I found a proof of the uniform boundedness theorem, which uses the following lemma:
If $T$ is a linear bounded operator from a normed space $X$ to a normed space $Y$. Then for any $x \in X$ and $r > 0$ we have $$ \sup_{x' \in B(x, r)} \|Tx' \| \geq r \| T \| \ .$$
Proof of the uniform boundedness theorem looks like this:
Suppose the theorem is false, i.e. for any $n > 0$ there is a mapping $T_n$ such that $\|T_n\| > 4^n$. Using the lemma above we can find a sequence $(x_n)$ in $X$ such that, $x_0 = 0$, $\|x_n – x_{n-1} \| < \frac{1}{3^n}$, for $n = 1,2, \dots$ and $\|T_n x_n \| \geq \frac{2}{3} \cdot \frac{1}{3^n} \| T_n \|$. The sequence $(x_n)$ is a Cauchy sequence and since $X$ is a Banach space $x_n \to x$ for some $x \in X$. Moreover $\| x – x_n \| \leq \sum_{k = n + 1}^{\infty} \frac{1}{3^{k}} = \frac{1}{2} \cdot \frac{1}{3^n} \ .$
And from that the author somehow deduce, that $\|T_n x \| \geq \frac{1}{6} \cdot \frac{1}{3^n} \| T_n \|$.
Can somebody please explain to me the very last step of the proof.
Best Answer
$\|T_n x\|=\|T_n(x-x_n)+Tx_n\|≥\big|||T_nx_n\|-\|T_n(x-x_n)\|\big|$. The second term has norm smaller than equal to $\frac12 3^{-n} \|T_n\|$, the first term has norm larger than $\frac23 3^{-n} \|T_n\|$, so the difference without the absolute values is positive and can be bounded by $(\frac23-\frac12)3^{-n}\|T_n\|=\frac16 3^{-n}\|T_n\|$.