Let $C[0,1]$ be the set of continous function on $[0,1]$, and $C^1[0,1]\subset C[0,1]$ be the subset of continuously differentiable functions on $[0,1]$. Equip $C[0,1]$ with the usual supremum metric. Define $\Phi: C^1[0,1]\to C[0,1]$ by $\Phi(f)=f'$. Is $\Phi$ continuous?
My thought is, as $d(f,g)\to 0$, $f$ and $g$ will eventually be the same, so $d(f',g')\to 0$ as well. So $\Phi$ should be continuous. But I could not prove it with the $\delta-\epsilon$ approach. Can anyone give a hint on the proof or a counterexample?
Consider the sequence of functions $f_n:[0,1]\to\mathbb{R}$ with $f_n(x)=\frac{x^n}{n}$. Then $(f_n)\subset C^1([0,1])$ and $\|f_n\|_\infty=\max_{x\in[0,1]}|f_n(x)|=\frac{1}{n}\to 0$. Hence $(f_n)$ converges to the constant-function $0$ in $C^{1}([0,1])$ with the $\sup$ norm. But $\Phi(f_n)(x)$=$f_n'(x)$=$x^{n-1}$. What is $\Phi(0)$ and does $(\Phi(f_n))_{n=1}^{\infty}$ converge to it in the $\sup$ norm?