For real function $f(x)=\frac{x^\beta}{1+\alpha x^\beta}$ with real $\alpha \ge 1$ and real $\beta>0$ I suppose that equation $$f(x) = sf(x+B)$$ for real $B$ with $0<B<1$ and real $s$ with $0<s\le 1$ has at most a single solution in $x$ within the interval $0\le x \le 1$. Any ideas, how this could be proven?
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1Consider $f(x+B)/f(x)$ and prove that it is strictly decreasing at $0< x \leq 1$ – Ivan Kaznacheyeu Jan 20 '22 at 11:54
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Moreover, interval can be extended to $0<x<\infty$ – Ivan Kaznacheyeu Jan 20 '22 at 11:57
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Great! That's it. Thanks a lot, Ivan. The derivative of $f(x+b)/f(x)$ is
$$\frac{\beta x^{-\beta - 1} (B + x)^{\beta - 1} (\alpha ( x^{\beta + 1}-(B + x)^{\beta + 1}) - B)}{(\alpha (B + x)^\beta + 1)^2}$$
See Wolfram alpha
All factors are always postive except $(\alpha ( x^{\beta + 1}-(B + x)^{\beta + 1}) - B)$ which is always negative given the restrictions on parameter values from above. Thus, $f(x+b)/f(x)$ is strictly decreasing and can intersect with constant $s$ at most once.
athanasius
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