Fredholm's theorem

Remarks

Let $T$ be the integral operator with kernel $K$; it is a compact operator on $L^2[a, b]$, and its adjoint is the integral operator with kernel $K^*$. For $\mu = 0$ the statements are trivial, and for $\mu \ne 0$ they are statements about $I - \mu T$ with $\mu T$ compact. One proof approximates $K$ in $L^2$ by a degenerate kernel $K_0$ with $\abs{\mu} \norm{K - K_0} < 1$, with operator $T_0$. Then $I - \mu (T - T_0)$ is invertible by a Neumann series, and the equation reduces to a finite linear system. The Fredholm alternative for matrices applies to that system.

Example: on $[0, 1]$ with $K(x, y) = xy$ and $\mu = 3$, the homogeneous equation has the solution $\varphi(x) = x$, and $K^* = K$. The equation $\varphi(x) - 3 \int_0^1 x y \varphi(y) \, dy = f(x)$ is solvable exactly when $\int_0^1 f(x) \, x \, dx = 0$; for instance $f = 1$ fails this, and indeed integrating $\varphi - 3x \int_0^1 y \varphi = 1$ against $x$ gives the contradiction $0 = 1/2$.

The square-integrability of $K$ matters because it makes $T$ compact. For the identity operator on $L^2[a, b]$, which is bounded but not compact and not given by a square-integrable kernel, the equation $\varphi - \varphi = 0$ has infinitely many independent solutions, against part 2.

Going deeper

For a student

These are Fredholm's results, obtained first with his determinants for continuous kernels and later extended to square-integrable kernels. In the language of operators, they say that $I - \mu T$ is a Fredholm operator of index $0$ when $T$ is compact. They also say that its image is the orthogonal complement of the kernel of its adjoint. Part 4 is the statement that the nonzero spectrum of a compact operator is discrete, as in the spectral theory of compact operators. They are the reason why boundary value problems that reduce to integral equations of the second kind have a solvability theory as clean as linear algebra.