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If the space $C^1 [0,1]$ is equiped with norm
$$ \Vert f\Vert_1=\sup_{t\in [0,1]}|f'(t)|$$ Is it complete?

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Is $\lVert f'\rVert_C=\sup_{t\in [0,1]}|f'(t)|$? – Davide Giraudo Sep 16 '12 at 13:37
yes you are right. – 89085731 Sep 16 '12 at 13:42
So we assume there is a left derivative at $1$ and a right derivative at $0$, right? – Davide Giraudo Sep 16 '12 at 13:42
This norm is a norm on $\mathbb{R} \times C^0[0, 1]$ both of which are known to be Banach. I think that's enough? – Karolis Juodelė Sep 16 '12 at 13:51
up vote 1 down vote accepted

Take $\{f_n\}$ a Cauchy sequence for $\lVert\cdot\rVert_1$. As $C[0,1]$ endowed with the supremum norm is complete, and $\{f'_n\}$ is Cauchy in $C[0,1]$ with this norm, we can find $g\in C[0,1]$ such that $f'_n$ converges to $g$ uniformly on $[0,1]$.

The sequence of real numbers $\{f_n(0)\}$ is Cauchy, hence it converges to a real number $l$. We can define $f(x):=l+\int_0^xg(t)dt$. It's a $C^1$ function, since it's a primitive a continuous function, and we have left/right derivatives at $1$/$0$. We check that $f_n\to g$ in the norm $\lVert\cdot\rVert_1$. We have $f(0)=l=\lim_{n\to +\infty}f_n(0)$ and $f'(x)=g(x)$ which is the uniform limit of $\{f'_n\}$ on $[0,1]$.

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Its not defined a norm on $C^1[0,1]$ since, for example $\|1\|=0.$

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