I got stuck on the second part of the following question:

Let $R = \{f:[0,1]\to\mathbb{R}\mid f\text{ is bounded, integrable and } f(x)\geq 0\forall x\in[0,1]\}$ be a metric space with the metric $d(f,g) = \sup_{x\in[0,1]}|f(x)-g(x)|$. Define the function $\Phi_f:[0,1]\to \mathbb{R}$ by $$\Phi_f(x) = \int_0^x \frac{1}{1+f(s)}ds.$$ Show that this defines a Lipschits continuous map $$\Phi: R\to R.$$ Explain why $\Phi$ has a fixed point.

I managed to show that $\Phi$ is indeed Lipschitz continuous as follows (which I'm not sure is correct, so if I have any mistakes please let me know) $$ \sup|\Phi_f(x)-\Phi_g(x)|=\sup|\int_0^x \frac{1}{1+f(s)}-\frac{1}{1+g(s)}ds|\\ \leq\int_0^x|\frac{g(s)-f(s)}{1+f(s)+g(s)+f(s)g(s)}ds|\\ \leq (x-0)\sup|f(x)-g(x)|\\ \leq \sup|f(x)-g(x)| $$

But since he Lipschitz contstant is 1, I can't use the Banach Contraction Theorem to show that $\Phi$ has a fixed point (Note: we haven't proved the mean-value theorem nor did we define differentiation, so I can't use either in proving the result above)

  • $\begingroup$ What is $a$ here? $\endgroup$
    – Gregory
    Commented Apr 17, 2020 at 15:50
  • $\begingroup$ I think in the integral you should have $0$ where you wrote $a$. $\endgroup$
    – pooper
    Commented Apr 18, 2020 at 4:31
  • $\begingroup$ It was supposed to be 0, fixed it now $\endgroup$
    – 14159
    Commented Apr 19, 2020 at 7:20

1 Answer 1


First of all,observe that $R$ is a closed convex subset of the Banach space $B[0,1]$,which is the space of all bounded functions on $[0,1]$ with sup-norm.Now,consider the collection of all such maps $\phi_{f}$ where $f$ is in $R$.Name this collection $S$.Now,you can show that any such $\phi_{f}$ is a continuous map on $[0,1]$.Using the fact that $f\geq 0$ for all $f$ in $R$,this collection can be shown to be bounded and equicontinuous.Hence,by Arzela-Ascoli theorem,$S$ is pre-compact.Now,as you have shown,the map $f\rightarrow \phi_{f}$ is Lipschitz continuous,defined on a closed convex subset of a Banach space,and from what I have shown above,it's image is contained in a compact set.Hence,by Schauder fixed point theorem,it has a fixed point.I hope it's all right.


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