1. ## Banach spaces

Hi,

The question: Let X, Y be Banach spaces, $\displaystyle T:X\to Y$ be a linear map. Prove that if $\displaystyle L\circ T$ is a continuous linear functional $\displaystyle X\to \mathbb{R}$ for every continuous linear functional $\displaystyle L: Y\to \mathbb{R}$, then $\displaystyle T$ is continuous.

The hypothesis of the problem (that X and Y both be Banach) suggest to me the Open Mapping Theorem, but we don't have T mapping onto Y, and T(X) may not be closed (which we need for it to be complete as a subspace of the complete space Y).

As always your help is much appreciated.

2. $\displaystyle L\circ T(B)$ is bounded in $\displaystyle \mathbb{R}$ for each $\displaystyle L\in Y^*$. Hence Banach Steinhauss implies $\displaystyle T(B)$ is bounded in $\displaystyle Y$.

3. Thanks for your response. I take it that B is the closed unit ball (to have completeness). However, the Banach Steinhaus theorem requires that, for each point, a family of operators all be bounded by some number. Although $\displaystyle L\circ T$ is bounded by some $\displaystyle M_L>0$ for each $\displaystyle L$, the supremum of all such $\displaystyle M_L$ may be infinite!

Are there any other suggestions?

Thanks again.

EDIT: Nevermind: it turns out to be a straightforward application of the CGT.

4. Originally Posted by james123
Thanks for your response. I take it that B is the closed unit ball (to have completeness). However, the Banach Steinhaus theorem requires that, for each point, a family of operators all be bounded by some number. Although $\displaystyle L\circ T$ is bounded by some $\displaystyle M_L>0$ for each $\displaystyle L$, the supremum of all such $\displaystyle M_L$ may be infinite!

Are there any other suggestions?

Thanks again.
You are right, the result is true even with no completeness. Being weakly bounded is equivalent to be norm bounded, it is a fundamental result in Functional Analysis, valid for locally convex spaces even non barrelled (spaces satisfying Banach Steinhauss). Your hypothesis clearly implies that T(B) is weakly bounded.