Show that this linear operator is surjective

1,917

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).

Share:
1,917

Related videos on Youtube

Tomas
Author by

Tomas

Updated on September 07, 2020

Comments

  • Tomas
    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.
      t.b. almost 10 years
      What did you try?
  • Rudy the Reindeer
    Rudy the Reindeer almost 10 years
    So how does this show that $T$ is surjective?
  • BaronVT
    BaronVT almost 10 years
    I filled in the missing details.
  • Rudy the Reindeer
    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
    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