Picard's criterion for first-kind equations

Remarks

Proof idea: if $K x = y$, then $y \perp \ker(K^*)$ and $\ip{y}{u_n} = \ip{x}{K^* u_n} = \sigma_n \ip{x}{v_n}$, so the sum in clause 2 equals $\sum_n \abs{\ip{x}{v_n}}^2 \le \norm{x}^2$ by Bessel's inequality. Conversely, clause 2 makes the series for $x^\dagger$ converge. Then $K x^\dagger = \sum_n \ip{y}{u_n} u_n$ is the orthogonal projection of $y$ onto the closure of $K(X)$, which is $\ker(K^*)^\perp$. Clause 1 makes it equal to $y$. The solution $x^\dagger$ lies in $\ker(K)^\perp$, so it has the smallest norm among all solutions.

Example: the Volterra operator $(V x)(t) = \int_0^t x(s) \, ds$ on $L^2(0, 1)$ is injective with dense range. Its singular system is $\sigma_n = \frac{2}{(2n - 1)\pi}$, $v_n(s) = \sqrt{2} \cos\big((n - \tfrac{1}{2}) \pi s\big)$ and $u_n(t) = \sqrt{2} \sin\big((n - \tfrac{1}{2}) \pi t\big)$. So $V x = y$ is solvable if and only if $\sum_n (2n - 1)^2 \abs{\ip{y}{u_n}}^2 < \infty$, a condition on the decay of the coefficients of $y$ that expresses the differentiability of $y$.

Going deeper

For a student

The criterion is the infinite-dimensional form of solving a linear system by the singular value decomposition, see singular value decomposition. It shows the ill-posedness of first-kind equations in a quantitative way: an error of size $\delta$ in the coefficient $\ip{y}{u_n}$ becomes an error $\delta / \sigma_n$ in the solution. The faster the singular values decay, the smoother the data must be, and the more ill posed the problem. Regularization methods such as truncated singular value expansion and Tikhonov regularization replace the factors $1/\sigma_n$ by bounded filter factors. The map $y \mapsto x^\dagger$ is the Moore–Penrose inverse of $K$ on its natural domain.