Application: Fredholm Operators

The most important class of operators with closed range are the Fredholm operators, where the four subspace picture is as clean as in finite dimensions, even in non-reflexive spaces.

Definition 1 (Fredholm operator)

Let $X, Y$ be Banach spaces and $T : X \to Y$ a bounded linear operator. Then $T$ is Fredholm if:

$\dim \ker(T) < \infty$ ,
$\operatorname{codim} \operatorname{Range}(T) < \infty$ (equivalently, $\dim(Y / \operatorname{Range}(T)) < \infty$ ).

The Fredholm index of $T$ is

\operatorname{ind}(T) = \dim \ker(T) - \operatorname{codim} \operatorname{Range}(T).

(1)

The index measures the net dimensional defect of $T$ : how many more dimensions are killed than are missed.

Theorem 1 (Properties of Fredholm operators)

Let $T : X \to Y$ be Fredholm. Then:

$\operatorname{Range}(T)$ is closed.
$T^* : Y^* \to X^*$ is Fredholm with $\operatorname{ind}(T^*) = -\operatorname{ind}(T)$ .
The four subspace relations hold with full equality:

\ker(T^*) = \operatorname{Range}(T)^\perp, \qquad \operatorname{Range}(T^*) = \ker(T)^\perp.

(2)

$X$ and $Y$ decompose as topological direct sums:

X = \ker(T) \oplus Z, \qquad Y = \operatorname{Range}(T) \oplus W,

(3)

where $T|_Z : Z \to \operatorname{Range}(T)$ is an isomorphism and $\dim W = \operatorname{codim} \operatorname{Range}(T)$ .

Proof 1

Remark 1 (Fredholm decomposition vs. Hilbert decomposition)

Example 1 (Shift operators on $\ell^p$ )

The shift operators on $\ell^p$ ( $1 \leq p < \infty$ ) illustrate the index:

Right shift $R(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$ : injective, range has codimension 1. So $\operatorname{ind}(R) = 0 - 1 = -1$ .
Left shift $L(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$ : surjective, kernel is $\operatorname{span}\{e_1\}$ . So $\operatorname{ind}(L) = 1 - 0 = 1$ .

Note $L = R^*$ (under the natural pairing of $\ell^p$ and $\ell^q$ ), confirming $\operatorname{ind}(L) = -\operatorname{ind}(R)$ . These are Fredholm on every $\ell^p$ , including the non-reflexive cases $p = 1$ and $p = \infty$ .

Theorem 2 (Stability of the Fredholm index)

Let $T : X \to Y$ be Fredholm.

Compact perturbation: If $K : X \to Y$ is compact, then $T + K$ is Fredholm with $\operatorname{ind}(T + K) = \operatorname{ind}(T)$ .
Small perturbation: There exists $\varepsilon > 0$ such that if $\|S\| < \varepsilon$ , then $T + S$ is Fredholm with $\operatorname{ind}(T + S) = \operatorname{ind}(T)$ .

In particular, the Fredholm index is a topological invariant: it is constant on connected components of the set of Fredholm operators.

The stability theorem is why the Fredholm index matters: it is computable (often by deformation to a simpler operator) and robust under perturbation. This makes it the central tool in index theory, where one relates the analytic index of a differential operator to topological data of the underlying manifold.

Remark 2 (The Fredholm alternative)

Solvability conditions¶

The solvability condition from Remark 2 is one of the most useful consequences of Fredholm theory in PDE applications. We state it explicitly.

Corollary 1 (Fredholm solvability condition)

Let $T : X \to Y$ be a Fredholm operator with $\ker(T^*) = \operatorname{span}\{\psi_1, \ldots, \psi_m\}$ . Then the equation $Tx = f$ has a solution if and only if

\psi_j(f) = 0, \qquad j = 1, \ldots, m.

(5)

When a solution exists, it is unique up to an element of $\ker(T)$ : the general solution is $x = x_p + \sum_{i=1}^n \alpha_i v_i$ , where $x_p$ is any particular solution, $\{v_1, \ldots, v_n\}$ is a basis of $\ker(T)$ , and the $\alpha_i$ are arbitrary.

