Here is the question:

Given $X$ and $Y$ are Banach spaces and $T_n$ : $X$$Y$ a sequence of bounded operators.

Show that the sequence $(|f(T_nx)|)$ is bounded for every x ∈ $X$ and every f ∈ $Y^∗$ implies the sequence $(T_nx)$ is bounded for every x ∈ $X$.

Here is the given proof:enter image description here

I would like know how to show $\|g_n\|$ $\geq$ $\|x_n\|$ in the last line. This is the only point where I get stuck.

  • $\begingroup$ Where do you see that $\|g_n\|$ $\geq$ $\|x_n\|$? You have the equality but it comes from the uniform boundedness thm. $\endgroup$
    – Falcon
    Jun 14 at 8:46
  • $\begingroup$ @Falcon Does ∥$g_n$∥ = ∥$x_n$∥ come from uniform boundedness theorem? I don't know how the theorem implies the identity and I thought the only result we get from the theorem is (∥$g_n$∥) being bounded. $\endgroup$
    – math noob
    Jun 14 at 9:03
  • 1
    $\begingroup$ The notation is misleading It is better to write: let $T_nx=y_n$ and $g_n(f)=f(y_n).$ Then $g_n$ is bounded for every $f\in Y^*, $ so $\|g_n\|$ are bounded. Then $\|g_n\|=\sup_{\|f\|\le 1}|f(y_n)|\le \|y_n\|.$ There exists $f\in Y^*$ such that $\|f\|=1$ and $f(y_n)=\|y_n\|$ (this is usually stated in textbooks as a consequence of the Hahn-Banach theorem). Thus $\|g_n\|=\|y_n\|.$ $\endgroup$ Jun 14 at 10:07
  • $\begingroup$ @Ryszard Szwarc I get it. Thank you very much. The consequence of the Hahn-Banach theorem that you mentioned is very useful. It appears in many other exercises. $\endgroup$
    – math noob
    Jun 14 at 11:41


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