Let $1 , $f\in L^{p}(0,\infty)$ 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.
When is Hardy's inequality a strict inequality?
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0@BombyxMori I do not understand how this new inequality proves that $\frac {p}{p-1}$ is the best constant – 2017-08-03
1 Answers
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 $\frac{p}{p-1} > \frac{1}{p-1}$.
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0@GuachoPerez That's explained in the last part of the question or am I missing something? – 2017-08-04