Proof 2

In PDE applications, the $\psi_j$ are typically known explicitly (from the symmetry or structure of the problem), and the solvability conditions take the form of integral constraints on the data.

Example 2 (Solvability of the Neumann problem)

Consider the Neumann problem for Poisson’s equation on a bounded Lipschitz domain $\Omega \subset \mathbb{R}^d$ :

-\Delta u = f \quad \text{in } \Omega, \qquad \frac{\partial u}{\partial n} = 0 \quad \text{on } \partial\Omega.

(6)

Formulated weakly, the operator $T = -\Delta_N : H^1(\Omega) \to H^1(\Omega)^*$ defined by

\langle Tu, v \rangle = \int_\Omega \nabla u \cdot \nabla v \, dx

(7)

is Fredholm with index zero:

Kernel. $\ker(T) = \operatorname{span}\{1\}$ : the constant functions, since $\int_\Omega |\nabla u|^2 \, dx = 0$ implies $u$ is constant.
Cokernel. By self-adjointness, $\ker(T^*) = \ker(T) = \operatorname{span}\{1\}$ , so $\operatorname{codim} \operatorname{Range}(T) = 1$ .

By Corollary 1, $-\Delta_N u = f$ has a solution if and only if $\psi(f) = 0$ for the single cokernel element $\psi = 1$ :

\int_\Omega f \, dx = 0 \qquad \text{(the data must have zero mean).}

(8)

This is the classical compatibility condition: by the divergence theorem, $\int_\Omega \Delta u\, dx = \int_{\partial\Omega} \frac{\partial u}{\partial n}\, dS = 0$ , so $\int_\Omega f\, dx = 0$ is necessary. Fredholm theory shows it is also sufficient. When satisfied, the solution is unique up to an additive constant.

The Fredholm index is $\operatorname{ind}(T) = 1 - 1 = 0$ , but $T$ is not invertible. To obtain a well-posed problem, we must simultaneously impose a constraint to select a unique solution (pin the constant) and add a degree of freedom to absorb right-hand sides that violate the compatibility condition.

The bordered operator¶

Definition 2 (Bordered operator)

Let $T : X \to Y$ be Fredholm with $\operatorname{ind}(T) = 0$ and $n = \dim \ker(T)$ . Let $\{\phi_1, \ldots, \phi_n\} \subset Y$ and $\{\ell_1, \ldots, \ell_n\} \subset X^*$ . The bordered operator is

\widehat{T} = \begin{pmatrix} T & \Phi \\ L & 0 \end{pmatrix} : X \times \mathbb{R}^n \to Y \times \mathbb{R}^n,

(9)

where $\Phi : \mathbb{R}^n \to Y$ maps $\alpha \mapsto \sum_{j=1}^n \alpha_j \phi_j$ and $L : X \to \mathbb{R}^n$ maps $x \mapsto (\ell_1(x), \ldots, \ell_n(x))$ .

The bordered system $\widehat{T}(x, \alpha) = (y, c)$ reads:

Tx + \sum_{j=1}^n \alpha_j \phi_j = y, \qquad \ell_j(x) = c_j, \quad j = 1, \ldots, n.

(10)

The idea is transparent: the functionals $\ell_j$ select a unique element of $\ker(T)$ (by prescribing the “coordinates” $c_j$ ), and the vectors $\phi_j$ span a complement to $\operatorname{Range}(T)$ , so the extra unknowns $\alpha_j$ allow us to absorb the component of $y$ outside the range.

Theorem 3 (Invertibility of the bordered operator)

Let $T : X \to Y$ be Fredholm with $\operatorname{ind}(T) = 0$ and $n = \dim \ker(T)$ . Let $\{v_1, \ldots, v_n\}$ be a basis of $\ker(T)$ and $\{\psi_1, \ldots, \psi_n\} \subset Y^*$ a basis of $\ker(T^*)$ . Then the bordered operator $\widehat{T}$ is boundedly invertible if and only if:

The $\ell_j$ detect the kernel: the $n \times n$ matrix $M_{ij} = \ell_i(v_j)$ is invertible.
The $\phi_j$ complement the range: the $n \times n$ matrix $N_{ij} = \psi_i(\phi_j)$ is invertible.

