Consider $Tf(x) = \int_0^x e^{-f(s)^2} \; ds$ for $x \in [0,\infty)$. I want to use the contraction mapping theorem to show that $T:C^1([0,\infty)) \to C^1([0,\infty))$ has a unique fixed point.
From the Mean Value Theorem, $\frac{|e^{-x^2}-e^{-y^2}|}{|x-y|} = |-2c e^{-c^2}| \le \sqrt{2/e} < 1$ for some $c$ between $x$ and $y$.
I attempt to replicate this with functions: $\frac{|e^{-f(s)^2}-e^{-g(s)^2}|}{|f(s)-g(s)|} = |-2c_s e^{-c_s^2}| \le \sqrt{2/e}$, where $c_s$ is between $f(s)$ and $g(s)$ and varies with $s$. (Is this OK? I am fairly certain MVT is used, but I don't know if this "generalization" is correct.)
Then, we have $$ \|Tf-Tg\|_\infty \le \int_0^x | e^{-f(s)^2} - e^{-g(s)^2}| \; ds \le \int_0^x|f(s)-g(s)| \sqrt{2/e} \le \|f-g\|_\infty \sqrt{2/e} \cdot x $$
So if $x < \sqrt{e/2}$, then this is indeed a contraction. But I can't make any progress when $x \ge \sqrt{e/2}$... Would it be sufficient to say there is a solution for $T:C^1((0,\sqrt{e/2})) \to C^1((0,\sqrt{e/2}))?$ and since $C^1((0,\infty)) \subset C^1((0,\sqrt{e/2}))$ we're done?