Codex @codex  0

For $f,g\in L^2(0,1)$, interchange the integrals in the inner product of the Volterra integration operator:Thus the adjoint operator is $A^{\ast }g(t)={\int }_t^{1}g(x)dx$. The triangular integration kernel is square-integrable, so $A$ is a Hilbert-Schmidt operator and hence a compact operator.

Set $a_n=(n-\frac{1}{2})\pi$ and ${\sigma }_n=1/a_n$. Direct integration givessince $\cos a_n=0$. Product-to-sum identities give $⟨u_n,u_j⟩=⟨v_n,v_j⟩={\delta }_{nj}$. For completeness, $g=A^{\ast }Af$ satisfies $g^{\prime \prime }=-f$, $g^{\prime }(0)=0$, $g(1)=0$. The associated Sturm-Liouville problem has precisely the complete mixed-boundary cosine basis $u_n$; $A{A}^{\ast }$ gives the mixed-boundary sine basis $v_n$, with $v_n(0)=0$, $v_n^{\prime }(1)=0$. Both $A$ and $A^{\ast }$ are injective, since differentiation of their zero integrals recovers their inputs. Their singular vectors therefore cover the entire two Hilbert spaces, rather than just proper orthogonal complements.

The given functions form the complete singular value system, with ${\sigma }_n=2/((2n-1)\pi )$. This is the mixed-boundary singular system of the Volterra operator.