Proof 3

By Theorem 1, the spaces decompose as

X = \ker(T) \oplus Z, \qquad Y = \operatorname{Range}(T) \oplus W,

(11)

where $T|_Z : Z \to \operatorname{Range}(T)$ is an isomorphism and $W$ has dimension $n$ .

Injectivity. Suppose $\widehat{T}(x, \alpha) = (0, 0)$ . The second component gives $\ell_j(x) = 0$ for all $j$ . The first gives $Tx = -\sum \alpha_j \phi_j$ . Applying each $\psi_i \in \ker(T^*) = \operatorname{Range}(T)^\perp$ to the first equation:

0 = \psi_i(Tx) = -\sum_{j=1}^n \alpha_j \psi_i(\phi_j) = -(N\alpha)_i.

(12)

Since $N$ is invertible by assumption (2), $\alpha = 0$ . Therefore $Tx = 0$ , so $x \in \ker(T)$ . Writing $x = \sum \beta_j v_j$ , we have $\ell_i(x) = \sum \beta_j \ell_i(v_j) = (M\beta)_i = 0$ . Since $M$ is invertible by assumption (1), $\beta = 0$ , hence $x = 0$ .

Surjectivity. Given $(y, c) \in Y \times \mathbb{R}^n$ , we must find $(x, \alpha)$ such that $Tx + \sum \alpha_j \phi_j = y$ and $\ell_j(x) = c_j$ .

Step 1. Decompose $y = y_R + y_W$ with $y_R \in \operatorname{Range}(T)$ and $y_W \in W$ . Since $\{\psi_i\}$ is a basis for $\operatorname{Range}(T)^\perp$ , the coordinates of $y_W$ relative to the $\phi_j$ are determined by $\psi_i(y) = \sum \alpha_j \psi_i(\phi_j)$ , i.e., $\alpha = N^{-1}(\psi_1(y), \ldots, \psi_n(y))^T$ . With this choice, $y - \sum \alpha_j \phi_j \in \operatorname{Range}(T)$ .

Step 2. Since $T|_Z$ is an isomorphism onto $\operatorname{Range}(T)$ , there exists a unique $z \in Z$ with $Tz = y - \sum \alpha_j \phi_j$ .

Step 3. Set $x = z + \sum \beta_j v_j$ where $\beta = M^{-1}(c_1 - \ell_1(z), \ldots, c_n - \ell_n(z))^T$ . Then $Tx = Tz = y - \sum \alpha_j \phi_j$ and $\ell_i(x) = \ell_i(z) + (M\beta)_i = c_i$ .

Bounded inverse. $\widehat{T}$ is a bounded bijection between Banach spaces, so the open mapping theorem gives a bounded inverse.

Remark 3

Example 3 (Bordered Neumann Laplacian)

Returning to the Neumann Laplacian $T = -\Delta_N$ with $n = 1$ :

Kernel basis: $v_1 = 1$ (the constant function).
Cokernel basis: $\psi_1 = \frac{1}{|\Omega|}\int_\Omega (\cdot)\, dx$ (the mean functional), since $T$ is self-adjoint.

Choose $\phi_1 = 1$ and $\ell_1(u) = \frac{1}{|\Omega|}\int_\Omega u\, dx$ . Then:

$M_{11} = \ell_1(v_1) = \frac{1}{|\Omega|}\int_\Omega 1\, dx = 1 \neq 0$ . ✓
$N_{11} = \psi_1(\phi_1) = \frac{1}{|\Omega|}\int_\Omega 1\, dx = 1 \neq 0$ . ✓

By Theorem 3, the bordered operator

\widehat{T} = \begin{pmatrix} -\Delta_N & 1 \\[4pt] \displaystyle\frac{1}{|\Omega|}\int_\Omega (\cdot)\, dx & 0 \end{pmatrix}

(13)

is invertible. The bordered system

-\Delta u + \lambda = f, \qquad \frac{1}{|\Omega|}\int_\Omega u\, dx = 0,

(14)

has a unique solution $(u, \lambda)$ for every $f \in H^1(\Omega)^*$ . The constraint pins the mean of $u$ to zero, and the Lagrange multiplier $\lambda = \frac{1}{|\Omega|}\int_\Omega f\, dx$ absorbs the mean of $f$ .

