Problem 1: Let $T$ be a compact, linear operator on a Hilbert Space $H$ and let $\lambda \neq 0$.
(i) Show that the range of $\lambda I - T$,
$$ \{g \in{H} : \exists f\in{H} \quad s.t. \quad (\lambda I - T)f = g \} $$
is a closed subspace of $H$.
\textit{Solution}: In the proof of Lax-Milgram, the coercivity condition of the bilinear functional $\phi$ allowed us to develop the inequality
$$||x|| \leq \frac{1}{K} ||Ax|| \quad \text{for some $K > 0$}$$
This allowed us to show that $Ran(A)$ was closed. Let $E_\lambda = Null(\lambda I - T)$. Since $E_\lambda$ is closed, we have the decomposition $H = E_\lambda \oplus E_\lambda^\perp$. Then,
$$\lambda I-T = (\lambda I -T)|_{E_\lambda}+(\lambda I-T)|_{E_\lambda^\perp} = 0|_{E_\lambda}+S$$
If we can show that
$$\exists c>0 \quad \text{s.t.} \quad c||f|| \leq ||Sf|| \quad \forall f\in{E_\lambda^\perp} \quad (1)$$
then taking $g_n \in{Ran(\lambda I-T)}$ and $g_n \longrightarrow g$ we get $g_n = 0|_{E_\lambda}(f_n) + S(f_n) \longrightarrow g$. So, $\{Sf_n\}$ converges, implying $\{Sf_n\}$ is Cauchy:
$$||Sf_n-Sf_m|| \longrightarrow 0 \quad \text{as} \quad n,m \longrightarrow \infty $$
By (1) and the linearity of $S$,
$$c||f_n-f_m|| \leq ||Sf_n-Sf_m|| \longrightarrow 0$$
So, $\{f_n\}$ is Cauchy. Since $H$ is complete, $\{f_n\}$ is convergent: $f_n \longrightarrow f$. Since $S$ is continuous, $Sf_n \longrightarrow Sf$. But, $Sf_n \longrightarrow g$ as well. So, $Sf = g$ i.e. $g\in{Ran(S)} = Ran(\lambda I-T)$.
\textit{Proof of (1):} Suppose to the contrary that $\forall M>0$, $\exists h_M \in{E_\lambda^\perp}$ s.t. $M||h|| > ||Sh||$. So, $M>||Su||$ for some $||u||=1$. Equivalently, $\exists \{u_n\} \subseteq E_\lambda^\perp$, $||u_n||=1$ s.t. $||Su_n||\longrightarrow 0$ i.e. $$||(\lambda I-T)u_n|| \longrightarrow 0$$
Since $E_\lambda \cap E_\lambda^\perp =\{0\}$ and $||u_n||=1$, $\{u_n\}$ cannot converge to an element in $E_\lambda$.
In other words, $\lambda$ is an approximate eigenvalue of $T|_{E_\lambda^\perp}$. Since $\lambda \neq 0$ and $T|_{E_\lambda^\perp}$ being the restriction of a compact operator is compact, $\lambda$ is an eigenvalue of $T|_{E_\lambda^\perp}$...which is a contradiction since all eigenvectors with eigenvalue $\lambda$ are contained in $E_\lambda$. $\square$
(ii) Show by example that $Ran(\lambda I-T)$ may not be closed if $\lambda = 0$.
\textit{Solution:} With $\lambda = 0$, we want to find an example of a linear, compact operator $T$ s.t. $Ran(T)$ is not a closed subset of $H$. Consider
$$H = l_2(\mathbb{N}) = \{(x_j):x_j \in{\mathbb{C}}, \sum_{j=1}^\infty|x_j|^2 < \infty \}$$
and let $$T(x_1,x_2,x_3,...) = (x_1,\frac{x_2}{2},...,\frac{x_n}{n},...)$$
Then, $T$ is compact since $T$ is the limit of a sequence of finite rank operators: $T_1(x_1,x_2,...) = (x_1,0,0,...)$, $T_2 = (x_1,\frac{x_2}{2},0,0,...)$ etc. and
$(T-T_n)x = (0,0,...,0,\frac{x_{n+1}}{n+1},\frac{x_{n+2}}{n+2},...)$. So,
$$||T-T_n|| = \sup_{||u||=1}||(T_n-T)u|| \leq \frac{1}{n} \longrightarrow 0$$
$T$ also happens to be symmetric since $\left\langle Tx,y \right\rangle=\sum_{j=1}^\infty \frac{x_j}{j}y_j=\sum_{j=1}^\infty x_j\frac{y_j}{j}= \left\langle x, Ty \right\rangle$. Now, $x=\{\frac{1}{j}\}_{j=1}^\infty \in{l_2}$ however, $y=\{1\}$ is s.t. $Ty=x$ yet $y \not \in{l_2}$. So, $Ran(T)$ is not all of $l_2$. However, given any $y=(y_1,y_2,...) \in{l_2}$ then $x^{(n)} = (y_1,2y_2,...,ny_n,0,0,...)$ is s.t. $Tx_n \longrightarrow y$. So, $\overline{Ran(T)}=l_2$.
(iii) Show that $\lambda I-T$ is invertible iff $\bar{\lambda} I -T^*$ is invertible.
\textit{Solution:} We will show the following circle of implications: $Ran(\lambda I-T)=H \Longrightarrow \bar{\lambda}I-T^*$ is invertible $\Longrightarrow \lambda I - T$ is invertible.
Assume that $\lambda I-T$ is surjecive i.e. $Ran(\lambda I -T)=H$. Let $h\in{Null(\bar{\lambda}I-T^*)}$ and let $f\in{H}$ be arbitrary. Since $Ran(\lambda I_T)=H$, there exists $g\in{H}$ s.t. $(\lambda I-T)g=f$. Then, $\left\langle f,h \right\rangle=\left\langle (\lambda I-T)g,h \right\rangle=\left\langle g,(\lambda I-T)^*h \right\rangle = \left\langle g,(\bar{\lambda} I-T^*)h \right\rangle = \left\langle g,0 \right\rangle = 0$. Since $f$ was arbitrary, this implies that $h=0$ i.e. $Null(\bar{\lambda}I-T^*) = \{0\}$. Thus, $\bar{\lambda} I-T^*$ is invertible.
Now let $S = \bar{\lambda} I-T^*$. We want to show that $S^*$ is invertible by showing that $S^*(S^{-1})^*=(S^{-1})^*S^* = I$. Let $f,g\in{H}$. Since $S^{-1}$ exists and is bounded,
$\left\langle f,(S^{-1})^*S^*g \right\rangle = \left\langle S^{-1}f,S^*g \right\rangle = \left\langle SS^{-1}f,g \right\rangle = \left\langle f,g \right\rangle = \left\langle S^{-1}Sf,g \right\rangle = \left\langle Sf,(S^{-1})^*g \right\rangle = \left\langle f,S^*(S^{-1})^*g \right\rangle$. Therefore, $S^* = \lambda I-T$ is invertible and, in particular, surjective. This completes the circle of implications.
