Question

Show that \[||x||_{\infty}\leq||x||_2\leq ||x||_1 \forall x\in \mathbb R^n\]

Answer 1

induction?

Answer 2

No ... instead consider the function in p
\[f(x; p) = \left( \sum_{i=1}^n |x_i|^p \right)^{1/p}\]
Now for a given x, show first that:
\[||x||_{\infty} \leq ||x||_p = f(x; p)\]
In fact it is also possible to show--but not necessary here to do so--that
\[\lim_{p \rightarrow \infty} ||x||_p = ||x||_{\infty}\]

Answer 3

this is what i want to show. how does defining the function help?

Answer 4

Because now you just need to differentiate f with respect to p and show it has the right sign.

Answer 5

partially of course.

Answer 6

ii will try it. thank you again

Answer 7

Try it; it's slightly messy as you need to consider a few different cases but you'll get the general result.


However, for your problem, it might be easier to consider this problem directly. E.g.,

||x||_inf = max | xi | = |xj | say.

Then

||x||_inf = | xj | \leq ||x||2
iff
|xj|^2 \leq |x1|^2 + |x2|^2 + .... + |xj|^2 + ... + |xn|^2, which is obvious

That's one part of the inequality.

Try the other inequality directly and I think you won't have a hard time convince yourself that
||x||2 \leq ||x||1