# Understanding the proof that $L_\infty$ norm is equal to $\max\{f(x_i)\}$

Attached is a proof I found. It is probably very basic, but I can not understand the marked thing. Why this term is zero? I hope someone can explain it for me. Edit(elaboration): A Norm is a function that takes a function $f$ and returns a number. Discreet norm's input is not the function itself but it's values at certain defined points. Each discreet norm has it's own set of points $\{x_i\}$ (and also weights $\{w_i\}$ ).

There are some conditions it should follow to be called a norm (you can google it).

Here, $f_i$ is short notation of $f(x_i)$, and the $L_p$ norm is defined as $$L_p \equiv (\sum{|f_i|^p w_i})^{1/p}$$

$|f_i|$ is simply absolute value. And of course $f$ should be defined at $\{x_i\}$ points.

If we send $p$ to infinity then we get the infinity norm $L_\infty$.

• Much more context has to be supplied before we can help you. – Ragib Zaman Feb 15 '12 at 13:50
• @Ragib I thought that the question was self contained. Can you elaborate about what is missing? – Artium Feb 15 '12 at 13:53
• What kind of objects are these $f_i$? If there are functions, what is their domain and codomain, and what does $|f_i|$ mean? What kind of quantity are the $w_i$ ? – Ragib Zaman Feb 15 '12 at 14:02
• @Ragib See my edit. – Artium Feb 15 '12 at 14:35
• @Artium: The marked term is not equal to zero but rather tends to zero as $p$ tends to infinity, assuming that $f_i$, which by definition is $f(x_i)$, is strictly smaller than $M$ (which is an assumption that seems to be made here). You could alternatively try to prove the limit above using the squeeze theorem: on the one hand the quantity in parentheses is $\geq w_mM^p$. On the other hand, it is dominated by $(w_1 + w_2+\cdots + w_n)M^p + w_mM^p = (1 + w_m)M^p$. – user2093 Feb 15 '12 at 14:46

## 2 Answers

Let $x\in R^n$ and $\|x\|_\infty=\max_{1\leq i\leq n}|x_i|$, write $\|x\|_p$ as $$\|x\|_p = \left(\sum_{i=1}^n|x_i|^p\right)^{1/p}=\|x\|_\infty\left(\sum_{i=1}^n\left(\frac{|x_i|}{\|x\|_\infty}\right)^p\right)^{1/p}$$

noting that $\left(\frac{|x_i|}{\|x\|_\infty}\right)^p\leq1$ for every $i$, with equality at least once and at most $n$ times, then $$\|x\|_\infty\leq\|x\|_p\leq \|x\|_\infty n^{1/p}$$ and because $n>0$ gives $\lim_{p\to\infty}n^{1/p}=1$, then $\lim_{p\to\infty}\|x\|_\infty$.

• Thank you very much indeed. – user85798 Nov 3 '15 at 5:35
• Can you explain the inequality where ||x||_infty \leq ||x||_p? – brdcastguy Feb 8 '17 at 21:12
• @brdcastguy Imagine we could re-write the p-norm of $x$ in terms of whichever component happens to be the $||x||_{\infty}$. This would say $||x||_{p} = (||x||_{\infty}^{p} + |x_2|^{p} + ...)^{1/p}$. Since the p-th root function is strictly increasing along the positive real axis, and we're only considering absolute values of input components, this means it is greater than $(||x||_{\infty}^{p})^{1/p} = ||x||_{\infty}$. This is for positive powers $p$, since we care about $p$ defining an $L_{p}$ norm. – ely Apr 9 '18 at 17:00

$(f_i/M)^p$ tends to zero becouse of exponent p tends to +infinite and $f_i/M<1$.

• This is part of one half of the proof. – robjohn Oct 17 '12 at 12:25