The proof I'm referring to can be found here.1
I have worked through this proof and am literally at the last line of the UBP proof but am stuck. I have two questions:
Firstly, how does he obtain the inequality $\| x - x_n \| \leq \frac{1}{2}3^{-n}$? I can get $\| x - x_n \| \leq (6)3^{-n}$ via geometric series (unless I've done the arithmetic wrong!). I suspect this will do, based on what follows, but I'd still like to know how to get this estimate.
Secondly, immediately after this, he has an inequality $\| T_n x \| \geq \frac{1}{6} 3^{-n} \| T_n \|$. Where does this inequality come from? Seems like I'm missing something simple here again.
In any case, thanks in advance!
1Sokal, Alan D., A really simple elementary proof of the uniform boundedness theorem, Am. Math. Mon. 118, No. 5, 450-452 (2011). ZBL1223.46022, MR2805031.
You start with $x_0$. The lemma tells you that for $r=1/3$, the supremum of $\|T_1x'\|$ over $x'\in B(0,1/3)$ is at least $\|T_1\|/3$. So you can surely find some $x_1\in B(0,1/3)$ such that $\|T_1x_1\|>{2\over 3}\|T_1\|/3$. (The term ${2\over 3}$ here is just a number less than $1$ but close to it). Now move on. The supremum of $\|T_2x'\|$ over $x'\in B(x_1,1/9)$ is at least $\|T_2\|/9$. So you can find some $x_2\in B(x_1,1/9)$ such that $\|T_2x_2\|>{2\over 3}\|T_2\|/9$. And so on.
Next, if you agree with $\|x_n-x\|<{1\over 2}3^{-n}$, then you have:
\begin{align}\left|\, \|T_nx\|-\|T_nx_n\|\,\right|&\leq \|T_nx-T_nx_n\|\leq\|T_n\|\cdot \|x_n-x\|\\ &< \|T_n\|{1\over 2}3^{-n}\end{align} so $$-\|T_n\|{1\over 2}3^{-n}+\|T_nx_n\| < \|T_nx\|$$ but $\|T_nx_n\|\geq {2\over 3}3^{-n}\|T_n\|$, so $$\|T_n\|{1\over 6}3^{-n}=\|T_n\|({2\over 3}-{1\over 2})3^{-n}<\|T_nx\|$$