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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?

user153582
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    another point of view: $C^1[a,b] \subset C^0[a,b]$ is dense in $C^0[a,b]$ (Stone weiestrauss). If $C^1[a,b]$ is also complete with sup norm, then $C^1[a,b] = C^0[a,b]$, which is not true. –  Nov 12 '15 at 14:46
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    For yet another point of view, you might consider giving $C^1([a, b])$ the inner product $\langle {f, g}\rangle = \int_a^b fg$ and considering the Riesz representation theorem. – anomaly Nov 12 '15 at 14:50
  • I believe the metric that you are looking for is defined as follows: $d(f, g) := \displaystyle \sup_{x \in [a, b]} |f(x) - g(x)| + \displaystyle \sup_{x \in [a, b]} |f'(x) - g'(x)|, \forall f, g \in C^1([a, b])$ – Newton Apr 05 '23 at 00:31

2 Answers2

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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.

Tomasz Kania
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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.

gerw
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