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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.

Pineapple Tacos
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3 Answers3

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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 $(1)$.

Macavity
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  • I only need to show it in $\mathbb{R}^2$. Does that make this simpler? – Pineapple Tacos Dec 02 '13 at 19:27
  • Not much - the steps remain the same, except $n=2$. The simplification in the end is perhaps a little easier. – Macavity Dec 02 '13 at 19:30
  • I have a question. Equation (1), that is Holder's inequality is trivially true for the case when $p=q=1$, and $a_i>0$ and $b_i>0$? I only see the formulation of this result for $p>1$. –  Apr 05 '14 at 10:09
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We want to show that $$ a_1b_1 + a_2b_2 \le (a_1^p+a_2^p)^{1/p} (b_1^q+b_2^q)^{1/q} $$ First note that the inequality is homogeneous in both $(a_1,a_2)$ and $(b_1,b_2)$, separately. Thus we can scale them both to halve the number of variables involved: dividing both sides by $a_1b_1$, we get $$ 1 + \frac{a_2}{a_1}\cdot\frac{b_2}{b_1} \le \Bigl(1+\Bigl(\frac{a_2}{a_1}\Bigr)^p\Bigr)^{1/p} \Bigl(1+\Bigl(\frac{b_2}{b_1}\Bigr)^q\Bigr)^{1/q} $$ Write $u=a_2/a_1$ and $v=b_2/b_1$; we then want to show $$ 1 + uv \le (1+u^p)^{1/p} (1+v^q)^{1/q} $$ Taking logarithms, we get the equivalent $$ \ln(1+uv) \le \frac1p \ln(1+u^p) + \frac1q \ln (1+v^q) $$ Writing $x=p\ln u$ and $y=q\ln v$, this is equivalent to $$ \ln(1+e^{x/p+y/q}) \le \frac1p \ln(1+e^x) + \frac1q \ln (1+e^y) $$ which asserts the convexity of $x\mapsto\ln(1+e^x)$. So, check the second derivative and you're done.

(This argument generalizes to give the inequality in this question, which was question A2 on the 2003 Putnam; see Kedlaya's archive for that solution and some other nice ones.)

Community
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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.

Raghav
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