Continuity of differentiation

Note that differentiation is linear, so the continuity of differentiation is equivalent to continuity at 0 (here, the 0 of the vector space $C^{\infty}([0,1],\R)$, that is, the constant function equal to 0). In other words: for every $\varepsilon >0$, there would exist a $\delta >0$, such that for every function $f:[0,1]\rightarrow \R$ of class $C^{\infty}$ satisfying $\|f\|\leq \delta$, we would have $\|f'\|\leq\varepsilon$.

However, this is false for any norm. To see this, it is enough to find functions such that $\|f'\|$ is arbitrarily larger than $\|f\|$. Natural candidates are the functions $f_{\lambda}(x)=e^{\lambda x}$. Indeed, we have:
$$||f_{\lambda}'||=||\lambda e^{\lambda x}||=|\lambda| \|e^{\lambda x}\|=|\lambda|\|f_{\lambda}\|.$$

Now, if we take $\varepsilon>0$, and fix $\delta >0$, the problem with the functions $f_{\lambda}(x)=e^{\lambda x}$ is that their norm is not necessarily less than or equal to $\delta$.
We can therefore simply normalize them and multiply by $\delta$ so that this is the case: we define
$$g_{\lambda}(x)=\frac{\delta e^{\lambda x}}{||e^{\lambda x}||}$$

and automatically their norm is exactly $\delta$; moreover, we still have the condition $\|g_{\lambda}'\|=|\lambda|\|g_{\lambda}\|$.
By taking $\lambda>\varepsilon/\delta$, we therefore obtain $\|g_{\lambda}\|\leq\delta$ but $\|g_{\lambda}'\|>\varepsilon$, which contradicts continuity.