送交者: street 于 2007-12-24, 11:39:54:
回答: 有学泛函的仁人志士吗?讨论一个定理 由 piapia 于 2007-12-23, 10:37:43:
now for any x \in X define
T^* such that for any f \in Y*
T^* f (x) = f(T(x))
because of the assumption on T, we know T^*f is continuous.
hence T^* \in Y^*. That is T^* is a continuous linear operator
on Y*. hence ||T^*||< \infty. We can not claim T^* is the conjugate
of T yet (to claim ||T^*||=||T|| < \infty), as we do not yet know T is continous
(which is to be proved.)
But, there exists x0 \in X, T x0 \ne 0.
Using Hahn-Banach, we know that there exists f0 \in Y* such that
||f0||=1 and f0(Tx0)=||Tx0||
Hence ||Tx0||= ||f0(Tx0)|| = ||T*f0(x0)||
\le ||T*||||f0|| ||x0||=||T^*||||x0||
Hence
||T|| <=||T^*|| <\infty. So T is continuous.