# Let f(x) be an increasing, strictly concave function with f(0)=0. I have to show that given x<y, f(y+epsilon)−f(x+epsilon)<f(y)−f(x), where epsilon is a small, positive number.

Let $f\left(x\right)$ be an increasing, strictly concave function with $f\left(0\right)=0$. Show that given $x, $f\left(y+\epsilon \right)-f\left(x+\epsilon \right), where $\epsilon$ is a small, positive number.
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partatjar6t9
Your guess for a homotopy inverse is correct. A homotopy between the identity and $\left(x,t\right)↦\left(d\left(s\left(x\right)\right),t\right)$ is given by the expression

where $u\in \left[0,1\right]$ is the parameter of the homotopy. Geometrically this shifts a point in the mapping torus of d∘s towards the right a distance of u units.
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evitagimm9h
Let $\alpha =\frac{ϵ}{y-x}$, which is between $0$ and $1$ for small positive $ϵ$
$\begin{array}{llll}f\left(y\right)& =& f\left(\left(1-\alpha \right)\left(y+ϵ\right)+\alpha \left(x+ϵ\right)\right)& >\left(1-\alpha \right)f\left(y+ϵ\right)+\alpha f\left(x+ϵ\right)\\ f\left(x+ϵ\right)& =& f\left(\left(1-\alpha \right)\left(x\right)+\alpha y\right)& >\left(1-\alpha \right)f\left(x\right)+\alpha f\left(y\right).\end{array}$
Sum up these $2$ inequalities, combine terms, and divide by $1-\alpha$, we get
$f\left(y\right)-f\left(x\right)>f\left(y+ϵ\right)-f\left(x+ϵ\right)$