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For a bounded linear operator $K:U\to V$ between Hilbert spaces, a least-squares solution of data $f$ is a minimizer of the residual norm:The minimum-norm least-squares solution is the one of smallest $U$-norm among these minimizers. Equivalently it is the unique least-squares solution in $\mathcal{N}(K{)}^{\perp }$. If the equation is consistent, this is the smallest-norm exact solution.

Let $P$ be the orthogonal projection onto $\overline{\mathcal{R}(K)}$. The orthogonal decomposition by a closed subspace givesThe first term has infimum zero by the definition of closure; that infimum is attained exactly if $Pf\in \mathcal{R}(K)$. This proves the least-squares existence criterion:They exist for every datum precisely when the range is closed. Once a solution exists, all solutions differ by elements of the closed null space, and projecting onto its orthogonal complement gives the unique minimum-norm solution.

For a counterexample to existence, take $K:{\ell }^2\to {\ell }^2$, $(Ku{)}_j=u_j/j$, and $f_j=1/j$. The range is dense, so the least residual infimum is zero; finite truncations realize residuals tending to zero. But an exact preimage would have every coordinate equal to one and would not belong to ${\ell }^2$. Thus no least-squares minimizer exists for this datum.