= Solution
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 means
$$
\langle Lv,v\rangle>0\qquad(v\ne0).
$$
A <uniformly positive definite symmetric operator> satisfies the stronger <coercive operator> condition
$$
\boxed{\langle Lv,v\rangle\geq\gamma\|v\|^2\quad\text{for some }\gamma>0.}
$$
In 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 $\langle Lv,v\rangle=\|v\|^2$, 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).
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