# When is Hardy's inequality a strict inequality?

Let $1<p<\infty$, $f\in L^{p}(0,\infty)$ and $$F(x)=\frac{1}{x}\int^{x}_{0}f(t)dt$$ Hardy's inequality states $$|F|_{p}\leq \frac{p}{p-1}|f|_{p}$$

To show the bound is sharp Rudin suggested to use $f(x)=x^{-1/p}$ on $[1,A]$ and $0$ otherwise. Then let $A$ be arbitrally large. But I am at loss how to show this rigorously.

So we have $F(x)=\frac{1}{x}\int^{x}_{1}\frac{1}{t^{1/p}}dt$ for $A\ge x\ge 1$, for $x<1$ this is $0$. For $x> A$ we have $$F(x)=\frac{1}{x}\int^{A}_{1}\frac{1}{t^{1/p}}dt$$

To fixed the formula we have $$\int^{x}_{1}\frac{1}{t^{1/p}}dt=\frac{p}{p-1}t^{\frac{p-1}{p}}|^{x}_{1}=\frac{p}{p-1}(x^{\frac{p-1}{p}}-1)$$

Putting together we have $$|F|_{p}=(\int_{1}^{A}F^{p}+\int_{A}^{\infty}F^{p})^{1/p}$$

and the first half is $$C\int^{A}_{0}\frac{1}{x^{p}}(x^{\frac{p-1}{p}}-1)^{p}=C\int^{A}_{1}(x^{-1/p}-1/x)^{p}$$ here $C=(\frac{p}{p-1})^{p}$. The second half is $$C \cdot D^{p} \cdot \int_A^\infty \frac{1}{x^p} \, dx = C \cdot \frac{1}{p-1} \cdot A \cdot \left(A^{-\frac{1}{p}}- \frac{1}{A} \right)^p$$ where $D=(A^{\frac{p-1}{p}}-1)$.

But now I do not know how to evaluate $$\int^{A}_{1} (x^{-1/p}-1/x)^{p}$$ though it is clear that $x^{-1/p}\ge \frac{1}{x}$ and hence this can be bound from above by $\log[A]$. If we ignore the second part, then first part of the integral is less than $$\frac{p}{p-1}\log[A]^{1/p}$$ whereas the right hand side $$\frac{p}{p-1}|f|_{p}=\frac{p}{p-1}(\int^{A}_{1}\frac{1}{x})^{1/p}=\frac{p}{p-1}\log[A]^{1/p}$$ But this ignored the second part. So I am wondering how to fix it.

What I want to show after fixing all the constants is $$[\int^{A}_{1}(x^{-1/p}-1/x)^{p}]+\frac{1}{p-1} \cdot A \cdot \left(A^{-\frac{1}{p}}- \frac{1}{A} \right)^p)\le \log[A]$$ Now since $A$ is very large the $\frac{1}{A}$ factor is small. So the left hand side become $$[\int^{A}_{1}(x^{-1/p}-1/x)^{p}]+\frac{K}{p-1}\le \log[A]$$ where $K$ is some constant that can be arbitrally close to 1 as we select $A$. So now all we need is to prove

$$\int^{A}_{1}(x^{-1/p}-1/x)^{p}\le Log[A]-\frac{1}{p-1}$$ where $A$ is a large enough constant.

This problem address the identical question following a different hint.

-
I just want to point out I learned a lot from the answer at here:math.stackexchange.com/questions/126889/… –  user32240 Dec 27 '12 at 17:43
I added a new hint. Hopefully there are no mistakes in my calculations. –  saz Dec 28 '12 at 8:44

Hint The calculation of the "second half" is not correct. It should be

$$C \cdot D^p \cdot \int_A^\infty \frac{1}{x^p} \, dx = C \cdot D^p \cdot \frac{1}{A^{p-1}} \cdot \frac{1}{p-1}$$

Putting this all together:

$$C \cdot D^p \cdot \int_A^\infty \frac{1}{x^p} \, dx = C \cdot \frac{1}{p-1} \cdot A \cdot \left(A^{-\frac{1}{p}}- \frac{1}{A} \right)^p$$

Nexe Hint Use the estimate $$\left(x^{-1/p}-\frac{1}{x} \right)^p = \left(x^{-1/p}-\frac{1}{x} \right) \cdot \underbrace{\left(x^{-1/p}-\frac{1}{x} \right)^{p-1}}_{\leq (x^{-1/p})^{p-1}}$$

(works fine since $p>1$). This gives $$\int_1^A\left(x^{-1/p}-\frac{1}{x} \right)^p \, dx \leq \log A - \frac{p}{p-1} \cdot \left(1- A^{\frac{1}{p}-1} \right)$$

and the second term converges to $\frac{p}{p-1} > \frac{1}{p-1}$.

-
Yes, that is my mistake. –  user32240 Dec 27 '12 at 19:26
It is $\frac{1}{p-1}$ though. –  user32240 Dec 27 '12 at 19:37
No? $\frac{d}{dA} \frac{1}{A^{p-1}} = \frac{d}{dA} A^{1-p} = (1-p) \cdot A^{-p}$. –  saz Dec 27 '12 at 20:47
$\int^{\infty}_{A} x^{-p}=\frac{1}{1-p}x^{-p}|^{\infty}_{A}=\frac{1}{p-1}A^{-p}$. –  user32240 Dec 27 '12 at 21:08
Ah, okay, sorry, because $A$ is the lower bound. Thanks... –  saz Dec 27 '12 at 21:10
show 1 more comment

I get

$$\int_{1}^{A} dx \: \left ( x^{-1/p} - x^{-1} \right )^{p} = (-1)^{p} \log{A} + \sum_{k=0}^{p-1} (-1)^{k} \binom{p}{k} \frac{1}{p-k \left (1- \frac{1}{p} \right ) -1} \left ( 1 - A^{-\frac{1}{p-k \left (1- \frac{1}{p} \right ) - 1}} \right )$$

Not sure what this says about your bound from above being $\log{A}$ as the sum following can either be positive or negative at first glance.

-
No this is too coarse. But thanks nevertheless. –  user32240 Dec 27 '12 at 15:34