# Proving Holder's inequality for sums

I want to prove the Holder's inequality for sums:

Let $p\ge1$ be a real number. Let $(x_{k})\in l_{p}$ and $(y_{k})\in l_{q}$ . Then, $$\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}y_{k}\vert\le\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}$$ with $q\in\mathbb{R}$ such that $\frac{1}{p}+\frac{1}{q}=1$ .

Inspired by a proof I've seen before, I attempted at a solution. I would like you to confirm the ideas and to answer the questions, which correspond to steps of the proof I don't know how to justify.

My attempt: Let $p>1$ be a real number. $Let (x_{k})\in l_{p}$ and $(y_{k})\in l_{q}$ .

If $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}=0$ or $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}=0$ the inequality is trivially true (Question1: Is it? Why?)

In case both are nonzero, we can define the sequences $(z_{k})$ and $(w_{k})$ with

$$z_{k}=\frac{x_{k}}{\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}}\ \text{ and }\ w_{k}=\frac{y_{k}}{\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}}$$

Now, $$\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)$$ (by Young's Inequality).

But $$\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)=\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert x_{k}\vert^{p}}{p\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)}+\frac{\vert y_{k}\vert^{q}}{q\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)}\right)\le1$$ (Question2: Is this last inequality true? Why? If it is, the result follows smoothly...)

So $\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le1$ . Multiplying both sides by the (positive) term $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\cdot\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}$ , it's done. $\square$

• You actually have inequality since $\frac{1}{p} + \frac{1}{q}=1$. By your construction, the $l^p$ norm of $z=(z_k)$ and $l^q$ norm of $w=(w_k)$ are both 1. Feb 17 '17 at 1:58
• Note the first trivially true thing happens because $\sum_n^\infty |x_n|^p$ means that $x_n$ must identically zero (just consider what an infinite sums means, and see how if any $x_n\neq 0$ this will be positive).
– Mark
Feb 17 '17 at 2:24
• @CheeHan because $$\overset{\infty}{\underset{l=1}{\sum}}\vert z_{l}\vert^{p}=\overset{\infty}{\underset{l=1}{\sum}}\frac{\vert x_{l}\vert}{\Big\vert\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\Big\vert}=\frac{\overset{\infty}{\underset{l=1}{\sum}}\vert x_{l}\vert}{\Big\vert\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\Big\vert}=1$$ and similarly $$\overset{\infty}{\underset{l=1}{\sum}}\vert w_{l}\vert^{q}=1$$ Right?
– Soap
Feb 17 '17 at 10:26
• @Simoes Yes, but of course that is assuming that $(x_k), (y_k)$ are not identically zero sequences. Feb 17 '17 at 18:04

For completeness I'll leave here my complete proof, using the suggestions in the comments.

Proof: Let $$p>1$$ be a real number. $$Let (x_{k})\in l_{p}$$ and $$(y_{k})\in l_{q}$$ .

If $$\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}=0$$ or $$\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}=0$$ the inequality is true: This is equivalent to $$\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}=0$$ or $$\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}=0$$ , and, if any term of one of these series is nonzero, the series would be nonzero (because all terms are zero or positive), so $$\vert x_{k}\vert^{p}=0$$ or $$\vert y_{k}\vert^{q}=0$$ for all $$k$$ . Hence, $$\vert x_{k}\vert^{p}\vert y_{k}\vert^{q}=0$$ for all $$k$$.

In case both are nonzero, we can define the sequences $$(z_{k})$$ and $$(w_{k})$$ with

$$z_{k}=\frac{x_{k}}{\left(\overset{\infty}{\underset{l=1}{\sum}}\vert x_{l}\vert^{p}\right)^{\frac{1}{p}}}\ \text{ and }\ w_{k}=\frac{y_{k}}{\left(\overset{\infty}{\underset{l=1}{\sum}}\vert y_{l}\vert^{q}\right)^{\frac{1}{q}}}.$$

Now, $$\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)$$ by Young's Inequality.

But $$\overset{\infty}{\underset{l=1}{\sum}}\vert z_{l}\vert^{p}=\overset{\infty}{\underset{l=1}{\sum}}\frac{\vert x_{l}\vert^p}{\Big\vert\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\Big\vert^p}=\frac{\overset{\infty}{\underset{l=1}{\sum}}\vert x_{l}\vert^p}{\Big\vert\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\Big\vert^p}=1$$ and similarly $$\overset{\infty}{\underset{l=1}{\sum}}\vert w_{l}\vert^{q}=1.$$

Hence $$\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)=\frac{1}{p}+\frac{1}{q}=1.$$

So $$\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le1$$ . Multiplying both sides by the (positive) term $$\left(\overset{\infty}{\underset{l=1}{\sum}}\vert x_{l}\vert^{p}\right)^{\frac{1}{p}}\cdot\left(\overset{\infty}{\underset{l=1}{\sum}}\vert y_{l}\vert^{q}\right)^{\frac{1}{q}}$$ , it's done. $$\square$$

• For completeness, let me just remark that the result for $p=\infty$ is also true with Holder conjugate $q=1$, and the proof is much simpler, $$\sum_n |a_n b_n | \le \|b\|_{\infty} \sum_n |a_n| = \|b\|_{\infty} \|a\|_1$$ Sep 25 '18 at 13:11