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I have been studying for an upcoming exam and this question was on a previous exam:

Prove that for $1 < p < \infty, \alpha > 1/p$

$$ \int_0^{\infty} x^{-\alpha p} \left| \int_0^x f(t)dt\right|^pdx \leq C_p \int_0^{\infty} |f(x)x^{1-\alpha}|^p dx. $$

So this is what I wrote down:

After a change in variables, $t \mapsto xt$, we get:

$$ \left( \int_0^{\infty} x^{-\alpha p} \left| \int_0^x f(t) dt \right|^p dx \right)^{1/p} = \left( \int_0^{\infty} \left| \int_0^x x^{-\alpha}f(t) dt \right|^p dx \right)^{1/p} = \left( \int_0^{\infty} \left| \int_0^1 x^{1-\alpha} f(xt) dt \right|^p dx \right)^{1/p}. $$

Minkowski's inequality for integrals yields:

$$ \left( \int_0^{\infty} \left| \int_0^1 x^{1-\alpha} f(xt) dt \right|^p dx \right)^{1/p} \leq \int_0^1 \left( \int_0^{\infty} |x^{1-\alpha}f(xt)|^p dx\right)^{1/p} dt. $$

Another change in variables $x \mapsto x/t$ yields:

$$\int_0^1 \left( \int_0^{\infty} |x^{1-\alpha}f(xt)|^p dx\right)^{1/p} dt = \int_0^1 \left( \int_0^{\infty} |t^{\alpha-1}x^{1-\alpha} t^{-1}f(x)|^p dx\right)^{1/p} dt = \int_0^1 t^{\alpha-2} \left( \int_0^{\infty} |x^{1-\alpha} f(x)|^p dx\right)^{1/p} dt.$$

And this last term has the obvious bound:

$$\int_0^1 t^{\alpha-2} \left( \int_0^{\infty} |x^{1-\alpha} f(x)|^p dx\right)^{1/p} dt \leq C_p \left( \int_0^{\infty} |x^{1-\alpha} f(x)|^p dx\right)^{1/p}$$

with $C_p = \int_0^1 t^{\alpha-2} < \infty$ since $\alpha > 1/p > 1$, and the result follows after taking the $p$-th power.

Now what confuses me is why they would put $\alpha > 1/p$ where I only used $\alpha > 1$. This makes me think I made a mistake. This property is later reinforced with the question: Is there a $p \in (1, \infty)$ such that this inequality remains true for $\alpha = 1/p$. Well, if I did everything correctly, the answer would be it holds for all $p \in (1,\infty)$. Will some one please give me a sanity check here?

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1 Answer 1

up vote 2 down vote accepted

There is a mistake in your second change of variable. You put the $t^{-1}$ inside the absolute value, and so you are taking a $p$ power of it when you shouldn't. Without that mistake, the exponent in your $C_p$ is $\alpha-1-1/p$, and you require $\alpha>1/p$ for $C_p$ to be finite.

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Yes! Thank you. :) –  breeden Jul 26 '12 at 4:14

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