= Solution
Assume the <self-adjoint> interpretation of <positive-definite operator> from the preceding part. For $v,w\in H$ and real $\varepsilon$, expand the <quadratic functional>:
$$
I(v+\varepsilon w)=I(v)+2\varepsilon\langle Lv-f,w\rangle
+\varepsilon^2\langle Lw,w\rangle.
$$
The <first variation> therefore vanishes in every direction precisely when $\langle Lv-f,w\rangle=0$ for every $w$, or \b[$Lv=f$]. This is its <Euler-Lagrange equation> and its <weak formulation>. For any weak solution $u$, write $v=u+w$. Self-adjointness and $Lu=f$ cancel the cross terms and give the exact identity
$$
\boxed{I(v)-I(u)=\langle L(v-u),v-u\rangle.}
$$
Strict positivity makes this difference positive unless $v=u$, so \b[every weak solution is the unique global minimizer], and conversely every minimizer is a weak solution. If $L$ is uniformly positive, the <Lax-Milgram theorem> additionally gives existence for every $f\in H$ and $\|u\|\leq\|f\|/\gamma$. The identity above then gives the quantitative gap $I(v)-I(u)\geq\gamma\|v-u\|^2$.
The hypotheses matter. On $\mathbb R^2$, let
$$
L=\begin{pmatrix}1&-1\\1&1\end{pmatrix},\qquad f=\begin{pmatrix}1\\0\end{pmatrix}.
$$
Then $\langle Lv,v\rangle=\|v\|^2>0$ for nonzero $v$, but $I$ is minimized at $v=f$, whereas $L^{-1}f=(1/2,-1/2)^T$. Thus real quadratic positivity without symmetry does not imply the requested variational assertion: the <symmetric part determines a real quadratic functional>.
Also, strict <self-adjoint> positivity does not imply existence for arbitrary $f$. On $H=\ell^2$, take $(Lv)_j=v_j/j$ and $f_j=1/j$. The operator is bounded, <self-adjoint> and strictly positive, and $f\in\ell^2$; a solution would have $u_j=1$, which is not in $\ell^2$. In fact the trial vectors with their first $N$ coordinates equal to one give $I(v)=-\sum_{j=1}^N1/j\to-\infty$. The printed conclusion about a weak solution is valid whenever that solution exists; an unconditional existence assertion needs uniform positivity.
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