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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!

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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.

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    Very nice. Notwithstanding the fact that the weighted $\ell^{p}$-space contains the usual $\ell^{p}$-space, it will still be isometrically isomorphic to the usual $\ell^{p}$-space via the diagonal map (multiplying/dividing by $p$-th roots of the weights 'coordinate-wise'). From a purely Banach-space viewpoint this will therefore not give anything essentially new.2011-01-16
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    and Theo:Thanks! Sorry, I forgot to notice another difference between Generalized mean and lp norm: each element of a sequence is taken absolute value in the latter but not in the former. So do your results still hold?2011-01-16
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    @steveO: We both understood the generalized mean as $\left(\sum_{i=1}^{n} \frac{1}{n} |x_{i}|^{p}\right)^{1/p}$. The generalized mean is only defined for non-negative numbers (you're in trouble with roots of negative numbers otherwise).2011-01-16
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    @Theo: Good point. On the other hand, passing from one to the other reverses what happens as $p$ varies. @steveO: We had all generalized the generalized mean to allow negative numbers as well. If you don't include absolute values for the negative numbers, then even defining the powers causes problems, and whatever you get isn't likely to be positive. The definition you saw is only for nonnegative numbers, and the definition we gave is a reasonable extension of that definition.2011-01-16
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    +1. "... Lp 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." Why did you write "almost never" instead of "never"?2012-12-28
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    @Tim: I don't remember, but I might have been trying to allow the exception of a measure space with measure $+\infty$ having no subsets of positive finite measure.2012-12-28
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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.

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    Thanks for the answer. What do you call the set of real sequences?2011-01-16
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    @steveO: There is an $\ell^{p}$-norm on sequences, but I can't make sense of the word mean of an *infinite* sequence.2011-01-16
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    @steveO: To be more precise: The sequence space $\ell^{p}$ is defined as the space of real sequences such that $\|x\|_{p} < \infty$. The inequality $\|x\|_{q} \leq \|x\|_{p}$ still holds (same proof) and implies that $\ell^{p} \subset \ell^{q}$. It is not hard to show that the inclusion is in fact strict: For example $\sum_{n=1}^{\infty} \frac{1}{\sqrt[p]{n}} \in \ell^{q} \smallsetminus \ell^{p}$ if $1 \leq p < q$.2011-01-16
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    @steveO: I'm not sure whether you were considering infinite or only finite sequences, but for infinite sequences you can take generalized weighted means with a sequence of positive weights adding up to one. You'll still get norms, and they'll still be increasing with $p$.2011-01-16
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    @Jonas:I am considering both cases: finite and infinite.2011-01-16
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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.

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    @Jonas Meyer: I edited it. Also, assuming that you didn't know the generalized mean was a norm....you could use the $\ell_p$ norm.2011-01-16
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    Thanks. I still don't quite understand the last sentence. The generalized mean *is* a norm (if $p\geq1$), however you view it. I'm guessing that you mean something along the lines of what Theo said, that one way to see that it is a norm is to note that it is the $l^p$ norm divided by $\sqrt[p]{n}$.2011-01-16
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    @Jonas Meyer: I was going along the OP thought process. I am given something called the generalized mean for the first time. It looks similar to the $\ell_p$ norm. What is the relationship between the two?2011-01-16