# Fixed point of an integral equation

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)

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

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.