1. ## Linear Operators

Let X be a complex Banach space and T in L(X, X) a linear operator. Let (T* f)(x)=f (Tx), (where x in X, f in X*) define a linear operator T* in L(X*, X*).

Assuming that T is continuous, show that T* is continuous with ||T*||<=||T||.

Would anybody be kind enough to offer a proof of this result or point me in the direction of a book where the result is proved?

Thanks

2. ## Re: Linear Operators

Originally Posted by Cairo
Assuming that T is continuous, show that T* is continuous with ||T*||<=||T||.
We need to show that ‖T* f‖ ≤ ‖T‖ ‖f‖ for all f ∈ L*. Since T* f is a functional, this in turn means that ‖(T* f) x‖ ≤ ‖T‖ ‖f‖ ‖x‖ for all x ∈ L. Now, ‖(T* f) x‖ = ‖f (T x)‖ ≤ ‖f‖ ‖T x‖ ≤ ‖f‖ ‖T‖ ‖x‖, as required.

3. ## Re: Linear Operators

Thank you so much for this. When you say f in L*, do you mean f in L(X*, X*)?

Is it possible that you could now prove that ||T**||=||T*||=||T|| by only assuming that T is continuous? That is, I can't assume that X or X* are reflexive. I've tried using the natural isometry but can't get it to work.

4. ## Re: Linear Operators

First of all, I noted that Wikipedia denotes the set of all linear functionals over X by X* (algebraic dual space), and the set of continuous linear functionals over X by X' (continuous dual space). When X is infinite-dimensional, some linear functionals may not be continuous (and therefore bounded), so in general X* is not a normed space. Since we are talking about about the space L(X*, X*) of continuous linear operators, which presupposes norms on the domain and the codomain, I'll assume that in this thread X* denotes the space of continuous linear functionals over X.

Originally Posted by Cairo
When you say f in L*, do you mean f in L(X*, X*)?
No, f is a continuous linear functional from X to complex numbers. Members of L(X*, X*) are operators on such functionals.

Originally Posted by Cairo
Is it possible that you could now prove that ||T**||=||T*||=||T|| by only assuming that T is continuous? That is, I can't assume that X or X* are reflexive.
Could you remind what reflexive means here?

6. ## Re: Linear Operators

OK, it has been some time since I did this topic. Apparently, proving ‖T*‖ ≥ ‖T‖ requires Hahn–Banach theorem, in particular, the third corollary on this Wikipedia page. Here is a proof from Wikibooks (2nd last theorem; note that it has several typos).

We will prove that ‖T*‖ > ‖T‖ - ε for every ε > 0. By the definition of ‖T‖, there exists an x ∈ X such that ‖x‖ = 1 and ‖T x‖ > ‖T‖ - ε. By the corollary from Hahn–Banach theorem, there exists a functional f ∈ X* such that ‖f‖ = 1 and f(T x) = ‖T x‖. Then ‖T*‖ ≥ ‖T* f‖ ≥ |(T* f) x| = |f(T x)| = ‖T x‖ > ‖T‖ - ε.

7. ## Re: Linear Operators

Thank you so much for this, emakarov.

To show that ||T**||=||T|| I've found a theorem (outlined below) which I think might help, but am unsure how to use it to show the required equality. Any ideas?

Theroem: Any normed linear space X is isometrically isomorphic to a subspace M of the bidual X**.

8. ## Re: Linear Operators

If ‖U*‖ = ‖U‖ for all continuous linear operators U, then instantiating U with T* we get ‖T**‖ = ‖T*‖, and instantiating U with T we get ‖T*‖ = ‖T‖.