? p-norm <= (p-1)norm <= <= 2-norm <= 1-norm?

[patiobarbecue]

1.
the p-norm for a vector x in R^n is defined usually:

|| x ||_p = (x_1^p + x_2^p + ... + x_n^p)^{1/p}

the question is to verify:

|| x ||_p <= || x ||_{p-1}

Homework Equations

I guess even more generally p-norm is a decreasing function in p for "any" x?

The Attempt at a Solution

Neither Cauchy, nor the more general Holder doesn't seem to apply.

 

[jbunniii]

Try using Jensen's inequality:

If \phi is a convex function, and y_1, \ldots, y_N lie in its domain, and \lambda_1, \ldots \lambda_N are positive real numbers such that

\sum_{n=1}^{N} \lambda_n = 1

then

\phi\left(\sum_{n=1}^{N} a_n y_n\right) \leq \sum_{n=1}^{N} a_n \phi(y_n)

For p = 2, try using y_n = x_n^2, \phi(x) = x^{1/2}, and a_n = 1/N. You will have to modify these choices appropriately for general p.

Also, your proposed generalization is correct: if 1 \leq q \leq p \leq \infty then ||x||_p \leq ||x||_q. The p = \infty case is handled separately (Jensen's inequality doesn't apply) but it's easy.

 

[cracking]

I think it's a proof for norm 2 and 3. For any n, n-1 it's the same essentially.

First consider R2.
Note (x^2+y^2)^1/2 > (x^3+y^3)^1/3 is equivalent to
[1+(y/x)^2]^1/2 > [1+(y/x)^3]^1/3
Assume y/x < 1 without losing generality, then from [1 + ...] > 1 one sees
[1+(y/x)^2]^1/2 > [1+(y/x)^3]^1/2 > [1+(y/x)^3]^1/3.

Next do an induction toward Rn.
(x1^2+x2^2+...+xn^2)^1/2 > (x1^3+x2^3+...+xn^3)^1/3 is equivalent to
(y^2+xn^2)^1/2 > (z^3+xn^3)^1/3, where y and z stands for (x1^2+...+xn-1^2)^1/2 and (x1^3+...+xn-1^3)^1/3 respectively.
By induction y > z. Hence (y^2+xn^2)^1/2 > (z^2+xn^2)^1/2 > (z^3+xn^3)^1/3. QED.

 

[patiobarbecue]

thanks, guys, it helps!

 

[TechMan]

||x||_p isn't a norm for p<1 ?

Hello people! I have a question about the p-norm.
We know that the p-norm is a true norm for p >= 1.
But I dont'n understand why it is not a true norm for 0 < p < 1.
I think it hold the three conditions for a norm include when 0 < p < 1.

Somebody can helpme??
Thanks in advance!