Show whether $N(g)=\|g\|_\infty + \|g'\|_\infty$ and $\|\cdot\|_\infty$ are equivalent on $C[0,1]$ Let $L$ be the linear subspace of $C[0,1]$ is the space of continously differentiable functions. 
I know I've got to show whether there exists an $a,b>0$ such that:
$$a\|x\|_\infty \leq \|x\|_\infty + \|x'\|_\infty \mbox{ for all } g\in L$$
$$\iff (a-1)\|x\|_\infty \leq \|x'\|_\infty$$
So i'm looking for a function which has an differential which tends to infinity for some $t_0$ but has a finite supremum. 
Is this function acceptable?
$$f_n(x)=\left\{\begin{array}{ll} 0 & \mbox{ for } 0\leq x \leq \frac{1}{2}-\frac{1}{n} \\ \frac{nx}{2} & \mbox{ for } \frac{1}{2}-\frac{1}{n} \leq x \leq \frac{1}{2}+\frac{1}{n} \\ 1 & \mbox{ otherwise}\end{array}\right.$$
 A: Note that trivially
$$\|g\|_\infty \le \|g\|_\infty + \underbrace{\|g'\|_\infty}_{\ge 0}$$
You must now see if there is some constant $C$ such that
$$\|g\|_\infty + \|g'\|_\infty \le C\|g\|_\infty$$
Note that on $C[0,1]$, $g'$ need not even exist, let alone be continuous. This leads to a very simple counter-example: $g(x) = \sqrt x$ with $\|g'\|_\infty = \|\frac1{2\sqrt \cdot}\|_\infty = \infty$

EDITED QUESTION
For $C^1[0,1]$ (i.e. $g'$ is continuous as well) we can see that The spline functions defined by
$$g_n(x) = \cases{0 & $x<\frac12 - \frac1n$ \\
\ast & $x\in \frac12 + [-\frac1n, \frac1n]$\\
1 & $x > \frac12 + \frac1n$}$$
Where $\ast$ is the $C^1$ (degree $2$) spline interpolation are all $C^1$ and $\|g_n\|_\infty = 1$ but
$$\|g_n'\|_\infty \ge \frac n2 \qquad \text{by MVT}$$
So there cannot be a bounding constant on $C^1$ either.
A: Let us take $f_n(x) = x^n\in C^1[0,1]$. Notice that since these are polinomial functions, they are bounded and $\lVert f_n \rVert_\infty =1$. But we have that $f_n' = nx^{n-1}$ and therefore $\lVert f_n' \rVert_\infty = n$ so the norms can't be equivalent since there is no $c>0$ such that $N(f_n) = n+1 \leq c\lVert f_n\rVert_\infty = c$ for all $n$.
Also, since all $f_n$ are analytic then this counterexample works for $C^{\infty}[0,1]$.
