16
$\begingroup$

For any $p > 1$ and for any sequence $\{a_j\}_{j=1}^\infty$ of nonnegative numbers, a classical inequality of Hardy states that $$ \sum\limits_{k=1}^n\left(\frac{\sum_{i=1}^ka_i}{k}\right)^p\le \left(\frac{p}{p-1}\right)^p \sum\limits_{k=1}^n a_k^p$$ for each $n\in N$.

There are now many many proofs of Hardy's inequality. Which proof is your favourite one, which would be the simplest proof? It is preferable if you could present the detailed proof here so that everyone can share it.

$\endgroup$
  • 7
    $\begingroup$ Check out the book "The Cauchy-Schwarz Master Class" by Michael Steele. $\endgroup$ – Jonas Teuwen Jun 9 '11 at 23:15
  • $\begingroup$ A good reference. $\endgroup$ – Sunni Jun 9 '11 at 23:56
  • $\begingroup$ I have posted a proof below. Note that it uses Minkowski's inequality on convolutions. $\endgroup$ – Amitesh Datta Jun 10 '11 at 3:55
  • 4
    $\begingroup$ Why is this community wiki? $\endgroup$ – Eric Naslund Jun 25 '11 at 23:39
9
$\begingroup$

Let $(\mathbb{R}^{+},\frac{dt}{t})$ be the multiplicative group of positive real numbers with the usual topology and Haar measure $\frac{dt}{t}$. Define functions $g:\mathbb{R}^{+}\to [0,\infty)$, $h:\mathbb{R}^{+}\to [0,\infty)$ by $g(x)=\left|f(x)\right|x^{1-\frac{b}{p}}$ and $h(x)=x^{-\frac{b}{p}}\chi_{[1,\infty)}(x)$. We will apply Minkowski's inequality to the convolution $F=g\star h$. Note that:

\begin{align*} F(x)= &\int_{0}^{\infty} \left|f(t)\right|t^{1-\frac{b}{p}}\;\;\frac{t^{\frac{b}{p}}}{x^{\frac{b}{p}}}\;\;\chi_{[1,\infty)}\left(\frac{x}{t}\right)\frac{dt}{t} \ =& \frac{1}{x^{\frac{b}{p}}}\int_{0}^{x} \left|f(t)\right| dt \ \end{align*}

if $x\in\mathbb{R}^{+}$. Furthermore,

$$ \left\|h\right\|_{L^1(\mathbb{R}^{+},\frac{dt}{t})}= \int_{1}^{\infty} t^{-\frac{b}{p}-1}dt=\frac{p}{b} $$

and

$$\left\|g\right\|_{L^p(\mathbb{R}^{+},\frac{dt}{t})}=\left(\int_{0}^{\infty} \left|f(t)\right|^p t^{p-b-1} dt \right)^{\frac{1}{p}} $$

Minkowski's inequality thus implies that

\begin{align*} \left(\int_{0}^{\infty} \left(\int_{0}^{x} \left|f(t)\right| dt\right)^p x^{-b-1} dx\right)^{\frac{1}{p}}\leq \frac{p}{b}\left(\int_{0}^{\infty} \left|f(t)\right|^p t^{p-b-1} dt\right)^{\frac{1}{p}} \end{align*}

Let us now redefine the functions $g:\mathbb{R}^{+}\to [0,\infty), h:\mathbb{R}^{+}\to [0,\infty)$ by $g(x)=\left|f(x)\right|x^{1+\frac{b}{p}}$ and $h(x)=x^{\frac{b}{p}}\chi_{(0,1]}(x)$. We will apply Minkowski's inequality to the convolution $F=g\star h$. Note that,

$$ F(x)= \int_{0}^{\infty} \left|f(t)\right|t^{1+\frac{b}{p}}\;\;\frac{x^{\frac{b}{p}}}{t^{\frac{b}{p}}}\chi_{(0,1]}\left(\frac{x}{t}\right)\frac{dt}{t} = x^{\frac{b}{p}}\int_{x}^{\infty} \left|f(t)\right| dt $$

if $x\in\mathbb{R}^{+}$. Furthermore,

$$ \left\|h\right\|_{L^1(\mathbb{R}^{+},\frac{dt}{t})}=\int_{0}^{1} t^{\frac{b}{p}-1} dt = \frac{p}{b} $$

and

$$\left\|g\right\|_{L^p(\mathbb{R}^{+},\frac{dt}{t})}=\left(\int_{0}^{\infty} \left|f(t)\right|^p t^{p+b-1} dt \right)^{\frac{1}{p}} $$

Minkowski's inequality thus implies that,

\begin{align*} \left(\int_{0}^{\infty} \left(\int_{x}^{\infty} \left|f(t)\right| dt\right)^p x^{b-1} dx\right)^{\frac{1}{p}}\leq \frac{p}{b} \left(\int_{0}^{\infty} \left|f(t)\right|^p t^{p+b-1} dt\right)^{\frac{1}{p}} \end{align*}

$\endgroup$
  • $\begingroup$ Try avoiding the {align*} environments, there is nothing crucial you cannot do without them and they cause horrible rendering problems. (You can see the modifications I made by clicking on the link which indicates the time since editing above my name.) $\endgroup$ – Did Jun 10 '11 at 6:02
  • $\begingroup$ @Didier Thanks! I will remember that in the future. $\endgroup$ – Amitesh Datta Jun 10 '11 at 6:19
2
$\begingroup$

I'd like to share this article. HARDY TYPE INEQUALITIES VIA CONVEXITY - THE JOURNEY SO FAR http://ajmaa.org/searchroot/files/pdf/v7n2/v7i2p18.pdf It is refreshing!!!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.