> [!definition] > > Let $E$, $F$ be [[Normed Vector Space|normed spaces]], and $T \in L(E, F)$ be a [[Bounded Linear Map|bounded linear map]]. $T$ is **Fredholm** if > 1. $\ker{(T)}$ is finite-dimensional. > 2. $T(E)$ is [[Closed Set|closed]][^1] and finite codimensional ($F/T(E)$ is finite-dimensional). > > The set $\text{Fred}(E, F)$ is the space of all Fredholm operators from $E$ to $F$. If $T \in \text{Fred}(E, F)$, then > $ > \text{Ind}(T) = \dim\ker T - \dim{F/T(E)} > $ > is its **index**. > [!theorem] > > Let $E$ be a [[Banach Space|Banach space]]. If $u: E \to E$ is [[Compact Operator|compact]], then $I - u$ is Fredholm. > > *Proof*. The identity restricted to $\ker(I - u)$ is equal to $u$, and thus compact. Since the kernel is locally compact, it is finite-dimensional. > > Let $T = I - u$ and [[Split Subspace|split]] $E = \ker(I - u) \oplus G$, where $T|_{G}$ is injective. To show that $T(E)$ is closed, it's sufficient to show that $T(G)$ is closed. To show that $T(G)$ is closed, it's sufficient to show that the inverse $(T|_G)^{-1}: T(G) \to G$ is continuous. > > Suppose that $(T|_G)^{-1}$ is not continuous, then there exists $\eps > 0$ and a sequence $\seq{x_n} \subset G$ such that $\norm{x_n} \ge \eps$ for all $n \in \nat$ and $Tx_n \to 0$ as $n \to \infty$. Since $u$ is compact, there exists a subsequence $n_k$ such that $ux_{n_k}$ is [[Cauchy Sequence|Cauchy]]. This implies that $x_{n_k} - ux_{n_k} \to 0$, and $x_{n_k}$ converges as well. This contradicts the fact that $G$ is a closed subspace and only intersects $\ker(I-u)$ at $0$. Thus $(T|_G)^{-1}$ is continuous, and $T(G)$ is closed. > > To show that $T(E)$ has finite codimension, suppose that it has infinite codimension. Construct a sequence > $ > TE = H_0 \subset H_1 \cdots \subset H_n \subset \cdots > $ > by adding one dimension inductively, such that $H_n$ is closed for all $n$. Now let $x_n \in H_n$ such that $\norm{x_n} = 1$ and the [[Quotient Norm|quotient norm]] $\norm{x_n}_{/H_{n-1}} \ge 1/2$. From here, let $n \in \nat$ and $k < n$, then > $ > \norm{ux_n - ux_k} = \norm{x_n - Tx_n - x_k + Tx_k} \ge 1/2 > $ > where $Tx_n - x_k + Tx_k \in H_{n - 1}$. Thus $\seq{ux_n}$ is not Cauchy, which contradicts the fact that $u$ is compact. > [!theorem] > > Let $E$ and $F$ be Banach spaces, then > 1. $\text{Fred}(E, F)$ is open in $L(E, F)$. > 2. The index is a continuous, hence constant on connected components. > > *Proof*. Let $S \in \text{Fred}(E, F)$, and $T \in L(E, F)$. Let $N = \ker S$ and split $E = N \oplus G$, then $S$ induces a [[Space of Toplinear Isomorphisms|toplinear isomorphism]] from $G$ to $S(E) = S(G)$. Split $F = S(G) \oplus H$, then the map > $ > G \times H \to F \quad (x, y) \mapsto Sx + y > $ > is an isomorphism. Since the space of isomorphisms is an open set, there exists $\eps > 0$ such that the mapping > $ > G \times H \to F \quad (x, y) \mapsto Tx + y > $ > is an isomorphism for all $T \in B(S, \eps)$. This implies that $T(G) \cap H = \bracs{0}$, thus $T|_G: G \to T(G)$ is an isomorphism as well, so $\dim \ker(T) \le \dim \ker(S)$, and $\dim F/T(G) \le \dim H$. > > Now for continuity, split > $ > E = G \oplus \ker(T) \oplus M > $ > since $G$ already has finite codimension. From here, $T$ induces an isomorphism > $ > G \oplus M \to T(G \oplus M) = T(G) \oplus T(M) > $ > so $\dim M = \dim T(M)$. Thus > $ > \begin{align*} > \text{ind}(T) &= \dim \ker (T) - (\dim H - \dim T(M)) \\ > &= \dim \ker (T) + \dim M - \dim H \\ > &= \dim \ker (S) - \dim H = \text{ind}(S) > \end{align*} > $ > [!theorem] > > Let $E$ be a Banach space and $u \in K(E, E)$. If $I - u$ is injective, then $I - u$ is an isomorphism. > > *Proof*. For each $t \ge 0$, $tu$ is compact, so $t \mapsto tu$ is continuous, and $\text{ind}(I - tu)$ is constant and identically zero. This implies that $\dim E/(I - u)(E) = 0$, so $I - u$ is an isomorphism by the [[Open Mapping Theorem|open mapping theorem]]. > [!theorem] > > Let $E, F, G$ be Banach spaces, $S \in \text{Fred}(E, F)$, and $T \in \text{Fred}(F, G)$. Then $TS \in \text{Fred}(E, G)$ with > $ > \text{ind}(TS) = \text{ind}(T) + \text{ind}(S) > $ > *Proof*. Denote $E_S = \ker(S)$, then > $ > \begin{align*} > \ind{S} &= (E_S:0) - (F: SE) \\ > \ind{T} &= (F_T:0) - (G:TF) \\ > \ind{TS} &= (E_{TS}:0) - (G:TSE) > \end{align*} > $ > given that > $ > \bracs{0} \subset \ker(S) \subset \ker{(TS)} > $ > and > $ > TSE \subset TF \subset G > $ > Hence > $ > \begin{align*} > (E_{TS}: 0) &= (E_{TS}: E_S) + (E_S: 0) \\ > (G: TSE) &= (G: TF) + (TF: TSE) > \end{align*} > $ > yielding that $TS$ is Fredholm. Using a lemma on $SE \subset F$, > $ > (F:SE) = (TF:TSE) + (F_T: SE \cap F_T) > $ > combining all the equations yields that > $ > \ind{TS} - \ind{T} - \ind{S} = (E_{TS}:S) - (SE \cap F_T: 0) > $ > Split $E_{TS} = E_S \oplus W$ where $W$ is finite-dimensional, then > $ > E = E_{TS} \oplus U = E_{S} \oplus W \oplus U > $ > so > $ > SE = S(W \oplus U) = SW \oplus SU > $ > Now, since $TS(E_{TS}) = 0$, $SW \subset F_T$. On the other hand, if $y = Sx \in SE$ and $TSx = 0$, then we can decompose $x = x_s + x_w \in E_S \oplus W$, where $Sx_w \in SW$. Thus $SE \cap F_T = SW$, so > $ > (SE \cap F_T: 0) = (W:0) = (SW:0) = (E_{TS}:E_S) > $ > since $S$ is an isomorphism on $W$. [^1]: While finite codimension implies that $T(E)$ is closed, sometimes it's useful to prove that $T(E)$ is closed and then work out the codimension.