Basic Fredholm Theory

Kreyszig Functional Analysis

Let $X$ be a normed space and $T:X\to X$ be a compact linear operator. Note the adjoint operator $T^{\times }:X^{\prime }\to X^{\prime }$ must then also be compact. Let $\lambda$ be a fixed nonzero complex number. We consider the solvability of (that is, the existence of a $x\in X$ or $f\in X^{\prime }$ satisfying) the following equations, where $y\in X$ and $g\in X^{\prime }$ are treated as givens:

(1) $T_{\lambda }x=Tx-\lambda x=y$ (2) $T_{\lambda }x=Tx-\lambda x=0$ (3) $T_{\lambda }^{\times }f=T^{\times }f-\lambda f=g$ (4) $T_{\lambda }^{\times }f=T^{\times }f-\lambda f=0$

Theorem 1: (1) is solvable if and only if $y$ satisfies $f(y)=0$ for all $f$ satisfying (4). Proof) This means $y\in T_{\lambda }(X)$ if and only if all functionals $f$ destroying $T_{\lambda }(X)$ also destroy $y$, which is intuitively trivial. If $y=T_{\lambda }x\in T_{\lambda }(X)$, for any $f$ such that $T_{\lambda }^{\times }f=0$, $f(y)=f(T_{\lambda }x)=(T_{\lambda }^{\times }f)x=0$. If $y\notin T_{\lambda }(X)$, as $T_{\lambda }(X)$ is closed, there exists a functional $f\in X^{\prime }$ such that $f(y)=\delta >0$ and $f(v)=0$ for all $0\in T_{\lambda }(X)$. Thus, $T_{\lambda }^{\times }f=0$ while $f(y)\ne 0$.

Theorem 2: For all values $y$ for which (1) is solvable, there is a solution of minimum norm $\tilde{x}$, and there exists a constant $c>0$ independent of $y$ such that

$$\Vert \tilde{x}\Vert \le c\Vert y\Vert$$

Proof) For any $y\in T_{\lambda }(X)$, the set of solutions of (1) can be represented as $x_0+\mathcal{N}(T_{\lambda })$ where $x_0$ is some arbitrary element satisfying $T_{\lambda }{x}_0=y$. It suffices to show there is a “projection” in this solution space minimally separated from the origin. We begin by taking

$$\delta =\underset{v\in \mathcal{N}(T_{\lambda })}{\inf }\Vert x_0+v\Vert$$

Let $p(v)=\Vert x_0+v\Vert$ and take a sequence $(v_n)$ in $\mathcal{N}(T_{\lambda })$ such that $p(v_n)\to \delta$. Clearly, $v_n$ is bounded; as $I=(T-T_{\lambda })/\lambda$, if we take a subsequence $(v_{n_k})$ such that $T{v}_{n_k}$ converges, $v_{n_k}$ is convergent to some point $v$ in $X$. By the continuity of $p$, this $v$ achieves $\Vert x_0+v\Vert =\delta$, and $\tilde{x}=x_0+v$ is our desired solution of minimal norm. A notable property of $\tilde{x}$ is that, for any $w\in \mathcal{N}(T_{\lambda })$, $\Vert \tilde{x}-w\Vert \ge \Vert \tilde{x}\Vert$ by the minimality of $\Vert \tilde{x}\Vert$. In other words, as $\tilde{x}$ is the “projection” of the origin onto the solution space, it is guaranteed to stay distant from the null space.

Assume there is no fixed constant $c>0$ such that $\Vert \tilde{x}\Vert \le c\Vert y\Vert$. We can then take a sequence $(y_n)\in T_{\lambda }(X)$ such that, if we let $(x_n)$ be their corresponding minimum-norm solutions, $\Vert x_n\Vert /\Vert y_n\Vert \to \infty$. As (1) is linear, we can scale all the vectors such that $\Vert x_n\Vert =1$, after which our constraint becomes $\Vert y_n\Vert \to 0$. Thus, there is a sequence $y_n\to 0$ in $T_{\lambda }(X)$ such that their minimum-norm solutions are all of norm $1$.

By compactness, there is a subsequence $(x_{n_k})$ such that $T{x}_{n_k}$ is convergent. As $T_{\lambda }{x}_n\to 0$, $x_{n_k}\to x$ for some $x\in X$. This $x$ must satisfy $T_{\lambda }x=0$ and thus $x\in \mathcal{N}(T_{\lambda })$. However, for any $x_n$, this implies $\Vert x_n-x\Vert \ge \Vert x_n\Vert =1$ and thus no subsequence could possibly converge to $x$, a contradiction.

