Consider the following norms on $C[0,1]$: for each $p\in [1,+\infty]$ $||x||_p=(\int_0^1|x(t)|^pdt)^{1/p}$ (this $p$-norm is induced from $L^p$ wich contains $C[0,1]$). Are they equivalent? Of course, $||x||_p$ is not equivalent to $||x||_{\infty}$ (it is easy to build a counterexample), but I also believe that $p$-norm is not equivalent to $q$-norm (when $p\ne q$). Really?
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Given $n\ge 1$, let $x_n(t)=t^n$, $\forall ~t\in[0,1]$. By definition, $x_n\in C[0,1]$, and for every $p\ge 1$, $$\|x_n\|_p=\left(\frac{1}{1+np}\right)^{\frac{1}{p}}.$$ It follows that when $p>q\ge 1$, $$\lim_{n\to\infty}\frac{\|x_n\|_p}{\|x_n\|_q}=\infty,$$ i.e. $\|\cdot\|_p$ and $\|\cdot\|_q$ are not equivalent on $C[0,1]$.
Hu Zhengtang
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Look at $p=1$ and $p=2$. Define functions $$x_m(t) = \left\{ \begin{array}{ll} t^{-1/2} & \text{if }\ \frac 1m \le t \le 1, \\ \sqrt{m} & \text{if }\ 0 \le t \le \frac 1m. \end{array} \right.$$
Then $\|x_m\|_1 \le 2$ for all $m$, but $\|x_m\|_2 \to \infty$.
Umberto P.
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