Ill-posedness of first-kind equations

Remarks

Proof idea: if $\norm{K x} \ge c \norm{x}$ for all $x$, then $K$ is injective and $K^{-1}$ is bounded on $K(X)$ with norm at most $1/c$. The identity of $X$ equals $K^{-1} K$, the product of a bounded and a compact operator, so it is compact. Then the closed unit ball of $X$ is compact, which contradicts Riesz's lemma since $X$ is infinite-dimensional.

The main application is to integral equations of the first kind. Let $k$ be continuous on $[0, 1]^2$ and $(K x)(t) = \int_0^1 k(t, s) \, x(s) \, ds$ on $X = Y = L^2(0, 1)$; this operator is compact. For $x_n(s) = \sin(n \pi s)$ one has $\norm{x_n}_2 = 1/\sqrt{2}$, but $(K x_n)(t) \to 0$ uniformly in $t$ by the Riemann–Lebesgue lemma and the uniform continuity of $k$. So the data $y + K x_n$ are close to $y$ while the solutions $x + x_n$ stay far from $x$.

The hypothesis that $X$ is infinite-dimensional is needed. On a finite-dimensional space every linear operator is compact, and an invertible matrix gives a well-posed equation.

Going deeper

For a student

The theorem explains why inverse problems modelled by first-kind equations, such as deblurring an image or recovering a source from far-field measurements, need regularization. Its quantitative form uses the singular values $\sigma_n \to 0$ of $K$: the error in the component along the $n$-th singular vector is multiplied by $1/\sigma_n$. Picard's criterion describes exactly which data lie in the range, and Tikhonov regularization gives stable approximate solutions. By contrast, a second-kind equation $x - K x = y$ with compact $K$ is well posed as soon as it is uniquely solvable, by the Fredholm alternative.