Processing math: 80%

existence of adjoints of bounded operators


Let be a Hilbert spaceMathworldPlanetmath and let T:𝒟(T) be a densely defined linear operatorMathworldPlanetmath.

Theorem - If T is bounded (http://planetmath.org/ContinuousLinearMapping) then its adjointPlanetmathPlanetmathPlanetmath T* is everywhere defined and is also bounded.

Proof : Since T is densely defined and bounded, it extends uniquely to a bounded (everywhere defined) linear operator on , which we denote by ˜T.

For each z, the function f: defined by f(x)=˜Tx,z defines a bounded linear functionalMathworldPlanetmath on . By the Riesz representation theoremMathworldPlanetmath there exists u such that

f(x)=x,u

i.e.

˜Tx,z=x,u.

Since ˜T extends T, we also have that for every z there exists u such that

Tx,z=x,ufor everyx𝒟(T).

We conclude that T* is everywhere defined. To see that it is bounded one just needs to check that

sup

where the last inequality comes from the Cauchy-Schwarz inequality and the fact that T is bounded.

Remark - This theorem shows in particular that bounded linear operators T: have bounded adjoints T*:.

Title existence of adjoints of bounded operators
Canonical name ExistenceOfAdjointsOfBoundedOperators
Date of creation 2013-03-22 17:33:44
Last modified on 2013-03-22 17:33:44
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 4
Author asteroid (17536)
Entry type Theorem
Classification msc 47A05
Synonym bounded operatorsMathworldPlanetmathPlanetmath have (bounded) adjoints