$\displaystyle \lim_{p \to \infty} \lVert x \rVert _p = \lVert x \rVert _{\infty}$ is uniform in $x$?

Question

There is some sense in saying that the limit of $\ell^p$ norms

$\displaystyle \lim_{p \to \infty} \lVert x \rVert _p = \lVert x \rVert _{\infty}$

is uniform in $x$?

Maybe with this definition

$(\forall \varepsilon > 0)(\exists M>0)(\forall x \in \ell^q) \big( p>M \ \Longrightarrow \ | \ \lVert x \rVert_p - \lVert x \rVert_{\infty} \ | < \varepsilon \big)$

where $q \in [1, \infty)$ is fixed, and $p$ take values in $(q, \infty)$.

This is true? If not, someone have a counterexample?

Thanks

You want to assume that $x$ is the unit ball of $l^q$ or something like that, or else it is evidently false. — uniquesolution, Sep 10 '15 at 18:31
it will be true on compacts of $\mathbb R^n$ ... hard to say about $\ell^p$ — user251257, Sep 10 '15 at 19:00

score 1 · Accepted Answer · answered Sep 11 '15 at 15:43

1

For $n \in \mathbb{N}$ define a sequence $x^n$ by

$$x_i^n := \begin{cases} 1, & i \leq n, \\ 0, & i>n. \end{cases}$$

Then $\|x^n\|_{\infty}=1$ for all $n \in \mathbb{N}$ and $$\|x^n\|_p = \left( \sum_{i=1}^n 1 \right)^{\frac{1}{p}} = n^{\frac{1}{p}}.$$

Consequently,

$$\big|\|x^n\|_p-\|x^n\|_{\infty} \big| = |n^{\frac{1}{p}}-1|.$$

This shows that we cannot expect uniform convergence of $\|x\|_p$ to $\|x\|_{\infty}$.

answered Sep 11 '15 at 15:43

saz

120,083

@uniquesolution But the OP was asking for one of these "trivial" counterexamples. – saz Sep 15 '15 at 05:26
@uniquesolution No need to get insulting. – saz Sep 15 '15 at 08:39

$\displaystyle \lim_{p \to \infty} \lVert x \rVert _p = \lVert x \rVert _{\infty}$ is uniform in $x$?

1 Answers1