# Math Help - Supremum Norm and P Norm.

1. ## Supremum Norm and P Norm.

Prove that $||x||_p \rightarrow ||x||_{\infty}$ as $p \rightarrow \infty$.
This is something that I thought I knew how to do until I actually sat down to do it.

So I have to show that:

$\lim_{p \rightarrow \infty} \left( \sum_{i=1}^n |x_i|^p \right)^{\frac{1}{p}}= \max \{|x_1|,...,|x_n| \}$

I tried doing a proof by induction, but it didn't really get me anywhere. Can anyone give me a hint or tip?

2. Are you sure that what's on the LHS is the p-norm?

3. Originally Posted by Showcase_22
This is something that I thought I knew how to do until I actually sat down to do it.

So I have to show that:

$\lim_{p \rightarrow \infty} \left( \sum_{i=1}^n |x_i|^p \right)^{\frac{1}{p}}= \max \{|x_1|,...,|x_n| \}$

I tried doing a proof by induction, but it didn't really get me anywhere. Can anyone give me a hint or tip?
$\max_{1\leqslant \ell\leqslant n}|x_\ell|\leqslant\left(\sum_{k=1}^{n}|x_k|^p\rig ht)^{\frac{1}{p}}\leqslant \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|^p\right)^{\frac{1}{p}}$

4. I mistyped it, sorry!

When I did my proof by induction it works well for $p=1$. The big problem I get is trying to manipulate the sum $\left( \sum_{i=1}^n|x_i|^{k+1} \right)^{\frac{1}{k+1}}$.

I tried using Young's inequality, but that just got really confusing =S

5. $

\max_{1\leqslant \ell\leqslant n}|x_\ell|\leqslant\left(\sum_{k=1}^{n}|x_k|^p\rig ht)^{\frac{1}{p}}\leqslant \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}}
$

This looks like a sandwich rule!

As $p \rightarrow \infty$ isn't $\left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}} \rightarrow 1$ since it's a constant raised to a decreasing power?

(I think I want it to tend to $\max_{1\leqslant \ell\leqslant n}|x_\ell|$).

6. Originally Posted by Showcase_22
$

\max_{1\leqslant \ell\leqslant n}|x_\ell|\leqslant\left(\sum_{k=1}^{n}|x_k|^p\rig ht)^{\frac{1}{p}}\leqslant \left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}}
$

This looks like a sandwich rule!

As $p \rightarrow \infty$ isn't $\left(n\max_{1\leqslant \ell\leqslant n}|x_\ell|\right)^{\frac{1}{p}} \rightarrow 1$ since it's a constant raised to a decreasing power?

(I think I want it to tend to $\max_{1\leqslant \ell\leqslant n}|x_\ell|$).
I forgot a $p$. Check again.

7. Now I see it! The RHS does tend to what I wanted it to.

Thankyou very much, you've been very helpful with regards to everything i've been stuck on.