5
$\begingroup$

I'm trying to proof Holder Inequality in metric spaces context. Here, we are in the $l^p$ space, every $x=(x_i)$ is a sequence such that $\sum |x_i|^p $ converges. The metric is given by $$d(x,y)=\left( \sum |x_i-y_i|^p \right) $$

First, I prooved the Young inequality.

Set $p\geq 1$, q is defined as $\frac{1}{p}+\frac{1}{q}=1$ then, for $\alpha,\beta$ positive real numbers, we have $\alpha \beta\leq\frac{\alpha^p}{p}+\frac{\beta^q}{q}$

No problem with that. Setting $\alpha=|x_i|$ and $\beta=|y_i|$ we have $$ |\sum x_iy_1|\leq \frac{1}{p}+\frac{1}{q}=1$$

if $\sum |x_i|^p=\sum|y_i|^q=1$.

I'm stucked here. How can I define a sequence in that space that converges, but the value is not 1 and get the Holder Inequality?

Thanks for the help.

$\endgroup$
2
  • 3
    $\begingroup$ It’s strange that the metric does not depend on y. $\endgroup$ Commented Jul 29, 2023 at 10:02
  • $\begingroup$ My mistake. Edited. $\endgroup$
    – ends7
    Commented Jul 30, 2023 at 4:04

1 Answer 1

6
$\begingroup$

Recall Holder's Inequality for $(x) \in l^p$, $(y) \in l^q$ for conjugate $p,q$ is $$\|xy\|_1 := \sum_{i=1}^\infty |x_i y_i| \leq \left( \sum_{i=1}^\infty |x_i|^p \right)^{1/p} \left( \sum_{i=1}^\infty |y_i|^q \right)^{1/q} =: \|x\|_p \|y\|_q.$$

So you're most of the way there by proving it for elements of unit norm. From here you just need a scaling argument.

Suppose $(x)$, $(y)$ are nonzero and as above. Define $(\tilde{x})$, $(\tilde{y})$ to be $(x/\|x\|_p)$, $(y/\|y\|_p)$ respectively. Then, by what you've done so far:

$$\sum_{i=1}^\infty |\tilde{x}_i \tilde{y}_i| \leq 1.$$

Then we get $$ \require{cancel} \begin{aligned} \sum_{i=1}^\infty |x_i y_i| &= \sum_{i=1}^\infty |\|x\|_p\tilde{x}_i \|y\|_q \tilde{y}_i| \\ &= \|x\|_p\|y\|_q \sum_{i=1}^\infty |\tilde{x}_i \tilde{y}_i| \\ &\leq \|x\|_p\|y\|_q \cdot 1 \\ &= \|x\|_p\|y\|_q \end{aligned}.$$

The case for when either $(x)$ or $(y)$ is $0$ should be trivial.

$\blacksquare$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .