Hardy's inequality dual with power weights

Question

We have that hardy inequality dual with power weights is

$$\hspace{20mm}\int_{0}^{\infty} \left( \frac{1}{x}\int_{x}^{\infty} f(t) dt \right)^{p} x^{\alpha} dx \leq \left( \frac{p}{\alpha+1-p} \right)^{p} \int_{0}^{\infty} f(x)^{p}x^{\alpha} dx \hspace{20mm}(1)$$

for $p \geq 1$ and $\alpha>p-1$

I want to prove that the conditions for $p \geq 1$ and $\alpha>p-1$ are essential for (1). I have thought that I could divide it into the following cases:

1. Case $0<p<1$ and $\alpha \geq p-1$. I consider the functions $f_{a}(x) = \mathcal{X}_{(a,a+1)}(x)$. So $\frac{1}{x} \int_{x}^{\infty} f_{a}(t)dx = \frac{1}{x}$ for $x\leq a$

$$\int_{0}^{\infty} \left( \frac{1}{x} \int_{x}^{\infty} f_{a}(t)dx \right)^{p}x^{\alpha}dx \geq \int_{0}^{a} x^{\alpha-p}dx = \frac{a^{\alpha-p+1}}{\alpha-p+1} \rightarrow \infty $$

when $a \rightarrow \infty$

2. Case $p\geq$ and $\alpha \leq p-1$. I consider the same function $f_{a}$.

$$\int_{0}^{\infty} \left( \frac{1}{x} \int_{x}^{\infty} f_{a}(t)dx \right)^{p}x^{\alpha}dx \geq \int_{0}^{a} x^{\alpha-p}dx $$

and how $ \alpha \leq p-1 $ then $ \alpha -p\leq -1 $then is divergent.

would it be alright?

Thanks in advance.

Answer 1

Your last calculation is incorrect. If $\beta \le -1$ the improper integral $\int_0^a x^\beta\, dx$ is divergent. Apply this to $\beta = \alpha - p$.