Show that this metric space isn't complete Consider the space $X$ of all continuously differentiable functions from $[0,1]$ to $\mathbb{R}^{2}$, such that $f(0) = a$ and $f(1) = b$, where $a,b \in \mathbb{R}^{2}$ and $f \in X$.  Define the following metric on $\delta(f_1 , f_ 2 )  = {\rm sup} \lVert f_1 (t)-f_2 (t) \rVert$, where $t \in [0,1]$.
How do I show that this metric is not complete? I was thinking something along the lines of having an absolute value function on each component of $f$ so that it converges to a function which is not differentiable at a certain point of $[0,1]$. But, I'm not able to come up with any good examples. Any hints/suggestions would be very appreciated.
 A: I'm changing the domain to $[-1, 1]$. Deal with it.
For a given $\varepsilon > 0$, consider the function
$$f(x) = |x|^{1 + \varepsilon}.$$
This function is continuously differentiable, with derivative
$$f'(x) = (1 + \varepsilon)|x|^\varepsilon.$$
So, with this in mind, define
$$f_n(x) = |x|^{1 + 1/n}.$$
I claim that $f_n(x) \to |x|$ uniformly over the given interval. We have,
$$\left||x|^{1 + 1/n} - |x|\right| = |x| \cdot \left||x|^{1/n} - 1\right|.$$
When $|x| \le 1$, we have $1 \ge |x|^{1/n} \ge |x|$, so we get
\begin{align*}
\left||x|^{1 + 1/n} - |x|\right| &\le |x|\left(1 - |x|^{1/n}\right) \\
&= \frac{|x|(1 - |x|)}{1 + |x|^{1/n} + |x|^{2/n} + \ldots + |x|^{(n-1)/n}} \\
&\le \frac{|x|(1 - |x|)}{1 + (n - 1)|x|}.
\end{align*}
I claim that
$$\frac{|x|(1 - |x|)}{1 + (n - 1)|x|} \le \frac{1}{n},$$
or equivalently,
\begin{align*}
n|x|(1 - |x|) \le 1 + (n - 1)|x| &\iff n|x| - nx^2 \le 1 + n|x| - |x| \\
&\iff n|x| + nx^2 + 1 \ge |x|,
\end{align*}
which is pretty evidently true, since $1 \ge |x|$, and the other terms are positive. Thus,
$$\Big|f_n(x) - |x|\Big| \le \frac{1}{n},$$
which is to say, $f_n$ converges to the absolute value function uniformly, even though it is not continuously differentiable.
Fun fact: the Stone-Weierstrass theorem says that every continuous function on a compact interval can be approximated uniformly by polynomial functions. So, it would be possible (though harder) to simply choose $f_n$ from polynomials, which are infinitely-differentiable.
