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For the quadratic variational principle for a symmetric positive operator below, use a symmetric operator $L$ on the real Hilbert space $H$. Strict positive definiteness meansA uniformly positive definite symmetric operator satisfies the stronger coercive operator conditionIn this variational setting, “positive definite” is often used for a symmetric operator with this uniform bound. We will state explicitly where the coercive operator bound is needed. For a bounded linear operator defined on all of $H$, symmetry means that the operator is self-adjoint. For an unbounded operator, positivity is imposed on its operator domain, and the variational formulation is made on its form domain.

Symmetry is essential in a real Hilbert space: positivity of the quadratic expression alone does not imply symmetry. For example, $L=I+K$ with nonzero real skew-symmetric matrix $K$ satisfies $⟨Lv,v⟩=\Vert v{\Vert }^2$, but its quadratic functional has derivative involving $I$, not $I+K$. A positive definite symmetric operator supplies both the positivity and symmetry needed in part (b).

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 givesThus vanishing of the first variation in every direction is precisely the weak equation $⟨Lu,w⟩=⟨f,w⟩$ 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:with equality only when $v=u$. Hence 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 $(1,1,\dots )$ does not. Thus strict positivity alone proves uniqueness and the minimizing property of a solution when one exists, not existence for all $f$.

Let $T_{\alpha }=A^{\ast }A+\alpha I$. For every $f\in X$,Thus $T_{\alpha }$ is a coercive operator. If it is invertible and $T_{\alpha }f=g$, the lower bound and the Cauchy-Schwarz inequality giveand therefore $\Vert T_{\alpha }^{-1}g\Vert \le {\alpha }^{-1}\Vert g\Vert$. Hence

A positive definite symmetric operator is uniformly positive definite if $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$ for some $\gamma >0$. For a bounded linear operator defined on a whole Hilbert space, this is a symmetric coercive operator, and the Lax-Milgram theorem supplies a unique solution to $Lu=f$ for every $f$. For a differential operator, the corresponding bounded coercive form is used on its form space.