Let $f, g:\mathbb{Z}\rightarrow [0,\infty )$ be two functions such that $f(n)=g(n)=0$ with the exception of finitely many integers $n$. Define $h:\mathbb{Z}\rightarrow [0,\infty )$ by \[h(n)=\max \{f(n-k)g(k): k\in\mathbb{Z}\}.\] Let $p$ and $q$ be two positive reals such that $1/p+1/q=1$. Prove that \[ \sum_{n\in\mathbb{Z}}h(n)\geq \Bigg(\sum_{n\in\mathbb{Z}}f(n)^p\Bigg)^{1/p}\Bigg(\sum_{n\in\mathbb{Z}}g(n)^q\Bigg)^{1/q}.\]
Problem
Source: Romania TST 4 2012, Problem 2
Tags: inequalities, function, inequalities proposed
dgrozev
20.05.2012 14:55
Let $S$ is finite, $S \subset \mathbb{Z}$ and $f(n)=g(n)=0,\, n \in \mathbb{Z} \setminus S$. We use a standart notation $\|f\|_{p} = \left( \sum_{i\in S} |f(i)|^p \right)^{\frac{1}{p}}, p < \infty;\, \|f\|_{\infty} = \max_{i\in S}|f(i)| $. We have: (1) $ \sum_{n\in\mathbb{Z}}h(n) \geq \|f\|_{\infty} \|g\|_1 $. Why? If $\|f\|_{\infty} = f(n_0) \, \Rightarrow \, h(n_0 + n) \geq \|f\|_{\infty} g(n)$ and summing up we get (1). In the same way: (2) $ \sum_{n\in\mathbb{Z}}h(n) \geq \|g\|_{\infty} \|f\|_1 $. From (1) and (2) follows: (3) $ \sum_{n\in\mathbb{Z}}h(n) \geq \frac{1}{q} \|f\|_{\infty} \|g\|_1 + \frac{1}{p} \|g\|_{\infty} \|f\|_1 $. Now we write: $\sum_{n\in S} f(n)^p \leq \|f\|_{\infty}^{p-1}\|f\|_1$, which implies: (4) $ \|f\|_{\infty}^{\frac{1}{q}} \|f\|_1^{\frac{1}{p}} \geq \|f\|_p $. (5) $\|g\|_{\infty}^{\frac{1}{p}} \|g\|_1^{\frac{1}{q}} \geq \|g\|_q $. Multiplying (4) and (5) we get: (6) $ \left( \|f\|_{\infty} \|g\|_1 \right)^{\frac{1}{q}} \left( \|g\|_{\infty} \|f\|_1 \right)^{\frac{1}{p}} \geq \|f\|_p \|g\|_q $. Now, (3) , (6) and weighted AM-GM imply: $ \sum_{n\in\mathbb{Z}}h(n) \geq \|f\|_p \|g\|_q $. Comment. The fact that $f,g$ vanish with the exception of finitely many integers is not so essential. Slightly restating the problem we can get rid of that requirement.