Continuity of a map between two metric spaces Consider the map $D:C^{1}([-1,1]) \rightarrow \mathbb{R}:f \mapsto f'(0)$, where $C^{1}([-1,1])$ is the vector space of functions from $[-1,1]$ to $\mathbb{R}$ that are at least once differentiable and have a continuous derivative. Also consider the $d_{\infty}$-metric on $C^{1}([-1,1])$ and the Euclidian metric on $\mathbb{R}$, where $d_{\infty}(f,g)= \sup \{ |f(x)-g(x) |$ where $x \in C^{1}([-1,1]) \}$.
What can we say about the continuity of this function? It holds that $D$ is continuous if and only if for every $f \in C^{1}([-1,1])$ and $\epsilon >0$ there exists a $\delta >0$ so that for every $g \in C^{1}([-1,1])$, $|f'(0)-g'(0)| < \epsilon$ if $\sup \{ |f(x)-g(x) |$ where $x \in C^{1}([-1,1]) \} < \delta$. Now I don't really see how to deduce anything on continuity. Can someone help?
 A: Consider $$f_n(x) = \frac1n \sin(nx)$$
We have $f_n \to 0$ in the space, but $D(f_n) = 1 \to 1 \neq D(0)$. 
This shows that $D$ is not continuous.
A: A little hint: imagine a sequence of periodic functions whose amplitude decreases but frequency increases as $n \to \infty$. 
A: the mapping $f \to f'(0)$ is the composite $E_0 \circ D$ where:
$$
D:C^{1}([-1,1]) \to C^0([-1,1])
$$
is the differentiation operator and
$$
E_0:C^0([-1,1]) \to \mathbb{R}
$$
is the evaluation at zero.
both maps are linear, and with the $d_{\infty}$ metric on its domain, $E_0$ is continuous. $D$ is an unbounded operator - it maps a unit ball round the origin in the domain to an unbounded set in the codomain. the examples given in previous answers illustrate how this can occur. Intuitively, for an $f \in C^{1}([-1,1])$ there is a $g=f' \in C^0([-1,1]) $ for which
$$
\forall x \in [-1,1] \quad f(x) = \int_{-1}^x g(t)dt
$$
the condition:
$$
\bigg|\int_{-1}^x g(t)dt \bigg| \le 1
$$
does not impose any upper bound on the value of $|g(x)|$ locally
