# Show that this linear operator is surjective

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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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### Tomas

Updated on September 07, 2020

• 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 years
What did you try?
• Rudy the Reindeer almost 10 years
So how does this show that $T$ is surjective?
• BaronVT almost 10 years
I filled in the missing details.
• Rudy the Reindeer almost 10 years
Nice, I'll have to read it later though, since I have to do something else right now.
• Tomas over 9 years
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