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Let X be the set of continuous real valued functions defined on $[0,\frac{1}{2}]$ with the metric $d(f,g):=\sup_{x\in[0,\frac{1}{2}]} |f(x)-g(x)|$.

Define the map $\theta:X\rightarrow X$ such that $$\theta (f)(x)=\int_{0}^{x} \frac{1}{1+f(t)^2} dt$$.

I need to show that $\theta$ is a contraction mapping and that the unique fixed point satisfies $f(0)=0$ and the differential equation $\frac{df}{dx}=\frac{1}{1+f(x)^2}$

So I'm pretty lost on this, I'm quite comfortable proving that things like $f(x)=1+\frac{1}{1+x^4}$ are contraction mapping but I'm a bit confused with this, so

$$d(\theta(f),\theta(g))=\sup|\int_{0}^{x} \frac{1}{1+f(t)^2}-\frac{1}{1+g(t)^2} dt|$$

and so this is: $$=\sup|\int_{0}^{x} \frac{(g(t)-f(t))(g(t)+f(t))}{(1+f(t)^2)(1+g(t)^2)} dt|$$

but I am unsure where to go from here, I know that I need to get this to be something like:

$$\leq\alpha\sup|f(x)-g(x)|$$ where $\alpha$ is the contraction constant?

Thanks very much for any help

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up vote 4 down vote accepted

you can show that the function $g(a,b)={{a+b}\over{(1+a^2)(1+b^2)}}$ has a maximum at $a=b=1/3$ with maximum value of $27/50$. The last integral is less than $ \sup \int_0^x 27/50 (g(t)-f(t)) dt$. Leading to the result.

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In fact, Picard-Lindelof Theorem tells you that what you really need is $H(x)=\dfrac{1}{1+x^2}$ being Lipshitz continuous in a neighbourhood of $0$.

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