Let $H$ be a Hilbert space and let $T: H \to H$ be a compact and self-adjoint operator. Let $$T = \sum_{\lambda}\lambda P_\lambda $$ be its spectral decomposition (which converges in the operator norm), where $P_\lambda$ is the orthogonal projection on the $\lambda$-eigenspace.
Let us define $$\cos(T)x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}T^{2n}x. $$
How can we show that $$\cos(T)x = \sum_{\lambda}\cos(\lambda) P_\lambda x? $$
The proof seems pretty easy to me:
\begin{align}
\cos(T)x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}T^{2n}x \\
&= \sum_{n=0}^\infty \sum_{\lambda} \frac{(-1)^n}{(2n)!} \lambda^{2n} P_\lambda x\\
&= \sum_{\lambda} \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \lambda^{2n} P_\lambda x\\
&= \sum_{\lambda} \cos(\lambda) P_\lambda x,
\end{align}
or, denoting by $(\lambda_i)_i$ the sequence of non-zero eigenvalues of $T$, writing this in a more explicit manner:
\begin{align}
\cos(T)x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}T^{2n}x \\
&= \lim_{k \to \infty} \sum_{n=0}^k \frac{(-1)^n}{(2n)!}T^{2n}x \\
&= \lim_{k \to \infty} \left( \sum_{n=0}^k \frac{(-1)^n}{(2n)!} \cdot \left( \lim_{l \to \infty} \sum_{i=1}^l \lambda_i^{2n} P_{\lambda_i} x \right) \right) \\
&= \lim_{k \to \infty} \left( \lim_{l \to \infty} \left( \sum_{n=0}^k \sum_{i=1}^l \frac{(-1)^n}{(2n)!} \lambda_i^{2n} P_{\lambda_i} x \right) \right)
\end{align}
However, I run into the following problem: what is the reason we can interchange the two limits in the last equality (if we can even do that)? I believe this is because the power series of $\cos$ converges on every compact subset of $\mathbb{C}$, and $\sigma(T)$ is a compact subset, but I am not sure.
Best Answer
Observe that $\oplus_{\mu}P_{\mu}$ is the identity.
\begin{split}\sum_{n=0}^{\infty}\sum_{\lambda}\frac{(-1)^{n}}{(2n)!}\lambda^{2n}P_{\lambda}x&=\sum_{\mu}P_{\mu}(\sum_{n=0}^{\infty}\sum_{\lambda}\frac{{(-1)}^{n}}{(2n)!}\lambda^{2n}P_{\lambda}x)\\&=\sum_{\mu}\sum_{n=0}^{\infty}\sum_{\lambda}\frac{{(-1)}^{n}}{(2n)!}\lambda^{2n}P_{\mu}P_{\lambda}x\\&=\sum_{\mu}\sum_{n=0}^{\infty}\frac{{(-1)}^{n}}{(2n)!}\mu^{2n}P_{\mu}x. \end{split}