Show that this linear operator is surjective
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 $Tx_n \to y$ $$ T(x_m)  T(x_n) > Cx_m  x_n $$ 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 (nontrivial) closed subspace (if it were, you could find a nonzero $f$ such that $f(y) = 0$ for all $y$ in the subspace, by HahnBanach).
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Tomas
Updated on September 07, 2020Comments

Tomas almost 2 years
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.

t.b. almost 10 yearsWhat did you try?


Rudy the Reindeer almost 10 yearsSo how does this show that $T$ is surjective?

BaronVT almost 10 yearsI filled in the missing details.

Rudy the Reindeer almost 10 yearsNice, I'll have to read it later though, since I have to do something else right now.

Tomas over 9 yearsThanks.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