= Solution
\b[The trial space must be a subspace of $H_0^1(0,1)$, not merely of $L^2(0,1)$.] The printed assumptions on the basis do not ensure square-integrable weak derivatives, and boundary traces are not defined for arbitrary <L2 space> functions. For example, $\varphi(x)=x^{1/4}(1-x)$ is continuous, vanishes at both endpoints and belongs to $L^2$, but $|\varphi'(x)|^2\sim x^{-3/2}/16$ is not integrable at zero. Assume the basis belongs to the <zero-boundary Sobolev space>, as it does for the hat functions in part (c).
Write $u_H=\sum_{j=1}^M c_j\varphi_j$. Differentiating the restricted energy with respect to $c_i$, or imposing the <Galerkin method> with test function $\varphi_i$, gives
$$
\boxed{\sum_{j=1}^M A_{ij}c_j=F_i,\qquad
A_{ij}=\int_0^1(\varphi_j'\varphi_i'+\varphi_j\varphi_i)\,dx,\qquad
F_i=\int_0^1x\varphi_i\,dx.}
$$
The derivative term is the usual diffusion <stiffness matrix> and the second is the <mass matrix>. The full matrix is real symmetric, and for every nonzero coefficient vector $c$,
$$
c^TAc=\int_0^1\left[\left(\sum_jc_j\varphi_j'\right)^2
+\left(\sum_jc_j\varphi_j\right)^2\right]dx>0,
$$
because a basis is <linearly independent>. Thus it is a <positive-definite matrix>, so the restricted <Ritz method> has a unique solution and minimum.
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