# Singular value decomposition of a specific integraloperator

I want to determin the singular value decomposition of the integral operator

$$L^2(0,1) \to L^2(0,1) ; f \mapsto Af(\cdot) = \int_0^\cdot f(y)dy.$$

$$A^*f(\cdot) = \int_\cdot^1 f(y) dy.$$
Hence $A^*Af(x) = \int_x^1 \int_0^y f(z)dzdy.$ Assuming that $f$ is smooth yields
$$\lambda f''(x) = A^*Af''(x) = \int_x^1\int_0^y f''(z)dzdy = -f(x) +f(1) - (1-x)f'(0).$$
Clearly, $\lambda_j = ((j-1/2)\pi)^{-2}$ and $v_j(x)= \sqrt{2}\cos((j-1/2)\pi x)$ are solutions of the above differential equation. Do more solutions exist? Since $A$ is injective the Eigenfunctions should span all of $L^2(0,1)$. We have $v_j(1)=0$ for all $j$, but this is merely a point and should not matter in $L^2(0,1)$. Moreover, the $v_j$ are even for some $j$ and odd for the other.