Let $(\Lambda_i)_{i\in I}$ a collection of linear operators from $X$ (Banach space) to $Y$ (Normed space).

Let $\alpha : X \rightarrow [0,\infty]$ be the function $\alpha(x):=\sup_{i \in I} \|\Lambda_ix\|$.

Let $A_n:=\{x \in X : \alpha(x)>n\}$

I want to prove in a direct way that this set is open for every $n\in \mathbb{N}$.

So I want to find an $r_{x_0}$ such that if $x_0 \in A_n$ then $\sup_{i \in I}\|\Lambda x\|>n$ for each $x$ in the open ball $B_{r_{x_0}}:=\{x\in X: \|x-x_0\|<r_{x_0}\}$ .

But I'm blocked. Can you give me a little help? Thank you!

  • 2
    $\begingroup$ Find an $i$ so that $\lVert \Lambda_i x_0\rVert > n$. Pick an $r_{x_0}$ based on that. $\endgroup$ – Daniel Fischer Jul 12 '14 at 22:12
  • $\begingroup$ I tried to take $x$ such that $\|x-x_0\|<\frac{1}{\|\Lambda_k x_0\|n}$ ($k$ being the index that you suggested) but I'm unable to arrive anywhere trying to fill the chain of unequalities $\sup_i \|\Lambda_i x\| > ... > n$. Another clue please? $\endgroup$ – Benzio Jul 12 '14 at 22:51
  • $\begingroup$ Pick $r_{x_0}$ so that $\lVert x-x_0\rVert < r_{x_0} \implies \lVert \Lambda_k x\rVert > n$. (For that specific $k$ you picked with $\lVert \Lambda_k x_0\rVert > n$.) $\endgroup$ – Daniel Fischer Jul 12 '14 at 23:00
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    $\begingroup$ Indeed. So you don't choose $1$ as the numerator, but something depending on $\Lambda_k x_0$. You want $$\lVert \Lambda_k x\rVert \geqslant \lVert \Lambda_k x_0\rVert - \lVert \Lambda_k\rVert\,\lVert x-x_0\rVert = n + (\lVert \Lambda_k x_0\rVert - n) - \lVert \Lambda_k\rVert\,\lVert x-x_0\rVert > n.$$ $\endgroup$ – Daniel Fischer Jul 12 '14 at 23:24
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    $\begingroup$ Note however that $$A_n = \bigcup_{i\in I} \{ x : \lVert \Lambda_ix\rVert > n\}$$ is a simpler way to see that $A_n$ is open: a union of open sets. $\endgroup$ – Daniel Fischer Jul 12 '14 at 23:33

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