Orthonormal system from spectral theorem is complete in $(\ker K)\perp$

Question

I'm preparing for the exam in Functional Analysis and got stuck in this remark of our script.

It says:

The orthonormal system $(x_n)_n$ from the Spectral Theorem is an orthonormal basis of $(\ker K)\perp = \overline{K(H)}$ where $K$ is the compact, self-adjoint operator and $H$ is Hilbert space.

Could someone please explain? Thanks!

Answer 1

If $(y_n)$ is an orthonormal set in a Hilbert space then there are several ways of characterizing completeness of this orthonormal set. One of the equivalent properties is vectors of the form $\sum\limits_{k=1}^{n} a_ny_n$ form a dense subset of the Hilbert space. In our case this property is obvious that $x_n \in K(H)$ for all $n$, $(x_n)$ is orthonormal and $Kx=\sum\limits_{k=1}^{\infty} \lambda_n \langle Kx, x_n \rangle x_n$ from the spectral theorem so $(x_n)$ is complete.

Answer 2

I dont't know how much theory background you have, so I will try to make it as elementary as possible.

Let $K$ be a nonzero (for $K=0$ it is trivial) compact self-adjoint operator on a Hilbert space $H$. Then by the Spectral Theorem, there is an orthonormal system $(x_n)_n$ and a system of nonzero numbers $(\lambda_n)_n$ (the eigenvalues of $K$ possibly without $0$), such that \begin{equation} \tag1 K(x) = \sum_n \lambda_n \langle x,x_n \rangle x_n. \end{equation}

First, lets point out that since $K$ is self-adjoint, we indeed have $$\overline{K(H)} = (\text{ker}K^*)^\perp = (\text{ker}K)^\perp.$$

Now consider the closed linear span $H'$ generated by the system $(x_n)_n$. Then by $(1)$ we have $K(H) \subset H'$, and also $\overline{K(H)} \subset H'$ as $H'$ is closed. Now, take $x \in H'$. Then $x = \sum_n \langle x, x_n \rangle x_n$ as it is equal to its ON projection to $H'$. Take $y_N = \sum_{n\leq N} \frac{\langle x,x_n \rangle}{\lambda_n}x_n \in H$. Then by orthogonality $$K(y_N) = \sum_m \lambda_m \left\langle \sum_{n\leq N} \frac{\langle x,x_n \rangle}{\lambda_n}x_n,x_m \right\rangle x_m = \sum_{m\leq N} \lambda_m \frac{\langle x,x_m \rangle}{\lambda_m} x_m \overset{N}{\longrightarrow} \sum_m \langle x, x_m \rangle x_m=x.$$

Hence, $x \in \overline{K(H)}$ and $\overline{K(H)} \supset H'$.