Compact, Self-Adjoint, Nonnegative operators have at least one Eigenvector

Question

This is a statement presented in my class and I am having trouble to understand the proof given by the Professor:

Let $T \in K(H)$, where $K(H)$ represents the space of compact operators on a Hilbert space $H$. If $T$ is self-adjoint and nonnegative, then there exists $v \in H$ such that $\| v \| = 1$, $Tv = \mu v$ and $|\mu| = \| T \|$.

Here is the proof presented in the class and I will make bold the part that I am confused about:

Let $\{ u_n \}_{n = 1} ^\infty \subseteq H$ such that $\| u_n \| = 1$ and $\| Tu_n \| \to \| T \| \equiv \lambda$. (To start off, why can we even assume such a sequence exists?) We then have by self-adjointness of $T$ that $$ ((\lambda^2 - T^2)u_n, u_n) = \lambda^2 - \| Tu_n \|^2 \to 0. $$ Now we have $$ \hat{T} = \lambda^2 - T^2 \in \mathcal{L}(H) $$ to be symmetric (I am not sure what symmetric here means. I do not think we have introduced symmetric operators anywhere. My guess is that the Professor meant self-adjoint.) and non-negative so $$ \| \hat{T}u_n \| \leq \| \hat{T} \|^{1/2}(\hat{T}u_n, u_n)^{1/2} $$ (This follows from a previous result, knowing $\hat{T}$ is self-adjoint). Therefore, we have $$ \| (\lambda^2 - T^2)u_n \| \leq \| \hat{T} \|^{1/2}((\lambda^2 - T^2)u_n, u_n)^{1/2} \to 0. $$ Thus $\lambda^2 u_n^2 - T^2(u_n) \to 0$. Since $T$ is compact, there exists subsequence (as $\| u_n \| = 1$) that $T(u_{n_k}) \to w \in H$. The Professor then left the rest of the proof as an exercise and I do not see how to complete the proof. What is the $w$ that we found here? I suppose it is the candidate for the eigenvector? But I am not seeing how $Tw = \mu w$ for $|\mu| = \| T \|$ and $\| w \| = 1$.

Answer 1

Let $\{ u_n \}_{n = 1} ^\infty \subseteq H$ such that $\| u_n \| = 1$ and $\| Tu_n \| \to \| T \| \equiv \lambda$. (To start off, why can we even assume such a sequence exists?)

By definition, $||T||=\sup_{u \in B_H } ||Tu||$, so take any sequence $u_n \in B_H$ that converges to that supremum.

Now we have $$ \hat{T} = \lambda^2 - T^2 \in \mathcal{L}(H) $$ to be symmetric (I am not sure what symmetric here means. I do not think we have introduced symmetric operators anywhere. My guess is that the Professor meant self-adjoint.)

The problem is that the domain of $T^2$, which is the range of $T$, is not $H$. When an operator is not defined on the entire space, you need to be more careful. An (unbounded) operator $A$ whose domain, $D(A)$, is a subspace of $H$ is said to be symmetric if $(Ax,y)=(x,Ay)$ for any $x,y \in D(A)$. [This is equivalent to $D(A) \subset D(A^*)$ and $A^* \mid_{D(A)}=A $]. Moreover, $A$ is said to be self-adjoint if it is symmetric and $D(A^*)=D(A)$. Symmetric operators are generally not self-adjoint.

Thus $\lambda^2 u_n - T^2(u_n) \to 0$. Since $T$ is compact, there exists subsequence (as $\| u_n \| = 1$) that $T(u_{n_k}) \to w \in H$. The Professor then left the rest of the proof as an exercise and I do not see how to complete the proof. What is the $w$ that we found here? I suppose it is the candidate for the eigenvector? But I am not seeing how $Tw = \mu w$ for $|\mu| = \| T \|$ and $\| w \| = 1$.

I would argue as follows. Since $Tu_n\to w$ (for some subsequence), so $(Tu_n)$ is Cauchy in $H$. Since $$ |λ^2 u_n-λ^2u_m| \le |λ^2 (u_n-u_m) - (T^2(u_n) -T^2(u_m))| +|T^2(u_n) -T^2(u_m)| , $$ we see that $(u_n)$ is also Cauchy in $H$. Let $u = \lim_n u_n$. Then $Tu = \lim_n Tu_n = w$. Since $λ^2 u_n -T^2 u_n \to 0$, we must have that $λ^2u=Tw= T^2u$. Hence, $λ$ or $-λ$ is an eigenvalue of $T$. [To see this, note that $$ (T-λ)(Τ+λ)u=0 .$$ If $(T+λ)u \neq 0$, then $λ$ is an eigenvalue of $T$, whereas, if $(T+λ)u=0$ then $-λ$ is an eigenvalue of $T$.] This shows that $\pm ||T||$ is an eigenvalue of $T$.