Let $I$ be bounded interval.Let $u \in W^{1,p}(I)$ for $1\le p\le \infty$ for any $q\in[1,\infty]$ prove that $W^{1,p} $norm are equivalent to $$|||u||| = \|u'\|_p + \|u\|_q$$
My attempt: first $\|u\|_{W^{1,p}} = (\|u\|_p^p+\|u'\|_p^{p})^{1/p}$ then it equivlalent to $\|u\|_p +\|u'\|_p$ by Holder inequality for $\ell_p$.
Secondly,we use Sobolev embedding $\|u\|_\infty\le C(\|u\|_p+\|u'\|_p)$. for any $q$,we have $\|u\|_q \le C\|u\|_\infty \le C'(\|u\|_p +\|u'\|_p)$ hence $|||u|||\le C(\|u\|_p +\|u'\|_p)$
Only need to prove the inequality $\|u\|_{W^{1,p}}\le C|||u|||$ then
To do this $$u(t)- u(t_0) = \int^t_{t_0}u'(t)dt$$
then $$|u(t)|-|u(t_0)|\le |\int^{t}_{t_0}u'(t)dt|\le \|u'\|_1$$
Hence $$|u(t)| \le \|u'\|_1 + |u(t_0)|$$ for all $t\in I$
So we have $\|u\|_\infty\le \|u'\|_1 + |u(t_0)|$
Finally integrate over $t_0$ we have $$|I|\|u\|_\infty \le |I|\|u'\|_1 + \|u\|_1$$
If $|I| \ne 0$ we can see $\|u\|_p \le C'\|u\|_\infty \le C(\|u\|_1 + \|u'\|_1)\le C|||u|||$
As daw's post shows we can only change the norm for zero order $u$ .
For $\|u'\|_q$ needs not to be bounded.
This is clearly false:
If $q>p$ then there are functions in $u\in W^{1,p}$ such that $u'\not\in L^q$, so $|||u|||=+\infty$, but $\|u\|_{W^{1,p}}<+\infty$. (To construct $u$ take $v\in L^q\setminus L^p$ and set $u(x)=\int_0^x v(s)ds$).
If $q<p$: Take $u\in W^{1,q} \setminus W^{1,p}$. Approximate $u$ by smooth functions $u_k \in C_c^\infty$ such that $u_k \to u$ in $W^{1,q}$. Then $\|u_k\|_{W^{1,q}} \to \|u\|_{W^{1,q}}$ buut $\|u_k\|_{W^{1,p}}\to\infty$.
What would work is if $|||\cdot|||$ would be defined as $\|u\|_{L^q} + \|u'\|_{L^p}$.