Is $C^1([a,b])$ with sup norm a complete metric space?

Question

Give $C^1([a,b])$ the sup metric induced from $(C^0([a,b]),||.||)$. I want to know if $C^1([a,b])$ is complete. My thinking is to show that $C^1([a,b])$ is a closed subset of $C^0([a,b])$.

Let $(f_n)$ be a sequence of $C^1([a,b])$ functions where $f_n\to f$ (sup-norm convergence). Since differentiability implies continuity, $(f_n)\in C^0([a,b])$, where sup convergence is the same as uniform convergence. The sequence $(f'_n)$ is also in $C^0([a,b])$.

I know that If $f_n$ is a sequence of differentiable functions with $f_n\to f$ uniformly and $f'_n\to g$ uniformly, then $f$ is differentiable and $f=g$.

So if $(f'_n)$ converges, it converges to $f'$. Since $(f'_n)\in C^0([a,b])$, $f'\in C^0([a,b])$, so $f\in C^1([a,b])$. So if I can show $f'_n$ converges in $C^0[a,b]$, then $C^1([a,b])$ is closed and the proof is done. But this only works if must $(f'_n)$ converge. Is this true, and if so why?

Answer 1

No, it is not true. Consider $$f_n(x) = |x|^{1+1/n}$$ on $(-1,1)$. You cannot control the derivatives with the sup-norm of the functions.

Answer 2

Take $f_n(x)=\frac{1}{\sqrt{n}}\sin(nx)$. Then $f_n\to 0$ uniformly but the sequence $(f_n^\prime)_{n=1}^\infty$ is unbounded.