Show that the sequence $(T_nx)$ is bounded for every x ∈ $X$.

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:

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.

• Where do you see that $\|g_n\|$ $\geq$ $\|x_n\|$? You have the equality but it comes from the uniform boundedness thm. Jun 14 at 8:46
• @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. Jun 14 at 9:03
• 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 $\|g_n\|=\|y_n\|.$ Jun 14 at 10:07
• @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. Jun 14 at 11:41