2
$\begingroup$

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,

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

$\endgroup$
4
$\begingroup$

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 $\varepsilon > 0$, $$n^{-1/p}(\max\{|x_1|\,\ldots, |x_n|\} - \varepsilon) \le g(p) \le \max\{|x_1|\,\ldots, |x_n|\} \quad (p \ge 1)$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Douglas Fir Jan 29 '15 at 6:41
  • $\begingroup$ @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 $\lim_{p\to \infty} g(p) = \max\{|x_1|\,\ldots, |x_n|\}$. $\endgroup$ – kobe Jan 29 '15 at 6:43
  • $\begingroup$ 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. $\endgroup$ – Douglas Fir Jan 29 '15 at 18:38
  • $\begingroup$ Sorry about that. Please see the edit. $\endgroup$ – kobe Jan 29 '15 at 18:58
  • $\begingroup$ How do we use Holder's inequality to get to that point? $\endgroup$ – jofl Jan 30 '15 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.