= Solution
For $f,g\in L^2(0,1)$, interchange the integrals in the <inner product> of the <Volterra integration operator>:
$$
\langle Af,g\rangle=\int_0^1\int_0^x f(t)\overline{g(x)}\,dt\,dx=\int_0^1f(t)\overline{\int_t^1g(x)\,dx}\,dt.
$$
Thus the <adjoint operator> is $A^*g(t)=\int_t^1g(x)\,dx$. The triangular integration kernel is square-integrable, so $A$ is a <Hilbert-Schmidt operator> and hence a <compact operator>.
Set $a_n=(n-\tfrac12)\pi$ and $\sigma_n=1/a_n$. Direct integration gives
$$
Au_n(x)=\sqrt2\int_0^x\cos(a_nt)\,dt=\sigma_n\sqrt2\sin(a_nx)=\sigma_nv_n(x),
$$
$$
A^*v_n(x)=\sqrt2\int_x^1\sin(a_nt)\,dt=\sigma_n\sqrt2\cos(a_nx)=\sigma_nu_n(x),
$$
since $\cos a_n=0$. Product-to-sum identities give $\langle u_n,u_j\rangle=\langle v_n,v_j\rangle=\delta_{nj}$. For completeness, $g=A^*Af$ satisfies $g''=-f$, $g'(0)=0$, $g(1)=0$. The associated <Sturm-Liouville problem> has precisely the complete mixed-boundary cosine basis $u_n$; $AA^*$ gives the mixed-boundary sine basis $v_n$, with $v_n(0)=0$, $v_n'(1)=0$. Both $A$ and $A^*$ 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.
\b[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>.
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