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Theorem: If a sequence of linear bounded operators $\{A_n\}_{n=1}^{\infty}$ is a Cauchy sequence in every point of the Banach space $E_x$, then the sequence of norms $\{\lVert A_n \rVert\}_{n=1}^{\infty}$ is bounded.

First, let the sequence $\{A_n\}_{n=1}^{\infty}$ be a Cauchy sequence in every point of the space $E_x$.

Lets assume the opposite, that the sequence of norms $\{\lVert A_n \rVert\}_{n=1}^{\infty}$ is not bounded. Then $\{\lVert A_n \rVert\}_{n=1}^{\infty}$ is not a bounded set for closed ball $\lVert x-x_0 \rVert \le \epsilon$.

Lets assume that $\lVert A_n x \rVert \le C$ for some constant $C$, for every n and for every x in the closed ball $\overline{B}(x_0, \epsilon)$.

Now, for every $\xi\in E_x$ the element

$x=\frac{\epsilon}{\lVert \xi \rVert}\xi+x_0$ belongs to that ball and therefore $\lVert A_n x \rVert \le C$ for $\forall n \in \mathbf N$.

From there we conclude that

$\frac{\epsilon}{\lVert \xi \rVert}\lVert A_n\xi \rVert - \lVert A_n x_0 \rVert \le \lVert \frac{\epsilon}{\lVert \xi \rVert} A_n\xi + A_n x_0 \rVert \le C$

I really don't understand the last inequality. I'm familiar with the triangle inequality and reversed triangle inequality, but I can't seem to convince myself of this one.

Thanks in advance!

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$\frac {\epsilon} {\|\xi\|} \|A_n\xi\|= \|(\frac {\epsilon} {\|\xi\|}A_n\xi+A_nx_0) -A_nx_0\|\leq \|\frac {\epsilon} {\|\xi\|}A_n\xi+A_nx_0 \|+\|A_nx_0\| =\|A_nx\|+\|A_nx_0\|$ which gives the inequality.

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