Arzela-Ascoli theorem and the Banach contraction principle

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?

• 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. – Guido A. Jun 23 at 13:03
• $C^\infty[0,1]$ is not a complete metric space (in the sup distance). But $C[0,1]$ is. – Nate Eldredge Jun 23 at 15:53

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