= Solution
Work first in a real <Hilbert space>. A <bounded linear operator> $L$ is \b[<self-adjoint>] when
$$
\langle Lu,v\rangle=\langle u,Lv\rangle\quad(u,v\in H).
$$
For the variational interpretation, \b[elliptic] means uniformly coercive: some $\gamma>0$ satisfies
$$
\langle Lv,v\rangle\geq\gamma\|v\|_H^2\quad(v\in H).
$$
This is <ellipticity of a bounded Hilbert-space operator>, expressed through its <quadratic form>, rather than the principal-symbol definition for an <elliptic differential operator>. In the nonsymmetric case this condition concerns the symmetric part of $L$ and does not itself imply <self-adjointness>.
A <positive-definite operator> is <self-adjoint> and \b[strictly positive]:
$$
\langle Lv,v\rangle>0\quad(v\ne0).
$$
Some variational conventions use “positive definite” for the stronger combination of self-adjointness and ellipticity; we distinguish strict from uniform positivity explicitly. A <uniformly positive definite symmetric operator> is coercive, whereas strict positivity alone in infinite dimension need not give a bounded inverse or solvability for every forcing. If positivity is instead defined only by the displayed real quadratic inequality, symmetry must be added separately for the next part. In a complex <Hilbert space>, use a <self-adjoint> operator, the real <quadratic form>, and $\operatorname{Re}\langle f,v\rangle$ in the linear term.
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