= Solution
First let $L$ be a <bounded linear operator> on $H$ satisfying the symmetric positive-definiteness convention in part (a). For any direction $w\in H$, expansion of the quadratic functional gives
$$
\begin{aligned}
I(v+\varepsilon w)-I(v)
&=2\varepsilon\{\langle Lv,w\rangle-\langle f,w\rangle\}
+\varepsilon^2\langle Lw,w\rangle,\\
DI(v)[w]&=2\langle Lv-f,w\rangle.
\end{aligned}
$$
Thus vanishing of the <first variation> in every direction is precisely the weak equation $\langle Lu,w\rangle=\langle f,w\rangle$ for every $w$. In this whole-space bounded-operator setting it is equivalent to $Lu=f$, the <Euler-Lagrange equation>.
If $u$ solves that equation, set $v=u+w$. The linear terms cancel:
$$
\boxed{I(v)-I(u)=\langle L(v-u),v-u\rangle\geq0,}
$$
with equality only when $v=u$. Hence \b[the weak solution is the unique global minimizer]. Conversely, any minimizer has zero <first variation>, and therefore solves the weak equation. For a symmetric <bounded bilinear form> $B$ on a form space $V$, exactly the same calculation gives $I(v)=B(v,v)-2\ell(v)$ and $B(u,w)=\ell(w)$ for every $w\in V$; it does not require an unbounded differential operator to map every $v\in V$ into $H$.
Existence for every $f$ requires an extra hypothesis if “positive definite” means only strict positivity. The <coercive operator> bound makes the form coercive, so the <Lax-Milgram theorem> supplies existence and uniqueness. Without that bound, the <diagonal operator on sequence space> $L(x_n)=(x_n/n)$ on $\ell^2$ is symmetric and strictly positive, but $f=(1/n)$ belongs to $\ell^2$ and its formal inverse $(1,1,\ldots)$ does not. Thus strict positivity alone proves uniqueness and the minimizing property of a solution when one exists, not existence for all $f$.
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