Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 326 / 1 / 2

Put $M=\mathcal{N}(K{)}^{\perp }$ for a bounded linear operator between Hilbert spaces. Its restriction $K{|}_M:M\to \mathcal{R}(K)$ is bijective. Define the Moore–Penrose inverse of an operator byThe inverse vanishes on $\mathcal{R}(K{)}^{\perp }$. For each datum in this domain, $K^{†}f$ is the minimum-norm least-squares solution; all least-squares solutions are $K^{†}f+\mathcal{N}(K)$. On their natural domains,The equivalent continuity condition isIf the range is closed, the inverse restriction is bounded by the bounded inverse theorem. Conversely, if $\Vert K^{†}f\Vert \le C\Vert f\Vert$ on its domain and $K{u}_n$ converges, replace $u_n$ by $P_M{u}_n$. The inequality makes these preimages a Cauchy sequence, whose limit maps to the limiting datum; the range is therefore closed. Boundedness here uses the data norm, not a graph norm that would conceal instability.

For the diagonal operator on sequence space,Equality holds at the first coordinate vector, so $\Vert K\Vert =1$ and $K$ is a bounded linear operator. Every diagonal entry is nonzero, giving the null space $\mathcal{N}(K)=\{0\}$. A datum has a preimage exactly when its weighted coordinates belong to the l2 sequence space. HenceFinite sequences lie in the range and are dense in ${\ell }^2$, so $\mathcal{R}(K{)}^{\perp }=\{0\}$ andThe sequence $f^{(n)}=e_n/n$ converges to zero in the data norm, but $K^{†}{f}^{(n)}=e_n$ has norm one. Thus $K^{†}$ is unbounded and discontinuous. In particular the dense range is not closed; the datum $f_j=1/j$ also proves that it is a proper subset of ${\ell }^2$.