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. enter image description here

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 $L_\infty$.

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    $\begingroup$ Much more context has to be supplied before we can help you. $\endgroup$ – Ragib Zaman Feb 15 '12 at 13:50
  • $\begingroup$ @Ragib I thought that the question was self contained. Can you elaborate about what is missing? $\endgroup$ – Artium Feb 15 '12 at 13:53
  • $\begingroup$ 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 $ ? $\endgroup$ – Ragib Zaman Feb 15 '12 at 14:02
  • $\begingroup$ @Ragib See my edit. $\endgroup$ – Artium Feb 15 '12 at 14:35
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    $\begingroup$ @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 $(w_1 + w_2+\cdots + w_n)M^p + w_mM^p = (1 + w_m)M^p$. $\endgroup$ – user2093 Feb 15 '12 at 14:46

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 $\lim_{p\to\infty}\|x\|_\infty$.

  • $\begingroup$ Thank you very much indeed. $\endgroup$ – user85798 Nov 3 '15 at 5:35
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    $\begingroup$ Can you explain the inequality where ||x||_infty \leq ||x||_p? $\endgroup$ – brdcastguy Feb 8 '17 at 21:12
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    $\begingroup$ @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. $\endgroup$ – ely Apr 9 '18 at 17:00

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

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    $\begingroup$ This is part of one half of the proof. $\endgroup$ – robjohn Oct 17 '12 at 12:25

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