I don't understand this application of Holder's inequality

Question

I don't understand this inequality and which are the Holder's conjugates:

$$\int_0^{1} \Biggl( \frac{1}{x^\alpha} \, \int_0^{x} f(t) \,\, dt \Biggr)^p \, dx \leq \, \int_{0} ^{1} \frac {1} {x^{p\alpha}} \, \int_{0}^{x} f^p(t) \,dt\, \, x^{p-1} dx $$

where $\alpha, x \in (0,1)$ , $f \in L^p(0,1) \quad \forall \, p \in \, [1,\infty) $ .

Answer 1

What Serg says is pretty true. I will assume $f$ to be positive (need to think about general case).

Note that for $p=1$ the claim is trivially true.

Now, let $p\in (1,\infty)$ with Hölder conjugate $q \in (1,\infty)$, i.e. $\frac{1}{p}+\frac{1}{q}=1$.

Then we have by Hölder

\begin{align*} \int_0^x f(t) ~\mathrm{d}t &\leq \left(\int_0^x 1^q~\mathrm{d}t\right)^{\frac{1}{q}}\cdot \left(\int_0^x f^p~\mathrm{d}t\right)^{\frac{1}{p}}\\ &=x^{\frac{1}{q}}\cdot \left(\int_0^x f^p~\mathrm{d}t\right)^{\frac{1}{p}}. \end{align*}

Raising this to the power of $p$ we obtain the claim as $\frac{p}{q}=p-1$.