Classical Hardy inequality from the modern form of the Hardy inequality.

Question

This is well known Hardy operator $$(Hf)(x)=\frac{1}{x}\int_{0}^{x}f(t)dt. (1)$$

We have the classical Hardy inequality $$\int_0^{\infty}\left(\frac{1}{x}\int_0^x f(t)dt\right)^pdx\leq \left( \frac{p}{p-1}\right)^p\int_0^{\infty}f^p(x)dx, (2)$$ which tells us about the boundedness of the Hardy operator in $L^p(0,\infty)$ (conditions on $p$ and $f$ are not impotant for this question).

During the last decades this inequality was extended to the form $$\left( \int_0^{\infty}\left( \int_0^x f(t)dt\right)^pu(x)dx\right)^{\frac{1}{p}}\leq C \left( \int_0^{\infty}f^p(x)v(x)dx\right)^{\frac{1}{p}}, (3)$$ which holds when $A<\infty,$ where $$A=\sup_{t>0}\left(\underbrace{ \int_t^{\infty}u(s)ds}_{T}\right)^{1/p}\left(\int_0^tv^{1-p^{'}}(s)ds\right)^{1/p^{'}}.$$

The inequlity (3) (if I understand correct) tells us about the boundedness of the operator $\int_0^xf(t)dt$ from $L^p_u(0,\infty)$ to $L^p_v(0,\infty)$

QUESTION 1: Does the condition on A mean that the operator $\int_0^xf(t)dt$ does not bounded in $L^p(0,\infty)$? (because of the term $T$).

QUESTION 2: How we get the classical Hardy inequality from the modern one? We show that the operator $\int_0^xf(t)dt$ is bounded from $L^p_{\frac{1}{x^p}}(0,\infty)$ to $L^p(0,\infty)$. In other words $$\left( \int_0^{\infty}\left( \int_0^x f(t)dt\right)^p\frac{1}{x^p}dx\right)^{\frac{1}{p}}\leq C \left( \int_0^{\infty}f^p(x)dx\right)^{\frac{1}{p}}.$$

And then say that it is equavalent to the boundedness of Hardy operator $$(Hf)(x)=\frac{1}{x}\int_{0}^{x}f(t)dt$$ in $L^p(0,\infty)$. Are my thoughts correct?

Answer 1

Answer to Question 1: The operator $\int_0^x f(t)\,dt$ is not bounded on $L^p(0,\infty)$ for $p>1$, but your reason is not correct. In this case, $u=v=1$ and thus $A=\infty$. However, what you invoked here is that "$A<\infty$ implies boundedness", which doesn't mean that "$A=\infty$ implies unboundedness"! So, you can't conclude anything if $A=\infty$. To show that $Tf(x)=\int_0^x f(t)\,dt$ is not bounded on $L^p(0,\infty)$ for $p>1$, we consider the function $f(t)=\frac1t \chi_{(1,\infty)}(t)$, then $f\in L^p(0,\infty)$ for $p>1$, but $Tf(x)=\log x\cdot\chi_{(1,\infty)}(x)\notin L^p(0,\infty)$. So, $T$ is not bounded on $L^p(0\infty)$.

Answer to Question 2: Your thoughts are correct. We only need to take $u(x)=x^{-p}$ and $v=1$, in which case we have $$\left(\int_t^{\infty}u(s)ds\right)^{1/p}\left(\int_0^tv^{1-p^{'}}(s)ds\right)^{1/p^{'}}\stackrel{\text{if }p>1}=C\left(t^{-p+1}\right)^{1/p}t^{1/p^{'}}=C t^{-1+\frac1p+\frac1{p'}}=C,\qquad t>0,$$ thus $A<\infty$. Note that here we need $p>1$ ("conditions on $p$ are not impotant for this question", no, $p>1$ is needed). Now it follows from the modern Hardy inequality that $$\left( \int_0^{\infty}\left( \int_0^x f(t)dt\right)^p\frac{1}{x^p}dx\right)^{\frac{1}{p}}\leq C \left( \int_0^{\infty}f^p(x)dx\right)^{\frac{1}{p}}.$$ This is the classcial Hardy inequality, expect that we cannot determine the best constant $\frac{p}{p-1}$. We cannot deduce this constant from the modern one.