Theorem 3: (3) is solvable if and only if $g$ satisfies $gx=0$ for all $x\in X$ satisfying (2). Proof) This intuitively means $g$ can be represented as the composition of $T_{\lambda }$ and a bounded linear functional iff $g$ kills $\mathcal{N}(T_{\lambda })$, which seems clear. If $g=T_{\lambda }^{\times }f$ for some $f$, for any $x\in \mathcal{N}(T_{\lambda })$, $gx=f{T}_{\lambda }x=0$. For the other direction, if $gx=0$ for all $x\in \mathcal{N}(T_{\lambda })$, we can define $f:T_{\lambda }(X)\to \mathbb{C}$ such that $f(T_{\lambda }x)=g(x)$. This is well-defined as if $T_{\lambda }{x}_1=T_{\lambda }{x}_2$, $x_1-x_2\in \mathcal{N}(T_{\lambda })$ and thus $g(x_1)=g(x_2)$. $f$ is clearly linear and must be bounded as, for any $y\in T_{\lambda }(x)$, there is some $\tilde{x}$ such that $T_{\lambda }\tilde{x}=y$ and $\Vert \tilde{x}\Vert \le c\Vert y\Vert$ for some fixed $c>0$. Thus,

$$\Vert f(y)\Vert =\Vert f(T_{\lambda }\tilde{x})\Vert =g(\tilde{x})\le c\Vert g\Vert \Vert y\Vert$$

By Hahn-Banach, there exists a functional $\tilde{f}\in X^{\prime }$ that is an extension of $f$. This $\tilde{f}$ satisfies $T_{\lambda }^{\times }\tilde{f}=g$, and we are done.

We briefly return to compact operator properties as Theorem 2 allows us to arrive at some interesting new conclusions.

Theorem 4: $T_{\lambda }(X)=X$ iff $\mathcal{N}(T_{\lambda })=\{0\}$. In this case, the inverse $R_{\lambda }(T)$ exists and, in fact, is in the space $B(X,X)$. Proof) Both $T_{\lambda }(X)=X$ and $\mathcal{N}(T_{\lambda })=\{0\}$ imply null space/range stabilization occurs beginning at $T_{\lambda }^0$, which implies the other statement. In this case, as $T_{\lambda }$ is bijective it has an inverse $R_{\lambda }(T)$ defined on the entire space $X$. The inverse must clearly be linear and, by Theorem 2, be bounded. Thus, $R_{\lambda }(T)\in B(X,X)$ and $\lambda \in \rho (T)$. We also provide a more basic proof of $T_{\lambda }(X)=X⟺\mathcal{N}(T_{\lambda })=\{0\}$ below, for completeness:

If $T_{\lambda }(X)=X$ and there is some $x\in X$ such that $T_{\lambda }x=0$, we can form an infinite chain $(x_n)$ where $x_1=x$ and $T_{\lambda }{x}_{n+1}=x_n$. For any $n$, $x_n\notin \mathcal{N}(T_{\lambda }^{n-2})$ but $x_n\in \mathcal{N}(T_{\lambda }^{n-1})$, and thus there is a contradiction as the null spaces must stabilize. This implies $\mathcal{N}(T_{\lambda })=\{0\}$. Alternatively, if $\mathcal{N}(T_{\lambda })=\{0\}$, the only solution of (2) is $0$ and thus by Theorem 3 (3) is solvable for all $g$. This means $T_{\lambda }^{\times }(X^{\prime })=X^{\prime }$, which implies $\mathcal{N}(T_{\lambda }^{\times })=\{0\}$ by the first part of this theorem. By Theorem 1, this means (1) is solvable for all $y\in X$, or $T_{\lambda }(X)=X$.

Theorem 5: For any linearly independent set $\{f_1,f_2,\dots ,f_m\}\in X^{\prime }$, there are elements $z_i\in X$ such that $f_j(z_i)={\delta }_{ji}$ where $\delta$ is the Kronecker delta. Proof) We first proceed by induction on $m$. The $m=1$ case is trivial, and assume this holds for $m-1$. It suffices to show there exists a $z_m$ such that $f_m(z_m)=1$ with $f_j(z_m)=0$ for all $j<m$. Let $z_1,z_2,\dots ,z_{m-1}$ be such that $f_j(z_i)={\delta }_{ij}$ for all $i,j\le m-1$.

Assume no such $z_m$ exists. Then, for any $v\in M=\{x\in X|f_1(x)=f_2(x)=\dots =f_{m-1}(x)=0\}$, $f_m(v)=0$. Furthermore, for any $x\in X$, note that $\tilde{x}=x-{\sum }_{i<m}{f}_i(x)z_i$ satisfies $f_i(\tilde{x})=0$ for all $i<m$ and thus $f_m(\tilde{x})=0$. Thus for any $x\in X$,

$$0=f_m(\tilde{x})=f_m(x-\sum _{i<m}{f}_i(x)z_i)=f_m(x)-\sum _{i<m}{f}_m(z_i)f_i(x)$$

This implies $f_m$ is in the span of $\{f_1,f_2,\dots f_{m-1}\}$, contradicting linear independence.

