Let $T: C^{1} \rightarrow C^{0}$ be the operator such that $T(f)(x) = f^{1}(x) $, is this operator uniformly continuous? $C^{1}$ is the space of continuously differentiable functions and $C^{0}$ the space of continuous functions, both defined on the real interval $[0,1]$. Both spaces equipped with $d(f,g) = \sup_{x \in [0,1]|f(x) - g(x)|}$ as metric.
$f^{1}(x)$ is the first derivative of f.
Now, is this operator uniformly continuous? It seems so but going about this demonstration became too confusing for me. Could you guys give me any hints, please?
 A: The continuity of the linear map
$f \mapsto f': C^1[0, 1] \to C[0, 1] \tag 1$
depends on the norms placed on $C^1[0, 1]$ and $C[0, 1]$.  If, as is typical, we define
$\Vert \cdot \Vert _1:  C^1[0, 1] \to \Bbb R \tag 2$
via
$\Vert f \Vert_1 = \displaystyle \sup_{x \in [0, 1]} \vert f(x) \vert + \sup_{x \in [0, 1]} \vert f'(x) \vert, \tag 3$
and
$\Vert \cdot \Vert_0:  C[0, 1] \to \Bbb R \tag 4$
by
$\Vert f \Vert_0 = \displaystyle \sup_{x \in [0, 1]} \vert f(x) \vert, \tag 5$
then the mapping (1) is in fact continuous, since it is a bounded linear map:
$\Vert f' \Vert_0 = \displaystyle \sup_{x \in [0, 1]} \vert f'(x) \vert \le \displaystyle \sup_{x \in [0w, 1]} \vert f(x) \vert + \sup_{x \in [0, 1]} \vert f'(x) \vert = \Vert f \Vert_1. \tag 6$
However, if we use the metric
$d(f, g) = \displaystyle \sup_{x \in [0, 1]} \vert f(x) - g(x) \vert \tag 7$
which is in fact the $\Vert \cdot \Vert_0$ norm, 
$d(f, g) = \Vert f - g \Vert_0, \tag 8$
on both $C[0, 1]$ and $C^1[0, 1]$, then $f \to f'$ is not continuous, since a function may be $\Vert \cdot \Vert$-small yet have an arbitrarily large derivative, for example
$f_k(x) = a\sin (kx) \tag 9$
is bounded in absolute value by $\vert a \vert$, but
$\vert f_k'(x) \vert = \vert ak\cos(kx) \vert \tag{10}$
becomes arbitrarily large as $k \to \infty$.
A: This function is not continuous. For you can take any function on the real line and rescale the parameter ($f(x) \mapsto f(rx)$ for large $r$) to make the derivative large without affecting the $C^0$ norm of the function. 
For a specific example, consider $f_n(x) = n^{-1} \sin (n^2x)$. As $n$ gets large, $f_n(x)$ goes to $0$ in $C^1(0)$ with the uniform norm, but $\frac{d}{dx}f_n(x) = n \cos(n^2x)$ diverges. 
A: I think this might be a bit long for a comment, so I post as an answer.
In reply to your intuition about continuity, linearity “simplifies” continuity in that you only have to check continuity at 0 to determine continuity on the whole (vector) space. This is because
$$\|T(f) - T(g)\| = \|T(f-g)\| = \|T(h)\|$$
since we can replace $f-g$ by any other vector $h$.
From that it's easy to check that continuity is equivalent to boundedness: there is a constant $M > 0$ such that
$$ \|T(f)\| \leq M\| f \|$$
for all $f$.
So to show that a (linear) operator between normed spaces is continuous, all you have to do is estimated a suitable $M$. On the other hand, to show discontinuity it is enough to pick a sequence of functions $(f_n)_n$ that violate the inequality, say
$$ \| T(f_n) \| \geq n \| f_n \| $$
(You need infinitely many: if you only had a finite number, then you could pick the biggest norm and you'd be bounded again.)
In general, derivative operators are not bounded on $C^1([0,1])$: just think of polynomials $(x^n)_n$. 
