# An easy application of the spectral theorem for a self-adjoint operator on $L^2$

Let $$(E,\mathcal E,\mu)$$ be a probability space, $$L^2_0(\mu):=\left\{f\in L^2(\mu):\int f\:{\rm d}\mu=0\right\}$$ and $$U:L^2(\mu)\to L^2(\mu)\;,\;\;\;f\mapsto\int f\:{\rm d}\mu=\langle 1,f\rangle_{L^2(\mu)}1.$$ Note that $$U$$ is a nonnegative self-adjoint linear operator on $$L^2(\mu)$$ with $$\mathcal N(U)=L^2_0(\mu)$$. Moreover, if $$(f_i)_{i\in I}$$ is an orthonormal basis of $$L^2(\mu)$$, then $$\sum_{i\in I}\langle Uf_i,f_i\rangle_{L^2(\mu)}=\sum_{i\in I}\left|\langle1,f_i\rangle_{L^2(\mu)}\right|^2=\left\|1\right\|_{L^2(\mu)}^2=1\tag1$$ by Parseval's identity. So, if $$|I|\le|\mathbb N|$$, then $$U$$ is trace-class (cf. When is $L^2(\mu)$ separable?)

If $$L^2(\mu)$$ is separable, we know that there is an orthonormal basis of $$L^2(\mu)$$ consisting of eigenvectors of $$U$$ (the only eigenvalues are $$0$$ and $$1$$). The reason is that in that case $$U$$ is compact by $$(1)$$. In the non-separable case, the general spectral theorem applies and yields a resolution of the identity. Actually, since our operator is bounded, there is a continuous linear, multiplicative and involutive $$\Phi:C(\sigma(U))\to\mathfrak L(L^2(\mu))$$ with $$\Phi(\operatorname{id})=U$$, $$\Phi(1)=\operatorname{id}_{L^2(\mu)}$$. How does $$\Phi$$ look like in our concrete situation? And how is it related to the resolution of the identity from the spectral theorem?

You know that $$U$$ is compact/Hilbert Schmidt/Trace Class, simply because it is rank-one. The Spectral Theorem is crazy overkill here, as $$U$$ is already a rank-one projection. The spectral decomposition of $$U$$ is
$$U=1\,U,$$ that is $$U$$ is the spectral projection of $$U$$ corresponding to the eigenvalue $$1$$.
The map $$\Phi$$, since $$\sigma(U)=\{0,1\}$$, is $$\Phi:\mathbb C^2\to \mathfrak L(L^2(\mu))$$ given by $$\Phi(a,b)=a\,(I-U)+b\,U.$$
• Thank you for your answer. I'm working in the real setting and $\Phi$ should be defined on the continuous real-valued functions on $\sigma(U)=\{0,1\}$. So, $\Phi$ should be defined differently. Oct 5, 2019 at 13:06
• @0xbadf00d No. What Martin means by $(a,b)$ is the function $f$ with $f(0) = a$ and $f(1) = b$. It is completely correct what he writes. Oct 5, 2019 at 13:28
• @amsmath Hm, I don't get it. What are $a,b$ then? Can you write $\Phi$ as a function $C(\sigma(U))\to\mathfrak L(L^2(\mu))$? Oct 5, 2019 at 14:02
• @0xbadf00d Puh, I told you what $a$ and $b$ are. But to make it even simpler: $\Phi(f) = f(0)(I-U) + f(1)U$. Oct 5, 2019 at 14:25
• @0xbadf00d, $C(\{0,1\},\mathbb R)=\mathbb R^{\{0,1\}}=\mathbb R^2$. Oct 5, 2019 at 14:43