I have the following problem. Its supposed to be very easy but haven't yet properly digested the topic so i struggle.
Let H be be a separable complex Hilbert space with dim H $=\infty$, and let $A\in L(H)$ be compact and self-adjoint.
Question 1.) Let $z_0\in\Bbb{C}$ and $\epsilon>0$. Suppose there exists $u\in H$, $||u||=1$, so that $||(z_0-A)u||\leq\epsilon$. Show that there exists a point $z\in\sigma(A)$ with $|z-z_0|\leq\epsilon$.
What i have tried:
From my understanding, for the lack of a better word, $z_0$ is an "almost" eigenvalue to the vector $u$. In the special case where $||(z_0-A)u||=0$ its even an eigenvalue and especially $z:=z_0$ fullfills the conditions. Now let $0<\epsilon'\leq||(z_0-A)u||\leq\epsilon$. I explored several dead ends here. One of the things that bothers me, is that $u$ is fixed.
Question 2.) Let $B\in L(H)$ be compact and self-adjoint. Using the first part, show that for all $z_0\in\sigma(A)$ there exists $z\in \sigma(A+B)$ with $|z-z_0|\leq ||B||_{L(H)}$.
What i have tried:
A+B is compact and self-adjoint. If i could show the existence of some $u$ with $||u||=1$ such that $||(z_0-(A+B))u||\leq ||B||_{L(H)}$, it follows from 1.). I would assume that i have to utilise the fact, that in this setting i can find eigenvalues and orthonormal base for the image of A+B.
If somebody could have a look i would be thankful.
For 1) take the spectral decomposition of $A$, i.e., $Ax_k = \lambda_k x_k$. Then $$ u = u_0+ \sum_k u_kx_k $$ with $u_k = \langle u,x_k\rangle$ and $u_0 \in \ker A$. Then $$ (z_0 - A)u = z_0 u_0 + \sum_k (z_0 - \lambda_k) u_kx_k, $$ and $$ \|(z_0 - A)u\|^2 = |z_0|^2 \|u_0\|^2 + \sum_k |z_0 - \lambda_k|^2 |u_k|^2 \le \epsilon^2. $$ Since $\|u_0\|^2 + \sum_k|u_k|^2 = \|u\|^2 =1$, at least one of the factors of non-zero coefficients $\|u_0\|^2$ and $|u_k|^2$ has to be $\le \epsilon^2$:
More precisely, one of the following statements has to be true:
$\|u_0\|^2 > 0$ and $|z_0|^2 \le\epsilon^2$
there exists $k$ such that $u_k\ne0$ and $|z_0 - \lambda_k|^2\le\epsilon^2$
In the first case, $0$ is the desired eigenvalue, in the second case it is $\lambda_k$.