Let $\mathcal{F}\subset C^\infty[0,1]$ be a uniformly bounded and equicontinuous family of smooth functions on $[0,1]$ such that $f'\in\mathcal{F}$ whenever $f\in\mathcal{F}$. Suppose that $$\sup\limits_{x\in[0,1]}|f'(x)-g'(x)|\leq\frac{1}{2}\sup\limits_{x\in[0,1]}|f(x)-g(x)|\mbox{ for all }f,g\in\mathcal{F}.$$ Show that there exists a sequence $f_n$ of functions in $\mathcal{F}$ that tends uniformly to $Ce^x$, for some real constant $C$.

Hint: Use the contraction principle. In order to apply the contraction principle you can use, without proof, the fact that if $X$ is a complete metric space, $A\subset X$ and $T\colon A\to X$ is uniformly continuous, then $T$ uniquely extends to a continuous map $\bar{T}\colon\bar{A}\to X$ defined on the closure $\bar{A}$.

So this clearly looks like a combination of Arzela-Ascoli theorem and Banach contraction principle.

The Arzela-Ascoli theorem says that a bounded and equicontinuous sequence of functions on a compact has a uniformly convergent subsequence, so we already get that $\mathcal{F}$ has a subsequence that convergences uniformly to something for free. Now it suffices to show that the limit is $Ce^x$.

$C^\infty[0,1]$ is a complete metric space and so $\bar{\mathcal{F}}$ (the closure of $\mathcal{F}$) will be a complete metric space too. Due to the given properties, the differentiation map $f\mapsto f'$ will be a contraction $T\colon\mathcal{F}\to\mathcal{F}$, so the differentiation $T\colon\bar{\mathcal{F}}\to\bar{\mathcal{F}}$ will have a unique fixed point by the contraction principle.

I have some issues with ending up the proof and writing it all down rigorously. The function $Ce^x$ is definitely not in $\mathcal{F}$, moreover, I'm not even sure it is in $\bar{\mathcal{F}}$. How am I supposed to use the contraction principle to show that this is indeed the limit of the uniformly convergent subsequence given by the Arzela-Ascoli theorem?

  • $\begingroup$ I'm missing something but I don't know what: the hypothesis says that $T : \overline{\mathcal{F}} \to \overline{\mathcal{F}}$ is Lipschitz with constant $1/2$, hence we have a unique fixed point $f_0 = Tf_0 = f_0'$. Hence $f_0 = Ce^x$ for some $C > 0$. Since $f_0 \in \overline{\mathcal{F}}$, we have a sequence in the family such that $f_n \to f_0$. Can you spot a mistake here? I should be using the equicontinuity and uniform boundedness somewhere. $\endgroup$
    – qualcuno
    Jun 23, 2019 at 13:03
  • $\begingroup$ $C^\infty[0,1]$ is not a complete metric space (in the sup distance). But $C[0,1]$ is. $\endgroup$ Jun 23, 2019 at 15:53

1 Answer 1


I finally came up with the appropriate solution, although without contraction principle (this hint seems to be misleading).

Pick any $f\in\mathcal{F}$ and define a sequence of function $f_n(x)=f^{(n)}(x)$ in $\mathcal{F}$. By Arzela-Ascoli, it has some convergent subsequence $\{f_{n_k}\}$ with the limit $h\in\mathcal{F}$. Note that $\{f'_{n_k}\}$ also converges uniformly to $h'$.

Therefore $\forall\varepsilon>0 ~\exists K\in\mathbb{N} ~\forall k\geq N$ holds $$\sup\limits_{x\in[0,1]}|f_{n_k}(x)-h(x)|<\frac{\varepsilon}{2}$$ $$\sup\limits_{x\in[0,1]}|f'_{n_k}(x)-h'(x)|<\frac{\varepsilon}{2}$$

Using triangle inequality and given inequality between functions and their derivatives we can show that $$\sup\limits_{x\in[0,1]}|h(x)-h'(x)|\leq\ldots<2\varepsilon+\left(\frac{1}{2}\right)^{n_k}\sup\limits_{x\in[0,1]}|f(x)-f'(x)|.$$

Therefore $h(x)=h'(x)=Ce^x$.


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