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I have the following problem:


Let $X$ denote the collection of all differentiable continuous functions $f : [0, 1] \rightarrow \Bbb R$ such that $f(0) = 0$ and $f'$ is continuous. For $f, g \in X$, let $$\rho(f, g) = \sup_{x \in [0, 1]} \left | f'(x) - g'(x) \right |.$$

Show that $\rho$ is a metric on $X$ and that $(X, \rho)$ is complete.


I'm trying to show that $(X, \rho)$ is complete, but I'm stuck at showing that

given $\{f_n\}$ is a Cauchy sequence, then by Cauchy criterion for uniform convergence $f_n$ converges uniformly to some function $f : [0, 1] \rightarrow \Bbb R$ (and since $f_n$ is continuous, then so is $f$),

we get

$$f \in X \text{ and }\rho (f_n, f) = \sup_{x \in [0, 1]} \left | f_n'(x) - f'(x) \right | \rightarrow 0$$

Uniform convergence of $f_n$ to $f$ doesn't imply uniform convergence of $f'_n$ to $f'$...

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If $\{f_n\}$ is Cauchy in the metric of $X$, you have that the sequence $\{f_n'\}$ is uniformly Cauchy, hence uniformly convergent. Since each $f_n'$ is continuous, there exists a continuous function $g : [0,1] \to \mathbb R$ so that $f_n' \to g$ uniformly on $[0,1]$. Let $f$ be the antiderivative of $g$ in $X$: $$ f(x) = \int_0^x g(t) \, dt.$$ Thus $f' = g$ and thus $$\rho(f_n,f) = \sup_{x \in [0,1]} |f_n'(x) - g(x)| \to 0$$ as $n \to \infty$.

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  • $\begingroup$ Could you explain why $f'_n$ is uniformly Cauchy? $\endgroup$ – Katestrophical Apr 29 '14 at 19:36
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    $\begingroup$ That is how the metric is defined. If $\{f_n\}$ is Cauchy in the metric of $X$, then $$ \lim_{m,n \to \infty} \rho(f_n,f_m) = 0$$ which is equivalent to $$\lim_{m,n \to \infty} \sup_{x \in [0,1]} |f_n'(x) - f_m'(x)| = 0.$$ Thus the Cauchy property is satisfied uniformly across $[0,1]$. $\endgroup$ – Umberto P. Apr 29 '14 at 19:41

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