We are supposed to show/disprove that two norms are equivalent on $C^1([0,1])$ by finding two constants $c,C>0$:
$$c\| f\|_\infty\le\| f\|_{C^1}\le C\| f\|_\infty$$
$$\| f\|_\infty=\max |f(x)|$$and
$$\| f\|_{C^1}=\|f\|_\infty+ \|f'\|_\infty$$
are the norms.
What changes if we limit ourselves to polynomials of some finite maximum degree $n\in \mathbb{N}$?
I'm pretty sure that the norms are not equivalent on all $C^1([0,1])$ functions (infinite-dimensional) but I'm having some difficulty finding an example. Do any come to mind?
On the polynomials they must be equivalent (finite-dimensional), but what are the constants? They will probably depend on $n$, but playing around with different upper bounds or the trusted triangle inequality I can't make progress.
Best Answer
The norms are not equivalent. The trick is to find a sequence which blows up in one norm while remaining small in the other norm. Here, the sequence $f_n(x) = x^n$, for $x \in [0,1]$ works. For all $n \ge 1$, we see that $$\|f_n\|_\infty = 1, \,\,\,\, \text{ but } \,\,\,\, \|f_n\|_{C^1} = 1 + n.$$ This shows that you will never have a constant $C > 0$, such that $\|f\|_{C^1} \le C \|f \|_\infty$ for all $f \in C^1[0,1]$.
To address your last comment: yes, the norms are equivalent on any set $\mathbb P_N[0,1]$ - the set of polynomials of degree at most $N$ - for exactly the reason you stated. However, the set $\mathbb P[0,1]$ of polynomials of arbitrary degree is not finite dimensional, and this shows that the norms are not equivalent even when restricted to $\mathbb P[0,1]$.