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I am particularly interested in the function $f(x)=e^{-1/x}$ (only over positive reals). This function is convex up to the inflection point at $x=1/2$, and then concave. So it has two convex parts, so to speak.

The function as a whole is not convex, so the Brouwer fixed point theorem does not hold overall for this function.

But can the Brouwer fixed point thm be applied "piecewise" to each of the convex parts? Such that we may conclude that the function $f(x)=e^{-1/x}$ has two fixed points, one in each of its convex parts?

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The Brouwer fixed point theorem doesn't ask that your function be convex, it asks that it's domain be compact/convex. The only constraint it places on the function is continuity. In fact, it generally is stated for functions from the unit ball of $\mathbb{R}^n$ to itself, where convexity of functions is ill-defined. However, your domain does not satisfy compactness. But luckily, since you're operating in $\mathbb{R}$ it suffices to look for where $e^{-\frac{1}{x}}$ intersects the 45-degree line. There is one fixed point (assuming $f(0)= 0)$, namely at zero.

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  • $\begingroup$ So, if I add a parameter $A$ such that $f(x;A)=Ae^{-1/x}$ or perhaps $f(x;A)=e^{-A/x}$, are there values of $A$ such that $f(x;A)$ crosses the 45 degree line and therefore has a fixed point, or perhaps 2 fixed points? (I guess I could just graph it to find out, but don't have a graph program at my fingertips just now.) $\endgroup$ – ben Jun 14 '17 at 14:08
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    $\begingroup$ Yep! A good easy online grapher is 'desmos.com/calculator' too. You can add parameter sliders etc. $\endgroup$ – Pete Caradonna Jun 14 '17 at 15:08
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Recall that $x$ is a fixed point of $f$ if $f(x)=x$. Since, in your example, $f(x)<x$ for all $x>0$, we see that its only fixed point is $x=0$. The reason is that BFPT applies if $f$ continuously maps a convex+compact set into itself. It is not true that $f([2/3,1])\subset [2/3,1]$, for example, even though $f$ is concave on $[2/3,1]$, so $BFPT$ might not apply to the "concave part" of $f$, so to speak.

However, since $0\le f(x)\le x$ for all $x\ge 0$, we see that $f([0,a])\subset[0,a]$ for all $a>0$, so BFPT applies here; and indeed, $x=0$ is a (the) fixed point.

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