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Assume the self-adjoint interpretation of positive-definite operator from the preceding part. For $v,w\in H$ and real $\epsilon$, expand the quadratic functional:The first variation therefore vanishes in every direction precisely when $⟨Lv-f,w⟩=0$ for every $w$, or $Lv=f$. 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 identityStrict positivity makes this difference positive unless $v=u$, so 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 $\Vert u\Vert \le \Vert f\Vert /\gamma$. The identity above then gives the quantitative gap $I(v)-I(u)\ge \gamma \Vert v-u{\Vert }^2$.

The hypotheses matter. On ${\mathbb{R}}^2$, letThen $⟨Lv,v⟩=\Vert v{\Vert }^2>0$ 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 $I(v)=-{\sum }_{j=1}^{N}1/j\to -\infty$. The printed conclusion about a weak solution is valid whenever that solution exists; an unconditional existence assertion needs uniform positivity.