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This is a follow-up to the question I posted earlier this week.

Consider, for a fixed sequence $(a_n)_n\in\ell^2$ the subspace $$C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}\subset\ell^2.$$ Is this set convex in $\ell^2$?

According to the book [An introduction to Nonlinear Analysis, by Martin Schechter, page 175] it is true, but there is no proof. Can someone help me out?

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Re your deleted question: I hope it was clear that the objections were only about what was probably a typo in the question: For $1 \leq p \lt \infty$ there is a function $f_0$ such that always $f = \sup_{n \in \mathbb{N}} f \wedge |f_0|$, provided that the measure space is $\sigma$-finite: it suffices to take $f_0$ such that $|f_0| \neq 0$ a.e. Assuming $\sigma$-finiteness is also necessary (up to allowing a purely infinite part): if you take counting measure on an uncountable set then there is no function having that property (since every $\ell^p$-function is supported on a countable set). –  t.b. Aug 1 '12 at 0:01

1 Answer 1

up vote 6 down vote accepted

Let $x=(x_n), y=(y_n) \in C$. We need to show that for all $t \in [0,1]$ we have $(1-t)x+ty \in C$. So let $t \in [0,1]$; then

$$(1-t)x+ty = ((1-t)x_n + ty_n)$$

So you need to prove that $\left| (1-t)x_n + ty_n \right| \le a_n$ for all $n \in \mathbb{N}$.

This follows immediately from the following properties of $\left| - \right|$:

  • $\left| a+b \right| \le \left| a \right| + \left| b \right|$
  • $\left| a b \right| = \left| a \right| \left| b \right|$

Also noting that if $t \in [0,1]$ then $1-t \ge 0$ and $t \ge 0$.

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