 Solution

In the variational convention needed here, a positive definite symmetric operator on a real Hilbert space satisfiesFor a bounded operator on the whole space, symmetry is self-adjointness. On a complex Hilbert space, the corresponding convention is Hermitian positivity; the quadratic expression is real. Strict positivity should be distinguished from uniformly positive definite symmetric operator coercivity, which requires a fixed $\gamma >0$ such that $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$.

Symmetry matters in a real space. If “positive definite” means only the inequality for the quadratic form, the next variational assertion is false: $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has $⟨Lv,v⟩=\Vert v{\Vert }^2>0$, but its skew part disappears from that form. Its Euler-Lagrange equation is $v=f$, not $Lv=f$. Thus the usual symmetric/Hermitian convention, or an explicit symmetry assumption, is required.