Compact Operators & Fredholm Theory

A compact operator may act on an infinite-dimensional space, but every bounded input sequence has an output subsequence that converges in norm. That loss of degrees of freedom produces matrix-like nonzero spectral behavior and a sharp existence alternative for second-kind equations.

Definition and sequence test #

A bounded operator $K:X\to Y$ is compact when it maps the unit ball to a relatively compact set. Equivalently, for every bounded sequence $(x_n)$ in $X$, the sequence $(Kx_n)$ has a norm-convergent subsequence.

Every finite-rank operator is compact. Operator-norm limits of compact operators are compact. If $K$ is compact and $A,B$ are bounded with compatible domains, then $AK$ and $KB$ are compact. Thus compact operators form a closed two-sided operator ideal.

The identity on $X$ is compact iff $X$ is finite-dimensional. A compact operator on an infinite-dimensional space cannot have a bounded inverse on its infinite-dimensional range when that range is closed and equal to the whole space; otherwise the identity would be compact.

Integral operators #

If $k\in C([a,b]^2)$, then

$$(Kf)(x)=\int_a^b k(x,t)f(t)dt$$

defines a compact map $C[a,b]\to C[a,b]$. Bounded inputs give uniformly bounded outputs, and uniform continuity of $k$ gives equicontinuity; then Arzelà–Ascoli applies. On $L^2$, every square-integrable kernel defines a Hilbert–Schmidt operator, hence a compact operator.

Weak-to-strong continuity #

If $x_n\rightharpoonup x$ and $K$ is compact, then

$$Kx_n\to Kx$$

in norm. Boundedness gives a relatively compact output sequence. Every convergent output subsequence must have weak limit $Kx$, and uniqueness forces all norm cluster points to be $Kx$; a subsequence contradiction finishes the argument.

Nonzero spectrum is discrete #

For a compact operator $K$ on an infinite-dimensional complex Banach space:

• every nonzero spectral value is an eigenvalue;
• its eigenspace is finite-dimensional;
• distinct nonzero eigenvalues can accumulate only at $0$;
• the spectrum is finite or countable, together with $0$.

Zero is special: it may or may not be an eigenvalue, but it lies in the spectrum whenever compact $K$ acts on an infinite-dimensional space and cannot be invertible.

If $K$ is compact and self-adjoint on a Hilbert space, there is an orthonormal basis of $\overline{\mathop{\mathrm{ran}}K}$ consisting of eigenvectors with real eigenvalues tending to zero when infinite. Together with an ONB of $\ker K$, this gives

$$Kx=\sum_n\lambda_n\langle x,e_n\rangle e_n$$

in norm. This is the compact self-adjoint spectral theorem; general bounded self-adjoint operators may require a projection-valued measure rather than an eigenbasis.

Fredholm alternative #

Let $K$ be compact on a Banach space. For $\lambda\ne0$, $K-\lambda I$ is injective iff it is surjective; when invertible, its inverse is bounded. Equivalently, for $I-K$ exactly one of these regimes occurs:

1. the homogeneous equation $x-Kx=0$ has only $x=0$, and $(I-K)x=y$ has a unique solution for every $y$;
2. the homogeneous equation has a nonzero finite-dimensional solution space, so the inhomogeneous equation is not solvable for every $y$.

On a Hilbert space,

$$\mathop{\mathrm{ran}}(I-K)=\ker(I-K^*)^\perp.$$

Therefore $(I-K)x=y$ is solvable exactly when

$$\langle y,z\rangle=0 \quad\text{for every }z\in\ker(I-K^*).$$

For $I-\lambda K$, replace the adjoint equation by $\ker(I-\overline\lambda K^*)$. Solutions, when they exist in the singular case, are nonunique up to $\ker(I-\lambda K)$.

Fredholm operators and index #

A bounded operator $A:X\to Y$ is Fredholm when its kernel is finite-dimensional, its range is closed, and $Y/\mathop{\mathrm{ran}}A$ is finite-dimensional. Its index is

$$\mathop{\mathrm{ind}}A=\dim\ker A-operatorname{codim}\mathop{\mathrm{ran}}A.$$

$I-K$ is Fredholm of index zero for compact $K$. Compact perturbations preserve the Fredholm index, which explains why a finite-dimensional obstruction is stable even though individual eigenvalues may move.

Diagnostic checklist #

• Prove relative compactness of outputs, not merely boundedness.
• Distinguish compact operator from an operator with compact domain.
• Exclude $\lambda=0$ from the matrix-like spectral claim.
• In a singular equation, compute the adjoint nullspace and test the forcing against it.
• Do not infer an eigenbasis for an arbitrary compact nonnormal operator.