Theorem 6: Let $n=dim\mathcal{N}(T_{\lambda })$ and $m=dim\mathcal{N}(T_{\lambda }^{\times })$. Then, $n=m$. Proof) Note both null spaces are finite dimensional as $T$ and $T^{\times }$ are compact. $n=0$ implies the only solution to (2) is $x=0$, which implies (3) is solvable for all $g$. This, in turn, implies $m=0$ by Theorem 4. The other direction works similarly. We now deal with the case where $n,m>0$.

Let $\{x_1,x_2,\dots ,x_n\}$ be a basis of $\mathcal{N}(T_{\lambda })$ and $\{f_1,f_2,\dots ,f_m\}$ be a basis of $\mathcal{N}(T_{\lambda }^{\times })$. we associate each $x_i$ with a functional $g_i\in X^{\prime }$ such that $g_i(x_j)={\delta }_{ij}$ ($\delta$ is the Kronecker delta) and associate each $f_i$ with a $z_i$ such that $f_i(z_j)={\delta }_{ij}$. Our intuition is that the $g_i$s should capture the set of functionals $f$ that don’t kill the null space $\mathcal{N}(T_{\lambda })$, i.e. the functionals not in $T_{\lambda }^{\times }(X^{\prime })$; for any $f\in X^{\prime }$

$$\tilde{f}=f-\sum _{i=1}^{n}f(x_i)g_i$$

satisfies $\tilde{f}(x_i)=0$ for all $1\le i\le n$, and thus $\tilde{f}\in T_{\lambda }^{\times }(X^{\prime })$. Therefore, the $g_i$s are a set of linearly independent functionals in $X^{\prime }$ such that $X^{\prime }=T_{\lambda }^{\times }(X^{\prime })\oplus Span\{g_1,g_2,\dots ,g_n\}$. Similarly, our intuition is that the $f_i$s are the functionals that kill $T_{\lambda }(X)$ so taking directions $z_i$ that the $f_i$s uniquely preserve should form a basis that “fills” the space “orthogonal to” $T_{\lambda }(X)$. Formally, $X=T_{\lambda }(X)\oplus Span\{z_1,z_2,\dots z_m\}$ as if $\exists v\notin T_{\lambda }(X)\oplus Span\{z_1,z_2,\dots ,z_m\}$, there is some functional $f\in X^{\prime }$ such that $f(v)=1$ and $f(w)=0$ for any $w\in T_{\lambda }(X)\oplus Span\{z_1,z_2,\dots ,z_m\}$. Then, $f\in \mathcal{N}(T_{\lambda }^{\times })$ and $f\ne 0$ but $f(z_i)=0$ for all $i$, a contradiction.

How can we leverage this structure to prove $n=m$? The key object is the operator

$$S_{\lambda }=T_{\lambda }+\sum _{i=1}^n{g}_i{z}_i$$

This operator should be a neat bijection from $X$ to $X$ as the $g_i$s pick up all the directions in $\mathcal{N}(T_{\lambda })$ neglected by $T_{\lambda }$ and then assign each of them a different direction $z_i$ that populate the space in $X$ outside $T_{\lambda }(X)$. Thus, if there is a dimension mismatch and $n\ne m$, it should be very apparent through $S_{\lambda }$.

Assume, FTSOC, that $n<m$. $S_{\lambda }$ must have an empty null space; for any $x$ such that $S_{\lambda }x=0$, as $T_{\lambda }(X)\oplus Span\{z_1,\dots ,z_m\}=X$, $T_{\lambda }x=g_1(x)=g_2(x)=\dots =g_n(x)=0$; this implies $x=0$. Note that the base operator $S=T+\sum g_i{z}_i$ is the sum of compact operators and thus is compact; thus, $\mathcal{N}(S_{\lambda })=\{0\}$ implies $S_{\lambda }(X)=X$. This is a contradiction as $z_{n+1}$ is clearly outside the range of $S_{\lambda }$.

Now assume $m<n$. We consider the same operator

$$S_{\lambda }x=T_{\lambda }x+\sum _{i=1}^m{g}_i(x)z_i$$

As $T_{\lambda }(X)\oplus Span\{z_1,\dots ,z_m\}=X$, $S_{\lambda }(X)=X$, which implies $\mathcal{N}(S_{\lambda })=\{0\}$. However, $S_{\lambda }{x}_{m+1}=0$, a contradiction.

Theorem 7: Let $T:X\to X$ be a compact linear operator on a normed space $X$. If $\lambda$ is a nonzero spectral value of $T$, $\lambda$ must be in the point spectrum. Proof) If $\lambda \in {\sigma }_p(T)$, we are done. Otherwise, the inverse $R_{\lambda }(T)$ must exist; this implies $\mathcal{N}(T_{\lambda })=\{0\}$ and thus $R_{\lambda }(T)\in B(X,X)$ by Theorem 4. This forces $\lambda \in \rho (T)$, a contradiction.