# In $C[(0,1)]$ if $p_1 <p_2$ then $\|{f}\|_{p_1} < \|{f}\|_{p_2}$

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 $\frac{p_2}{p_2-p_1}$:
Taking the $p_1$-th root gives $\|f\|_{p_1} \le \|f\|_{p_2}$.
• 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! – tremblay Sep 16 '17 at 18:24