TITLE: Why is there no $C>0$ such that $C||f||_q \leq ||f||_p$?
QUESTION [1 upvotes]: Show that for $1\leq p <q \leq \infty$, there is no $C>0$ such that
  $C||f||_q \leq ||f||_p$ for all $f\in C([0,1])$.
Definition:
Let $a,b\in \mathbb{R}$ with $a<b$ and for $f\in C([a,b])$, $1\leq p
 <\infty$ let
$$||f||_p=\left(\int_{a}^{b}|f(x)|^p \, dx\right)^{1/p} \text{and }  |
|f||_\infty := \max\limits_{x\in [a,b]}|f(x)|$$

There's a hint that tells me to compute $||f_n||_p$ and $||f_n||_q$ with  $f_n : [0,1]\to \mathbb{R} , t\mapsto f_n(t)=t^n$ mit $n\in \mathbb{N}$, what exactly do I have to compute? I don't see why this helps...

REPLY [2 votes]: The hint is saying that you can show that $C$ has to be $0$ (a contradiction) by assuming that $C||t^n||_q \le ||t^n||_p$ for all $n$. For example, if $q = \infty$, we have
$$ C||t^n||_{\infty} = C \le ||t^n||_p = \frac{1}{(1+pn)^{1/p}} \xrightarrow{n\to\infty} 0. $$
Therefore there is no $C > 0$ such that $C||t^n||_{\infty} < ||t^n||_p$ for all $n$.
Now do the same thing but with $q < \infty$.
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