Proof 4

Solving the bordered system by block elimination¶

The bordered system can be solved using $T$ and $T^*$ as black boxes — we never form the bordered matrix, only call the forward and adjoint solvers. Consider the general bordered system

\begin{pmatrix} T & \phi \\ \ell & 0 \end{pmatrix} \begin{pmatrix} x \\ \lambda \end{pmatrix} = \begin{pmatrix} f \\ c \end{pmatrix},

(15)

where $T : X \to Y$ is Fredholm with index zero, $\dim\ker(T) = 1$ , $v \in \ker(T)$ , and $\psi \in \ker(T^*)$ . The algorithm is a walk through the four subspace decomposition:

X = \ker(T) \oplus Z, \qquad Y = \operatorname{Range}(T) \oplus \ker(T^*).

(16)

Step 1: Project $f$ onto $\ker(T^*)$ to determine $\lambda$ . The right-hand side $f \in Y$ decomposes as $f = f_R + f_\perp$ with $f_R \in \operatorname{Range}(T)$ and $f_\perp \in \ker(T^*)$ . Apply $\psi \in \ker(T^*) = \operatorname{Range}(T)^\perp$ to the first row: $\psi(Tx + \lambda\phi) = \psi(f)$ . Since $\psi(Tx) = (T^*\psi)(x) = 0$ , this kills the $\operatorname{Range}(T)$ component entirely and reads off the $\ker(T^*)$ component:

\lambda = \frac{\psi(f)}{\psi(\phi)}.

(17)

The denominator $\psi(\phi) \neq 0$ is the transversality condition from Theorem 3: $\phi$ is not in $\operatorname{Range}(T)$ , so $\psi$ sees it. This works for any $f$ — the multiplier $\lambda$ absorbs whatever component of $f$ lies outside $\operatorname{Range}(T)$ .

Step 2: Solve in $\operatorname{Range}(T)$ for a particular solution. By construction, $f - \lambda\phi$ has no $\ker(T^*)$ component: Step 1 chose $\lambda$ precisely so that $f - \lambda\phi \in \operatorname{Range}(T)$ . The restriction $T|_Z : Z \to \operatorname{Range}(T)$ is an isomorphism (Theorem 1), so the black-box solver returns the unique $x_p \in Z$ with

T x_p = f - \lambda \phi.

(18)

This is one call to the forward solver. The solution $x_p$ lies in the complement $Z$ of $\ker(T)$ , but a general solver may return $x_p$ only up to an element of $\ker(T)$ — the next step handles this.

Step 3: Project onto $\ker(T)$ using the constraint. The general solution of the first row is $x = x_p + \alpha v$ , decomposed along $X = \ker(T) \oplus Z$ . The parameter $\alpha$ is the $\ker(T)$ component, which $T$ cannot see. The second row $\ell(x) = c$ determines it:

\alpha = \frac{c - \ell(x_p)}{\ell(v)}.

(19)

The denominator $\ell(v) \neq 0$ is the detection condition from Theorem 3: $\ell$ is nontrivial on $\ker(T)$ . The solution is $x = x_p + \alpha v$ , $\lambda$ from Step 1.

Example 4 (Block elimination for the Neumann Laplacian)

For the Neumann Laplacian $T = -\Delta_N$ with $v = 1$ , $\psi = \ell = \frac{1}{|\Omega|}\int_\Omega (\cdot)\, dx$ , and $\phi = 1$ :

Step 1. Project onto $\ker(T^*) = \operatorname{span}\{1\}$ : $\lambda = \psi(f)/\psi(\phi) = \frac{1}{|\Omega|}\int_\Omega f\, dx$ , the mean of $f$ . This extracts the incompatible component of $f$ .
Step 2. Solve in $\operatorname{Range}(T)$ : $-\Delta_N u_p = f - \lambda$ . The right-hand side has zero mean by construction, so this is a compatible Neumann problem for any $f$ .
Step 3. Project onto $\ker(T) = \operatorname{span}\{1\}$ : $\alpha = (c - \ell(u_p))/\ell(1) = c - \frac{1}{|\Omega|}\int_\Omega u_p\, dx$ , and $u = u_p + \alpha$ .

Setting $c = 0$ pins the mean of $u$ to zero.