= Solution
Work with real functions and $f\in L^2(0,1)$. To give the variational problem a definite domain, choose the homogeneous <Dirichlet boundary conditions> used in the next part. The natural form space is the <zero-trace Sobolev space> $H=H_0^1(0,1)$, with norm $\|v\|_H=\|v'\|_2$. The <Sobolev trace operator> is well defined because one-dimensional $H^1$ functions have absolutely continuous representatives. For example $v(0)=\int_0^1v(x)dx-\int_0^1(1-x)v'(x)dx$, so its endpoint trace is bounded in the $H^1$ norm; the other endpoint has the analogous estimate. The zero-trace space is closed in the Hilbert space $H^1$, and the <Poincaré inequality> $\|v\|_2\leq\pi^{-1}\|v'\|_2$ makes the chosen derivative norm equivalent to its full norm. Thus $H$ is a Hilbert space.
Differentiability of $p,q$ implies continuity on the compact interval. Hence there are finite constants $p_{\max},q_{\max}$ and a positive $p_{\min}$ with $p_{\min}\leq p\leq p_{\max}$ and $0\leq q\leq q_{\max}$. Define the symmetric <bilinear form>
$$
a(u,v)=\int_0^1[p(x)u'(x)v'(x)+q(x)u(x)v(x)]dx.
$$
The <Cauchy-Schwarz inequality> and <Poincaré inequality> give all the estimates required by the <Lax-Milgram theorem>:
$$
|a(u,v)|\leq\left(p_{\max}+\frac{q_{\max}}{\pi^2}\right)\|u\|_H\|v\|_H,\qquad a(v,v)\geq p_{\min}\|v\|_H^2,\qquad|\langle f,v\rangle|\leq\frac{\|f\|_2}{\pi}\|v\|_H.
$$
That theorem states that a continuous coercive bilinear form on a Hilbert space and a bounded linear functional determine a unique weak solution. Completeness, continuity, <coercivity> and boundedness have now all been verified. Therefore there is a unique $u\in H$ with $a(u,v)=\langle f,v\rangle$ for every $v\in H$.
The energy version of the given <variational problem> is $I(v)=a(v,v)-2\langle f,v\rangle$. Its exact variation is
$$
I(v+w)-I(v)=2a(v,w)-2\langle f,w\rangle+a(w,w).
$$
Boundedness of $a$ makes the last term quadratic in $\|w\|_H$, so the <Fréchet derivative> is $DI(v)[w]=2a(v,w)-2\langle f,w\rangle$. Its vanishing for every test function is precisely
$$
\boxed{\int_0^1[pu'v'+quv]dx=\int_0^1fv\,dx\quad\text{for every }v\in H_0^1(0,1).}
$$
Taking compactly supported smooth tests and integrating in the distributional sense shows that this is the <Euler-Lagrange equation> $-(pu')'+qu=f$. Conversely that weak equation gives the displayed stationary condition. The stationary point is actually the unique global minimizer, because
$$
\boxed{I(v)-I(u)=a(v-u,v-u)\geq p_{\min}\|v-u\|_H^2.}
$$
Equality occurs only when $v=u$. This proves the variational equivalence, including existence and uniqueness rather than only a formal derivative calculation.
For precision about the literal <L2 inner product> $\langle Lv,v\rangle$, use the <flux-domain realization of a one-dimensional elliptic operator>
$$
D(L)=\{v\in H_0^1(0,1):pv'\in H^1(0,1)\}.
$$
For such $v$, weak <integration by parts> and its zero endpoint values give $\langle Lv,v\rangle=a(v,v)$. On the larger form space $H$, the same energy is interpreted by the weak operator pairing. The minimizer belongs to $D(L)$ automatically: its weak equation gives $(pu')'=qu-f\in L^2$, while $pu'\in L^2$. Thus its equation also holds as an equality of $L^2$ functions. This domain does not assume bounded $p'$, which mere differentiability would not supply. If $p\in C^1$ one may equivalently use $D(L)=H^2\cap H_0^1$, but that stronger coefficient assumption is not needed for the proof here.
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