT_{k}}x-\sum_{n}e^{i\lambda_{n}}a_{n}e_{n}\right\|^{2} = \sum_{n}\left|e^{i\lambda_{n}}-\exp\left(i\lambda_{n}\frac{k^{2}}{\lambda_{n}^{2}+k^{2}}\right)\right|^{2}|a_{n}|^{2}\|e_{n}\|^{2} $$ The exponential terms on the right are uniformly bounded and converge to $0$ as $k\rightarrow\infty$ for each fixed $n$. And $$ \|x\|^{2}=\sum_{n}|a_{n}|^{2}\|e_{n}\|^{2} < \infty. $$ So you can apply the Lebesgue dominated convergence theorem to see that the right side converges to $0$ as $k\rightarrow\infty$. And that's what you want.