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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? –  t.b. Sep 6 '12 at 5:38

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up vote 1 down vote accepted

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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So how does this show that $T$ is surjective? –  Matt N. Sep 6 '12 at 6:52
    
I filled in the missing details. –  BaronVT Sep 6 '12 at 7:13
    
Nice, I'll have to read it later though, since I have to do something else right now. –  Matt N. Sep 6 '12 at 7:15
    
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 definition –  sun Sep 8 '12 at 0:12

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