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Theorem: Let $V$ be a Banach Space and $W$ be a normed linear space. Let $I$ be an arbitrary indexing set and for each $i\in I$, let $T_i\in L(V,W)$. Then either,

(1)there exists $M>0$ such that $\|T_i\|\le M$ for all $i\in I$ or

(2) $\sup_{i\in I}$ $\|T_i(x)\|=\infty$, for all $x$ belonging to some dense $G_{\delta}$ set in $V$.

This is how the proof goes by,

For each $x$, define $\phi(x)=\sup_{i\in I}\|T_i(x)\|$

Define $V_n=\{x\in V| \;\phi(x)>n\}$.

$V_n$ is open for each $n$.

Assume now that there exists $N$ such that $V_N$ fails to be dense in $V$.

Then there exists $x_0\in V$ and $r>0$ such that $x+x_0\notin V_N$ if $\|x\|<r.$

This implies that $\phi(x+x_0)\le N$ for all such $x$ and so, for all $i\in I$ $\|T_i(x+x_0)\|\le N$.

Thus if $\|x\|\le r/2$, we have , for all $i\in I$, $\|T_i(x)\|\le \|T_i(x+x_0)\|+\|T_i(x_0)\|\le 2N$

It follows from this that, for all $i\in I$, $\|T_i\|\le \frac{4N}{r}$

I do not understand how the above highlighted argument holds. Can someone please explain it to me.

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    $\begingroup$ When you wrote the inequality "$\|T_i(x)\|\le \|T_i(x+x_0)\|+\|T_i(x_0)\|\le N$", did you actually mean to write $2N$ instead of $N$ on the RHS. (That would explain why it is $\frac{2N}{r/2}$ and not $\frac{N}{r/2}$.) $\endgroup$ – Martin Sleziak Oct 22 '17 at 5:31
  • $\begingroup$ Oops! It was a typo, thanks for pointing it out @MartinSleziak $\endgroup$ – Naive Oct 22 '17 at 5:33
  • $\begingroup$ "$\sup_{i\in I}$ $\|T_i\|=\infty$, for all $x$" that sentence does not make sense because there is no $x$ in this sentence. $\endgroup$ – miracle173 Oct 22 '17 at 11:37
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Note that \begin{align*} \|T_i\| = \sup_{\|x\|\le 1}\|T_i(x)\| = \sup_{\|x\|\le 1}\frac{2}{r}\left\|T_i\left(\frac{rx}{2}\right)\right\| \le \frac{2}{r}2N = \frac{4N}{r}, \end{align*} where the inequality follows from the previous line and the fact that $\|rx/2\|\le r/2$.

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