Consider the compact operator $S: L^2[0,1] \to L^2[0,1]$ given by $(L f)(t) = \int_0^t f(s) ds$. We know the adjoint of this operator is defined by $(S^*f)(t) = \int_t^1 f(s) ds$.
Since $S S^*$ is a compact, self-adjoint operator, it has countable eigenvalues $(\lambda_i)$, with eigenspaces $(E_i)$. The closure of the direct sum of its eigenspaces is equal to $L^2[0,1]$, i.e. $\overline{\oplus_i E_i} = L^2[0,1]$ (c.f. Direct sum of eigenspaces of a compact operator has finite codimension). Hence the $(v_i)$ form a basis of $L^2[0,1]$.
One can also explicitly compute the orthonormal eigenfunctions $(v_i)$ of $S S^*$ (c.f. Eigenvalues of the operator $(Tu)(x)=\int_0^x (\int_t^1 u(s)ds)dt.$). However, an element $h \in R(SS^*)$ satisfies $h(0) = 0$ and $h'(1) = 0$, i.e. these functions are restricted at the boundary. Any linear combination $\sum \mu_i v_i$ satisfies these boundary conditions as well. However, this is not the case for general functions $f \in L^2[0,1]$.
I am trying to resolve these seemingly contradictory statements. Does this mean that the $(v_i)$ are a basis for $L^2[0,1]$, despite being tied down at the boundary? And does this then indeed imply that the $L^2$-norm of an arbitrary function $f \in L^2[0,1]$ can be written as $\|f\|_{L^2} = \sum_i \langle f, v_i\rangle^2$?
The non-zero trigonometric functions that satisfy $$ -f''=\lambda f, \;\; f(0)=0,\; f'(1)=0 \;\;\;\; (\dagger) $$ form a complete orthogonal subset of $L^2[0,1]$. The reason for this is that $Lf = -f''$ is self-adjoint on the domain $\mathcal{D}(L)$ consisting of all twice absolutely continuous functions $f$ that satisfy $f(0)=0$, $f'(1)=0$. You can check that it is symmetric on this domain because, for all $f,g$ are in $\mathcal{D}(L)$, the following holds: \begin{align} \langle Lf,g\rangle-\langle f,Lg\rangle&=\int_0^1(-f'')g+f(g'')dx \\ &= \int_0^1(fg'-f'g)'dx \\ &=(fg'-f'g)|_{0}^{1}=0. \end{align} The same is true if you use a complex inner product. It takes a little more to show that $L$ is self-adjoint on $\mathcal{D}(L)$, but it is true.
The values of $\lambda$ for which there is a non-zero solution of $(\dagger)$ are $$ \lambda_n = (n+1/2)^2\pi^2. $$ and the corresponding eigenfunctions are $$ f_{n}(x)=\sin((n+1/2)\pi x),\;\;\; n=1,2,3,\cdots. $$ The normalized eigenfunctions form a complete orthonormal basis of $L^2[0,1]$; so you can expand any $f\in L^2$ in an $L^2$ convergent series $$ f = \sum_{n=1}^{\infty}\frac{\langle f,f_n\rangle}{\langle f_n,f_n\rangle} f_n $$ Every basis element $f_n$ satisfies $f_n(0)=f_n'(1)=0$, which is what you want. The series converges in $L^2[0,1]$.