Which of the following metric spaces are complete?
(a) The space $C^1[0, 1]$ of continuously differentiable real-valued functions on $[0, 1]$ with the metric $d(f, g) = \max_{t∈[0,1]}|f(t) − g(t)|$.
(b) The space of all polynomials in a single variable with real coefficients, with the same metric as above.
(c) The space $C[0, 1]$ with the metric $d(f, g) =∫_0^1 |f(t) − g(t)| dt$.
I can only say that (c) is not complete but no idea about (b) and (c)
HINT for (a) and (b): Use the Weierstrass approximation theorem and the function $f(x)=\left|x-\frac12\right|$.
This image basically conveys what Brian is saying:
$f_1,f_2,f_3,...$ is a sequence of differentiable functions converging to $|x|$, but of course the limit $|x|$ is not differentiable at $0$. The convergence may be uniform.
(image credit: Michael Spivak, Calculus, p.470)
I assume that you know that a space $(X,d)$ is complete if $\{f_n\}_{n=1}^\infty \subset X$ and $d(f_n,f)\to 0$ as $n\to\infty$, then $f\in X$.
(a) Can you think of a sequence $\{f_n\}_{n=1}^\infty \subset C^1[0,1]$ and candidate limit function $f$ such that $d(f_n,f):=\max_{t\in[0,1]}|f_n(t)-f(t)|\to 0$ as $n\to \infty$ but $f\not\in C^1[0,1]$?
The key here is that it's $C^1[0,1]$, not $C[0,1]$.
(b) If you can make (a) fail using real polynomials in one variable, you kill two birds with one stone.
(c) Take a look at $f_n(t)=t^n$ and $$f(t)=\begin{cases} 0, &0\le t<1,\\ 1, &t=1.\end{cases}$$ Examine $\int_0^1 |f_n(t)-f(t)|\,dt$ as $n\to \infty$.
Hint: For $(a)$, note that, uniform convergence does not prserve differentiability.
