# Da Prato's notion of Symmetric Operator

For anyone who's familiar with G. Da Prato's books on infinite dimensional analysis, I was wondering if someone could clarify something. In, for instance, "An Introduction to Infinite Dimensional Analysis" and "Second Order Partial Differential Equations in Hilbert Spaces", Da Prato introduces symmetric operators on Hilbert spaces as covariance operators for Gaussians. Since these operators are bounded, they are self-adjoint. In other places, he just mentions symmetric operators, without making it clear if he means them to be bounded. I am wondering, does Da Prato mean for his symmetric operators to all be self-adjoint?

EDIT: Per someone's query, I'm looking at "Second Order Partial Differential Equations in Hilbert Spaces", coauthored with Zabczyk, and on p. 14 of my copy, in the statement of Proposition 1.2.8, he begins "Assume $M$ is a symmetric operator..." Now, why is this important in this context, in his proof of this proposition, he begins "Let $(g_n)$ be an orthonormal basis for the operator $Q^{1/2} M Q^{1/2}$, and let $(\gamma_n)$ be the sequence of the corresponding eigenvalues."

The question is, how can the authors assert the existence of this orthogonal spectral decomposition. The operator $Q$ is symmetric and trace class, on Hilbert space $H$, so it is clearly self-adjoint. It's then clear that $Q^{1/2}$ will be self-adjoint too. So if $M$ were self-adjoint, $Q^{1/2} M Q^{1/2}$ would be self-adjoint, and we would be done. But, all that's state is that $M$ is symmetric, leaving things, to me, to be ambiguous.

Perhaps there's a bit of functional analysis I'm missing that gives the orthogonal spectral decomposition of $Q^{1/2} M Q^{1/2}$ with $M$ only symmetric, but I don't know it.

• Would you ask for more specific places? Cite a place where you cannot decide? Mar 20, 2015 at 20:43