In $C[(0,1)]$ if $p_1 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 $\|f\|_{p_1} \le \|f\|_{p_2}$.
