Let $f(x)$ be an increasing, strictly concave function with $f(0)=0$. Show that given $x<y$, $f(y+\epsilon )-f(x+\epsilon )<f(y)-f(x)$, where $\epsilon $ is a small, positive number.

inurbandojoa
2022-11-20
Answered

Let $f(x)$ be an increasing, strictly concave function with $f(0)=0$. Show that given $x<y$, $f(y+\epsilon )-f(x+\epsilon )<f(y)-f(x)$, where $\epsilon $ is a small, positive number.

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partatjar6t9

Answered 2022-11-21
Author has **8** answers

Your guess for a homotopy inverse is correct. A homotopy between the identity and $(x,t)\mapsto (d(s(x)),t)$ is given by the expression

$((x,t),u)\mapsto \{\begin{array}{ll}(x,t+u)& \text{if}t+u\le 1\\ (d(s(x)),t+u-1)& \text{if}t+u1\end{array}$

where $u\in [0,1]$ 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.

$((x,t),u)\mapsto \{\begin{array}{ll}(x,t+u)& \text{if}t+u\le 1\\ (d(s(x)),t+u-1)& \text{if}t+u1\end{array}$

where $u\in [0,1]$ 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.

evitagimm9h

Answered 2022-11-22
Author has **5** answers

Let $\alpha =\frac{\u03f5}{y-x}$, which is between $0$ and $1$ for small positive $\u03f5$

$\begin{array}{llll}f(y)& =& f((1-\alpha )(y+\u03f5)+\alpha (x+\u03f5))& >(1-\alpha )f(y+\u03f5)+\alpha f(x+\u03f5)\\ f(x+\u03f5)& =& f((1-\alpha )(x)+\alpha y)& >(1-\alpha )f(x)+\alpha f(y).\end{array}$

Sum up these $2$ inequalities, combine terms, and divide by $1-\alpha $, we get

$f(y)-f(x)>f(y+\u03f5)-f(x+\u03f5)$

$\begin{array}{llll}f(y)& =& f((1-\alpha )(y+\u03f5)+\alpha (x+\u03f5))& >(1-\alpha )f(y+\u03f5)+\alpha f(x+\u03f5)\\ f(x+\u03f5)& =& f((1-\alpha )(x)+\alpha y)& >(1-\alpha )f(x)+\alpha f(y).\end{array}$

Sum up these $2$ inequalities, combine terms, and divide by $1-\alpha $, we get

$f(y)-f(x)>f(y+\u03f5)-f(x+\u03f5)$

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