Trying to compare generalized mean and $l_p$ norm p is a non-zero real number. The forms of generalized mean with exponent p and $l_p$ norm seem similar except whether to take mean before taking p-root. I am trying to compare them together.
Can generalized mean with exponent p become a norm on the real sequence space?
Given a sequence, generalized mean with exponent p is nondecreasing with respect to p. Does $l_p$ norm also non-decrease with p?
Can someone explain why? Thanks in advance!
 A: If $1\leq p\leq\infty$, then yes, both are norms.
The ordinary $\ell^p$ norm can decrease with $p$. E.g., Consider $(1,1)$ in $\mathbb{R}^2$. (And as Theo shows in his answer, the $\ell^p$ norm is always nonincreasing.)
Both norms can be seen as $L^p$ norms on $\mathbb{R}^n$ considered as functions on an $n$-point measure space.  For the generalized mean, each point has measure $\frac{1}{n}$, so that the space is normalized to have measure $1$, making it a "probability space".  For the ordinary $\ell^p$ norm, each point has measure $1$.  It is true in general for probability spaces that the $L^p$ norm increases with $p$.  This almost never holds for spaces of measure greater than $1$, as can be seen by taking a characteristic function of a set of finite measure greater than $1$.  On the other hand, having the $L^p$ norm decrease with $p$ is also atypical, and happens only when there are no sets of positive measure less than $1$ (again consider a characteristic function).
Since I now see that you want to consider infinite sequences, I'll say a few words on generalized means for that case.  For each sequence of weights $w_1,w_2,\ldots$ consisting of positive numbers summing to $1$, you can define the generalized mean with exponent $p$ by $(w_1|x_1|^p+w_2|x_2|^p+\cdots)^{1/p}$.  The set of all sequences $(x_1,x_2,\ldots)$ with finite mean forms the $L^p$ space of a countable measure space with total measure $1$, if you like thinking of it that way.  It will contain the ordinary $\ell^p$ space.
A: For $p \geq 1$ the generalized mean defines a norm, because it is the $\ell^{p}$-norm only up to a factor $\sqrt[p]{n}$.
However, if $p \lt 1$, the generalized mean (and also the $\ell^{p}$-expression) don't define a norm because the set $\|x\|_{p} \leq 1$ is not convex: if $x_{i} \geq 0$ and $y_{i} \geq 0$ for all $i$ then $\|x + y\|_{p} \geq \|x\|_{p} + \|y\|_{p}$!
If $p \leq q$ then $\|x\|_{q} \leq \|x\|_{p}$. To see this, note that both sides of the inequality are invariant by multiplication with a positive real number, so we may take without loss of generality an $x$ with $\|x\|_{p} = 1$. Then $\|x\|_{q}^{q} = \sum_{j = 1}^{n} |x_{j}|^{q} \leq \sum_{j = 1}^{n} |x_{j}|^{p} = 1$ this is because for $t \leq 1$ and $p \leq q$ we have $t^{q} \leq  t^{p}$.
I don't understand what you ask about the sequence space.
A: The power mean is defined as (for $p \geq 1$): $$M_p(a_{1}, \dots, a_n) = \left(\frac{1}{n} \sum_{i=1}^{n} a_{i}^{p} \right)^{1/p}$$.
The $\ell_p$ norm is defined as follows (for $p \geq 1$): $$ \left(\sum_{i=1}^{n} |x_i|^{p} \right)^{1/p}$$ If we let $x_i = \frac{a_i}{\sqrt[p]{n}}$ then we can view the generalized mean as a norm. 
