Proving Holder's inequality using Jensen's inequality

Question

Let $p$ and $q$ be positive reals such that $\frac{1}{p}+\frac{1}{q} = 1$, so that $p,q$ in $(1,\infty)$.

For $\vec a$ and $\vec b \in \mathbb{R}^2$ prove that $|\vec a \cdot \vec b | \leq ||\vec a||_p|| \vec b||_q$.

A hint was posted for using Jensen's inequality to use $\phi(x) = ln(1 + e^x)$. But I don't know how I'd work that in.

Answer 1

Here's first proving an easier version: Note $\phi(x) = -\log x $ is convex, on $x > 0,$ and hence convexity (= Jensen) yields $$ -\log(tx + (1-t)y) \leq -t\log x - (1-t)\log y, $$ let $x = u^p, y = v^q,$ and $t = 1/p,$ where $u,v > 0.$ You then easily get, $$ uv \leq \frac{u^p}{p} + \frac{v^q}{q}. $$

Now, if $\|a\|_p = \|b\|_q = 1,$ then we see $$ |\sum_{i=1}^n a_i b_i| \leq \sum_{i=1}^n |a_i||b_i| \leq \sum_{i=1}^n \frac{|a_i|^p}{p} + \sum_{i=1}^n \frac{|b_i|^q}{q} = 1. $$ For general vectors, just normalize.

Answer 2

It is easy to get from Jensen to Young to Holder. However if you really want to do directly, note it is sufficient to show: $$ \sum_{k=1}^n \lvert a_k \rvert \lvert b_k \rvert \le \left(\sum_{k=1}^n \lvert a_k \rvert^p \right)^{\frac1p}\left(\sum_{k=1}^n \lvert b_k \rvert^q \right)^{\frac1q} \tag{1}$$ for $\lvert a_k \rvert > 0$ (why?).

As $x^q$ is convex in $(0, \infty)$, by Jensen inequality we have $\displaystyle \left(\sum_{k=1}^n w_k x_k\right)^q \le \sum_{k=1}^n w_k x_k^q$ for $x_k, w_k >0$ and $\sum_k w_k = 1$.

Using $w_k = \dfrac{|a_k|^p}{\sum_k |a_k|^p}$ and $x_k = \dfrac{|a_k||b_k|}{w_k}$ in the above form of Jensen Inequality, we can get $(1)$.