Property of spectrum of an operator: $\sigma(T)=\sigma_{app}(T)\cup\sigma_{comp}(T)$

I am reading ''class notes on operator theory on hilbert spaces'' by John Petrovic (http://homepages.wmich.edu/~petrovic/courses/Math6780/m678notesch1.pdf) And in page 39 it gives us the following exercise:

If $H$ is a Hilbert space and $T\in\mathcal{L}(H)$ prove that $\sigma(T)=\sigma_{app}(T)\cup\sigma_{comp}(T)$

where: $$\sigma(T)=\lbrace\lambda\in\mathbb{C}|T-\lambda I\text{ not invertible }\rbrace$$ $$\sigma_{app}(T)=\lbrace\lambda\in\mathbb{C}|T-\lambda I\text{ not bounded below }\rbrace$$ $$\sigma_{comp}(T)=\lbrace\lambda\in\mathbb{C}|\overline{\text{Im}(T-\lambda I) }\neq H\rbrace$$ Trivially $(\sigma_{app}(T)\cup\sigma_{comp}(T))\subset \sigma(T)$. Here is my attemp for the other inclusion:

I know that if an operator is bounded below then it is injective, so if $\lambda\in\sigma(T)\setminus\sigma_{ap}(T)$ then $T-\lambda I$ is injective but not surjective (otherwise it would be bijective). And I want to prove that $\lambda\in\sigma_{comp}(T)$, but the fact that $T-\lambda I$ is not surjective is not enough to say that $\overline{\text{Im}(T-\lambda I)}\neq H$. What shall I do next?

Thanks for your help.

Solution 1:

Suppose $S\in \mathcal L (H)$, range of $S$ is dense and there exists $c>0$ such that $\|Sx\| \geq c\|x||$ for all $x$. Then range of $S$ is closed: If $Sx_n \to y$ then $(S(x_n))$ is Cauchy and $\|x_n-x_m\| \leq \frac 1 c \|Sx_n-Sx_m\| \to 0$. Since $H$ is complete, $(x_n)$ convregs to some $x$ and so $y=\lim Sx_n=Sx$.

Since the range of $S$ is closed and dense $S$ is surjective and since $\|Sx\| \geq c\|x||$ for all $x$ we see that $S$ is invertible. Apply this argument to $S=T-\lambda I$.