I am trying to prove that for every $1<p_1 <p_2<\infty$ we have $\|{f}\|_{p_1} < \|{f}\|_{p_2}$ where

$$\|f\|_p = \left(\int_0^1 |f(x)|^pdx\right)^{\frac{1}{p}}$$

I have the intuition that I should use Hölder inequality which states that for all $p,q>0$ with $\frac{1}{p}+\frac{1}{q}=1$ we have

$$\left|\int_0^1f(x)g(x)\right|\leq \left(\int_0^1 |f(x)|^pdx\right)^{\frac{1}{p}}\left(\int_0^1 |g(x)|^qdx\right)^{\frac{1}{q}}$$

This would mean I need to chose a clever $p$ and $q$ which are in fonction of $p_1$ and $p_2$. I've tried some of them, but I can't seem to find some that work.


You can use the Hölder inequality for $\frac{p_2}{p_1}$ and $\frac{p_2}{p_2-p_1}$:

\begin{align}\|f\|_{p_1}^{p_1} &= \int_{[0,1]}|f|^{p_1} \\ &= \int_{[0,1]}|f|^{p_1} \cdot 1 \\ &\stackrel{\text{Hölder}}{\leq} \left(\int_{[0,1]}\big(|f|^{p_1}\big)^\frac{p_2}{p_1}\right)^\frac{p_1}{p_2}\underbrace{\left(\int_{[0,1]} 1^\frac{p_2}{p_2-p_1}\right)^{1 -\frac{p_1}{p_2}}}_{=1} \\ &= \left(\int_{[0,1]}|f|^{p_2}\right)^\frac{p_1}{p_2} \\ &= \|f\|_{p_2}^{p_1} \end{align}

Taking the $p_1$-th root gives $\|f\|_{p_1} \le \|f\|_{p_2}$.

  • $\begingroup$ Wow, yes that totally works. When I think about it now, I totally see how I could've guessed it, with $p_1$ being smaller than $p_2$ and the fact that $p$ had to be greater than 1, but I had no idea before. Thank you! $\endgroup$ – tremblay Sep 16 '17 at 18:24

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