# Proof Involving Generalized Mean

Let $x=(x_1,...,x_n) \in \mathbb R^n$ and

$$g(p)=\sqrt[p]{\frac{1}{n}\sum_{k=1}^{n} |x_k|^p)}$$

1. Using Hölder's inequality, show that $g(p)$ is increasing on $(0,\infty)$.

For a sequence with positive weights $w_k$ with sum $\sum w_k=1$, we define the weighted generalized mean as $g(p)=(\sum_{k=1}^{n}w_k|x_k|^p)^\frac{1}{p}$ To prove (1), we need to show that for any $p<q$, the following inequality holds:

$$\sqrt[p]{(\sum_{k=1}^{n}w_k|x_k|^p)} \leq \sqrt[q]{(\sum_{k=1}^{n}w_k|x_k|^q)}$$

I'm stuck after this point. Most of the helpful notes I've found online use Jensen's inequality for this proof, so any help regarding Hölder's inequality would be appreciated. Since we are asked to use Hölder's, I would assume that only the case for positive $p$ and $q$ needs to be proved.

1. Find $\lim_{p \rightarrow \infty}g(p)$.

I know that it's supposed to be $\max(x_1,...,x_n)$, but I have no idea how to show this.

Hölder's inequality can be written as:

Let $1<p,q<\infty$ be conjugate exponents ($\frac{1}{p}+\frac{1}{q}=1$), and $x=(x_1,...x_n) \in \mathbb R^n$, $y=(y_1,...,y_n) \in \mathbb R^n$. Then,

$$\sum_{k=1}^{n}|x_k||y_k| \leq \sqrt[p]{\sum_{k=1}^{n}|x_k|^p}\cdot\sqrt[q]{\sum_{k=1}^{n}|y_k|^q}$$

If $0 < p < q < \infty$, use Holder's inequality with conjugate exponents $\frac{q}{p}$ and $\frac{q}{q-p}$ to get $$\sum_{k = 1}^n |x_k|^p \le n^{1 - \frac{q}{p}} \left(\sum_{k = 1}^n |x_k|^q\right)^{\frac{q}{p}}$$ Rearrange the inequality to obtain $g(p) \le g(q)$.
To prove that $\lim_{p \to \infty} g(p) = \max\{|x_1|\,\ldots, |x_n|\}$, show that for every $\varepsilon > 0$, $$n^{-1/p}(\max\{|x_1|\,\ldots, |x_n|\} - \varepsilon) \le g(p) \le \max\{|x_1|\,\ldots, |x_n|\} \quad (p \ge 1)$$
• Sorry, I should have been more clear. $\max(x_1,...,x_n)$ is not provided in the question, so I cannot assume that the limit equals that; the question simply asks to find said limit. I discovered the answer from online research. – Douglas Fir Jan 29 '15 at 6:41
• @wearyrowboat note that I did not write $\max(x_1,\ldots, x_n)$, but $\max\{|x_1|\,\ldots, |x_n|\}$. If you prove the inequality I mentioned, then you can argue by the squeeze theorem that $\lim_{p\to \infty} g(p) = \max\{|x_1|\,\ldots, |x_n|\}$. – kobe Jan 29 '15 at 6:43
• I am having trouble understanding why $\max\{|x_1|,...,|x_n|\} \leq g(p)$. I assume that I have to use the result from part (1), but I do not know how to commence. – Douglas Fir Jan 29 '15 at 18:38