# A proof to Holder Inequality

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.

• It’s strange that the metric does not depend on y. Commented Jul 29, 2023 at 10:02
• My mistake. Edited. Commented Jul 30, 2023 at 4:04

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.

$$\blacksquare$$