# Maximum uniqueness

Consider the function $g:\left(0,1\right)\rightarrow\mathbb{R}$ defined by $$g\left(x\right)=\left(1-x\right)\left(1-\frac{1}{1+f\left(x\right)}\right),$$ where $f\left(x\right)$ is a continuously differentiable function that is positive and strictly increasing with $x\in\left(0,1\right)$. Can one claim that if $g$ has a maximum, this maximum is unique?

Paul

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For example if $$f(x) = 0.1 + x \text { if } 0 \le x \lt 0.1$$ $$f(x) = 1.15 + x \text { if } 0.1 \le x \lt 0.4$$ $$f(x) = 4.6 + x \text { if } 0.4 \le x \le 1$$ then I think you will find there are maxima at $x = 0.1 \text{ and } 0.4$ when $f(x) = 1.25 \text{ and } 5$ respectively and $g(x)=0.5$.

It would not be difficult to make $f(x)$ continuous.

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Thanks for all answers. I forgot to mention that $f$ is continuously differentiable. –  Paul Smith Jul 27 '11 at 22:53
Taking my example, any continuous differentiable function $f(x)$ passing through the points $(0,0.1), (0.1,1.25), (0.25,1.4), (0.4,5), \text{ and } (1,5.6)$ will produce at least two local maxima for $g(x)$, since it must go up then down then up and then down. With some further manipulation it can be made to have at least two global maxima. –  Henry Jul 27 '11 at 22:58
@Paul: When adding conditions to the question please be sure to edit them into the question itself so everyone can see it. –  Asaf Karagila Jul 27 '11 at 23:04

How about $$f(x) = x + \frac{{1 - \cos (100x)}}{{100}}.$$ Note that $f$ is strictly increasing on $(0,1)$, since $f'(x) = 1+\sin(100x)$. In turn, since $f(0+)=0$, $f$ is positive on $(0,1)$. However, $g$ seems to have quite many local maxima (and minima).

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As numerical results show, there exists a constant $a \approx 17.68909557710036$ for which $g$ has two global maxima, where $$f(x) = x + \frac{{1 - \cos (ax)}}{a}.$$ As in the answer above, $f$ is strictly increasing and positive on $(0,1)$. –  Shai Covo Jul 28 '11 at 2:51
(If you are satisfied with local maxima, then you can be satisfied with the original example.) –  Shai Covo Jul 28 '11 at 2:58
I thank you all for your helpful answers. –  Paul Smith Jul 28 '11 at 11:22