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For vector p-norm defined as $(∑_{i=1}^n x_i^p )^{\frac{1}{p}}$ for any $p\ge 1$ and vector ${\bf{x}}=\{x_1,...,x_n\}$. The following proves it is decreasing with respect to $p$ by taking derivative (you don't need to read the whole proof, just have a look),

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However, I am thinking if there is another approach without using the derivative. Is there any proof for monotonicity of p-norm without using derivatives? The upper bound can be proved by Holder's inequality by Relations between p norms

We have plenty of inequalities that lead to the definition of p-norm: Young's inequality, Jensen's inequality, Holder's inequality, Minkowski’s inequality. Maybe there is a proof using those inequalities?

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  • $\begingroup$ Correct me if I'm wrong, but I believe there is an error in the proof from your book. Specifically, the derivative looks to be incorrect. I believe it should be $\frac{f'(r)}{f(r)} = -\frac{1}{r^2}\ln(a_1^r+\cdots+a_n^r)+\cdots$. The error is the lack of the natural log in the book proof. $\endgroup$
    – BSplitter
    Jan 16, 2018 at 20:53
  • $\begingroup$ Also, I believe the second to last line is incorrect as well. It appears that numerator of the natural log argument should be $a_i$ instead of $a_i^r$, unless I'm missing something. $\endgroup$
    – BSplitter
    Jan 17, 2018 at 22:37

3 Answers 3

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In Help show $(∑_{i=1}^n |x_i | )^p≥∑_{i=1}^n |x_i |^p $ using common inequalities (like Holder's inequality) you find a proof for the special case $$\|x\|_1 \ge \|x\|_p.$$ Now, replace $x$ by $|y|^r$ and you find $$\|y\|_r^{1/r}=\||y|^r\|_1 \ge \||y|^r\|_p = \|y\|_{rp}^{1/r}.$$ Taking the $r$th root you have $$\|y\|_r \ge \|y\|_{r p}$$ and this settles the general case.

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  • $\begingroup$ This is a very nice proof! $\endgroup$
    – Ralph B.
    Oct 2, 2016 at 0:16
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I remember proving that inequality by the derivative route on the blackboard in a homework session in my measure theory class, with great difficulty and notational pain.

My professor sat through it all, then at the end said: "Yes, very good. Now imagine we scale $x$ so that the $p$-norm is $1$. Does that make things easier?"

I cried.


In case you want the details:

Notice first that the $p$-norm of $x$ behaves well with scaling: $|t x|_p = |t| \cdot |x|_p$ for any $t \in \mathbb{R}$. The same holds for the $q$-norm, so to prove our inequality we may scale $x$ so that $|x|_p = 1$.

Now, if some $x_j = 1$, then all the other are zero, and $|x|_q = 1$ as well.

Otherwise, all $x_j < 1$. Since $p \leq q$, we then have that $|x_j|^q \leq |x_j|^p$, so $$ \sum_{j=1}^n |x_j|^q \leq \sum_{j=1}^n |x_j|^p = 1. $$ Finally, taking the $q$-th root of something smaller than 1 gives something smaller than 1, so we conclude that $|x|_q \leq 1 = |x_p|$.

Whether this counts as avoiding calculus depends on whether you believe that $x^p$ is decreasing in $p$ for fixed $x < 1$. To really prove that (or even define $x^p$ in general), you need the exponential function and logarithm, and to know that $\exp$ is monotone.

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It's enough to show $(a^p + b^p)^{1/p} \ge (a^q + b^q)^{1/q}$ for $a,b>0$ and $p \le q$, then the general case follows by induction.

By dividing both sides by $a$ and letting $x = (b/a)^p$ the equation becomes $$ (1 + x)^{1/p} \ge (1 + x^{q/p})^{1/q} $$ which is equivalent to $$ (1 + x)^{q/p} \ge 1 + x^{q/p} $$ Now it holds in general that if $x \ge 0$ and $r \ge 1$, then $(1 + x)^r \ge 1 + x^r$. This is clear for integer $r$ by the binomial theorem and for general $r$ by differentiating with respect to $x$, but I'm not sure how one might show it in general without calculus. Suggestions are welcome.

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  • $\begingroup$ Why did the inequality sign flip? Is it because the initial $\le$ should have been $\ge$? $\endgroup$
    – BSplitter
    Jan 17, 2018 at 22:55
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    $\begingroup$ @BlakeSplitter Just a typo, fixed :) $\endgroup$
    – arkeet
    Feb 27, 2018 at 2:32

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