 Solution

Work first in a real Hilbert space. A bounded linear operator $L$ is self-adjoint whenFor the variational interpretation, elliptic means uniformly coercive: some $\gamma >0$ satisfiesThis 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 strictly positive: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 $\mathrm{Re}⟨f,v⟩$ in the linear term.