Hardy’s Inequality

Edited by Leun Kim

[latexpage]
Theorem 1. (Hardy’s Inequality for Sums)

Let $p >1$ and $ a_n \geqslant 0$ for all $n$. If $ \{ a_n \}_1^{\infty} \in \ell^p $, then $ \left \{ \frac{1}{n} \sum_{k=1}^n a_k \right \}_{1}^{\infty} \in \ell^p $ and  $$ \sum_{n=1}^\infty \left( \frac{a_1 + \cdots + a_n}{n} \right)^p \leqslant \left( \frac{p}{p-1} \right)^p \sum_{n=1}^\infty a_n^p . $$

Proof. Note that if it holds for $a_1 >0$ then it also holds for $a_1 \geqslant 0$. Let $a_1 >0$ and define $ \alpha_n := \frac{a_1 + \cdots a_n}{n}, \; \alpha_0 = 0$. Then \begin{eqnarray} \alpha_n^p – \frac{p}{p-1} a_n \alpha_n^{p-1} &=& \alpha_n^p – \frac{p}{p-1} ( n \alpha_n – (n-1) \alpha_{n-1} ) \alpha_n^{p-1} \\ &=& \alpha_n^p \left( 1 – \frac{np}{p-1} \right) + \frac{(n-1)p}{p-1} \alpha_n^{p-1} \alpha_{n-1} \\ & \leqslant & \alpha_n^p \left( 1 – \frac{np}{p-1} \right) + \frac{n-1}{p-1} \left( (p-1) \alpha_n^p + \alpha_{n-1}^p \right) \\ &=& \frac{1}{p-1} \left( (n-1) \alpha_{n-1}^p – n \alpha_n^p \right). \end{eqnarray} Thus we have $$ \sum_{n=1}^N \alpha_n^p – \frac{p}{p-1} \sum_{n=1}^N a_n \alpha_n^{p-1} \leqslant – \frac{N a_N^p}{p-1} \leqslant 0. $$ So by Hölder’s Inequality, $$ \sum_{n=1}^N \alpha_n^p \leqslant \frac{p}{p-1} \sum_{n=1}^N a_n \alpha_n^{p-1} \leqslant \frac{p}{p-1} \left( \sum_{n=1}^N a_n^p \right)^{1/p} \left( \sum_{n=1}^N \alpha_n^p \right)^{(p-1)/p} $$ or $$ \sum_{n=1}^N \alpha_n^p \leqslant \left( \frac{p}{p-1} \right)^p \sum_{n=1}^N a_n^p $$ and by letting $N \to \infty$, we obtain Theorem 1.

 
Theorem 2. (Hardy’s Inequality for Integrals)

Let $p >1, \; f : (0, \infty) \to \mathbb R$ and $ f \geqslant 0$. If $f \in L^p (0, \infty)$ then $\frac{1}{x} \int_0^x f(t) dt \in L^p (0, \infty)$ and $$ \int_0^\infty \left( \frac{1}{x} \int_0^x f(t)dt \right)^p dx \leqslant \left( \frac{p}{p-1} \right)^p \int_0^\infty f^p (x) dx. $$

Proof. Define $F(x) := \int_0^x f(t) dt$ and let $ 0 < \epsilon < M$ then $$ \int_{\epsilon}^M \left( \frac{F(x)}{x} \right)^p dx = \frac{\epsilon^{1-p} F^p ( \epsilon)}{p-1} – \frac{M^{1-p} F^p (M)}{p-1} + \frac{p}{p-1} \int_{\epsilon}^M x^{1-p} F^{p-1}(x) f(x) dx . $$ Note that by Hölder’s Inequality, $$ \epsilon^{1-p} F^p (\epsilon) = \epsilon^{1-p} \left( \int_0^{\epsilon} f(t) dt \right)^p \leqslant \epsilon^{1-p} \int_0^{\epsilon} f^p (t) dt \left( \int_0^{\epsilon} dt \right)^{p-1} = \int_0^{\epsilon} f^p (t) dt \to 0 $$ as $\epsilon \to 0$. Thus by Hölder’s Inequality again, \begin{eqnarray} \int_0^M \left( \frac{F(x)}{x} \right)^p dx & \leqslant & \frac{p}{p-1} \int_0^M x^{1-p} F^{p-1} (x) f(x) dx \\ & \leqslant & \frac{p}{p-1} \left( \int_0^M \left( \frac{F(x)}{x} \right)^p dx \right)^{1-1/p} \left( \int_0^M f^p (x) dx \right)^{1/p} \end{eqnarray} or w.l.o.g, if $f$ is not null, $$ \int_0^M \left( \frac{F(x)}{x} \right)^p dx \leqslant \left( \frac{p}{p-1} \right)^p \int_0^M f^p(x) dx $$ and by letting $M \to \infty$, we obtain Theorem 2.

References.
[1] G. H. Hardy, J. E. Littlewood, G. Polya (1934). Inequalities.

 

 
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.



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