Let $(X,d)$ be metric space with $d(f,g)=\sup |f(x)-g(x)|$ where $X$ is the set of continuous function on $[0,1/2]$. Show $\Phi:X\rightarrow X$ $$\Phi(f)(x)=\int_0^x \frac{1}{1+f(t)^2}dt$$ has a unique fixed point $f(0)=0$.

b) Show that it satisfies $\frac{df}{dx}=\frac{1}{1+f(x)^2}$

My attempt:

I assumed $[0,1/2]$ is complete. I need to show that:

$$d(\Phi(f)(x),\Phi(g)(x))=\sup \Big|\int_0^x \frac{1}{1+f(t)^2}dt-\int_0^x \frac{1}{1+g(t)^2}dt\Big|\leq\alpha d(f,g)$$


$$\sup \Big|\int_0^x \frac{1}{1+f(t)^2}dt-\int_0^x \frac{1}{1+g(t)^2}dt\Big|=\sup \Big|\int_0^x \frac{1}{1+f(t)^2+g(t)^2}dt\Big|$$

Now ideally I need to somehow change this integral to include $d(f,g)=\sup|f(x)-g(x)|$, but I don't know how.

Then I would integrate, and get the integration constant to be $<1$, which will be my contraction constant.


A useful approach for these sorts of estimates is to use the mean value theorem.

(a) Let $g(x) = {1 \over 1+x^2}$, then $g'(x) = -{2x \over (1+x^2)^2 }$ and $g''(x) = 2 { 3 x^2 -1 \over (1+x^2)^3 }$. It is not hard to show that $|g'|$ has a maximum of $L={3 \sqrt{3} \over 8} <1 $ at $x = \pm { 1\over \sqrt{3}}$.

Hence $|g(x)-g(y)| \le L |x-y|$ for all $x,y$.

Then \begin{eqnarray} |\Phi(f_1)(x)-\Phi(f_2)(x)| &\le& \int_0^x |{1 \over 1+f_1(t)^2} - {1 \over 1+f_2(t)^2} | dt \\ &\le& \int_0^x L|f_1(t)-f_2(t)| dt \\ &\le& \int_0^x L \|f_1 - f_2\|_\infty dt \\ &\le& { 1\over 2} L \|f_1 - f_2\|_\infty \end{eqnarray}

It follows that $\Phi$ has a unique fixed point $\hat{f}$, and it is clear that $\hat{f}(0) = (\Phi(\hat{f})(0) = \int_0^0 {1 \over 1+\hat{f}(t)^2} dt = 0$.

(b) follows from the fundamental theorem of calculus.

  • $\begingroup$ I had another read, thanks it's very clear. Do we always use $\infty$ metrics to solve such mappings? Do you know any other interesting example of fixed points? $\endgroup$ – GRS May 9 '16 at 22:38
  • $\begingroup$ This is a popular application as it is provides a nice proof of the existence and uniqueness of solutions to ODEs. A nice extension can be used to show continuous dependence on initial conditions and other parameters. The $\sup$ norm is typical in these applications. $\endgroup$ – copper.hat May 9 '16 at 23:20

Your final equality is incorrect. But more generally, a hint that might prove helpful:

$$\bigg\| \int_0^x h(t) dt \bigg\|_\infty \le \int_0^x \|h(t)\|_\infty$$

Try and use that to get the $d(f,g)$ you're looking for!


I'm unable to comment due to low reputation, so I'll have to use this instead.

Have you taken a look at Contraction Mapping question?


Of course $[0,1/2]$ is complete, since it is a closed subset of $\mathbb R$. But what is needed for the problem is that $X=C([0,1/2])$ is complete, which is also true.

Your last equality is incorrect. Use the following inequality: $$ \Bigl|\frac{1}{1+f^2}-\frac{1}{1+g^2}\Bigr|=\frac{|f^2-g^2|}{(1+f^2)(1+g^2)}\le\frac{|f|+|g|}{(1+f^2)(1+g^2)}\,|f-g|. $$ Since $$ \sup_{t>0}\frac{t}{1+t^2}=\frac12, $$ we see that $$ \Bigl|\frac{1}{1+f^2}-\frac{1}{1+g^2}\Bigr|\le d(f,g). $$

  • $\begingroup$ What happens to $\frac{|f|+|g|}{(1+f^2)(1+g^2)}$? If I undestand correctly, $|f-g|$ is the $d(f,g)$ term. I can split $\frac{|f|+|g|}{(1+f^2)(1+g^2)}=\frac{|f|}{(1+f^2)(1+g^2)}+\frac{|g|}{(1+f^2)(1+g^2)}$, but then it's still not in the $\frac{1}{1+t^2}$ form $\endgroup$ – GRS May 9 '16 at 16:59
  • $\begingroup$ $$\frac{|f|}{(1+f^2)(1+g^2)}\le\frac{|f|}{1+f^2}\le\frac12$$ $\endgroup$ – Julián Aguirre May 9 '16 at 19:38

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