L two operator conventions for weak mixing

Let $H$ be a closed complex $L^2$ subspace with the first-variable-linear pairing of The complex $L^2$ pairing is well-defined and satisfies Cauchy–Schwarz. All operators below map $H$ to $H$ and are complex-linear. An operator $A$ is bounded if $\mathrm{\parallel }Af\mathrm{\parallel }\le C\mathrm{\parallel }f\mathrm{\parallel }$ for some finite $C\ge 0$ and all $f\in H$; its norm is $\mathrm{\parallel }A\mathrm{\parallel }={\sup }_{\mathrm{\parallel }f\mathrm{\parallel }\le 1}\mathrm{\parallel }Af\mathrm{\parallel }$. It is compact if it is bounded and every bounded sequence $(f_n)$ has a subsequence $(f_{n_j})$ for which $(A{f}_{n_j})$ converges in norm to an element of $H$.

An adjoint $A^{\ast }$ is a bounded operator satisfying $⟨Af,g⟩=⟨f,A^{\ast }g⟩$ for all $f,g\in H$. Such an operator, if it exists, is unique: subtract two proposed identities and set $f$ equal to the difference of their values at $g$. Positive definiteness makes that difference zero. Existence is not assumed by this definition.

The operator $A$ is self-adjoint if $⟨Af,g⟩=⟨f,Ag⟩$ for all $f,g$. It is positive if $⟨Af,f⟩$ is real and nonnegative for every $f$. Later positive self-adjoint assertions impose both conditions explicitly.

An isometry preserves the norm; for linear operators it also preserves the pairing. Indeed expansion of $\mathrm{\parallel }f+g{\mathrm{\parallel }}^2$ gives $2\mathrm{Re}⟨f,g⟩=\mathrm{\parallel }f+g{\mathrm{\parallel }}^2-\mathrm{\parallel }f{\mathrm{\parallel }}^2-\mathrm{\parallel }g{\mathrm{\parallel }}^2$, and expansion of $\mathrm{\parallel }f+ig{\mathrm{\parallel }}^2$ gives $2\mathrm{Im}⟨f,g⟩=\mathrm{\parallel }f+ig{\mathrm{\parallel }}^2-\mathrm{\parallel }f{\mathrm{\parallel }}^2-\mathrm{\parallel }g{\mathrm{\parallel }}^2$. Applying both identities before and after the isometry proves the assertion. A unitary is a surjective linear isometry.

A linear subspace $E$ is invariant for $U$ if $U(E)\subseteq E$. Write $f\perp E$ when $⟨f,e⟩=0$ for every $e\in E$, and $E^{\perp }=\{f\in H:f\perp E\}$. These conventions allow $H=\{0\}$, $E=\{0\}$ and the zero operator. In the zero space the operator norm is zero because the unit ball is $\{0\}$. No infinite selection or assertion of an orthonormal basis enters these definitions.