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Let $E$ be a Banach space, and $T$ is a linear operator on $E$, furthermore,it's assumed that $\sup_{||x||=1}|f(T(x))|<\infty,\forall f\in E^*;$ $\inf_{||x||=1}\sup_{||f||=1}|f(T(x)|>0;$ and if $\forall x\in E$,$f(T(x)=0$ , then $f=0$.

Under these conditions, please show that $T$ is surjective.

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    What did you try?2012-09-06

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From the first condition, you can deduce that $T$ is bounded, using uniform boundedness principle:

$ \sup_{||x||=1}|f(T(x))|<\infty,\forall f\in E^* $ $ \implies ||f(T)||<\infty,\forall f\in E^* $ $ \implies ||T||<\infty $ first step by definition of $||\cdot||$, second by UBP (there's probably a quicker way to show this, but it's been a while since FA).

The second hypothesis will allow you to conclude the operator is bounded below, and in turn, that the range is closed: if $Tx_n \to y$ $ |T(x_m) - T(x_n)| > C|x_m - x_n| $ so $x_n \to x$ by completeness. Since $Tx_n \to Tx$, we have $Tx = y$.

The third hypothesis prevents the range from being a (non-trivial) closed subspace (if it were, you could find a non-zero $f$ such that $f(y) = 0$ for all $y$ in the subspace, by Hahn-Banach).

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    Thanks.indeed,for the boundedness of the operator,a direct use of UBP,may get that $\{Tx;||x||=1\}$is bounded,so$T$is bounded by definition2012-09-08