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Whenever we are given real numbers $v_1<\cdots<v_n$ and $\mu\in(v_1,v_n),$ is it true there are always $a,b\in\mathbb R$ such that $f(a)<0<f(b),$ where $$f(x)=\sum_{k=1}^n(v_k-\mu)2^{-v_kx}\ \ ?$$ This is a final piece of puzzle I'm missing in order to resolve some optimization problem, which has taken me more than a week so far. The goal is to show that $f(x)=0$ has a solution.

I don't know how to proceed. E.g. an obstacle for me is that we can have $f(x)\to-\infty$ as well as $f(x)\to0$ for $x\to\infty$ depending on the particular values $v_k$. That makes it kinda messy.

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Hints: to make things simpler you can reduce the proof to the case when $\mu=0$. To do this just consider $v_n'=v_n-\mu$ and $\mu'=0$. The new sum will differ from the original sum by a factor of $e^{\mu x}$ which is always positive, so it is enough to prove the result when $\mu =0$. Now just split the sum into the part with $v_i <0$ and the one with $v_i >0$. Look at the limits of the two parts as $x \to \infty$ and the limits as $x \to -\infty$. You should now be able to complete the argument.

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  • $\begingroup$ Thank you, I had a similar idea of splitting the sum but was not able to finish it, because I did not "centralize" the function. Now, it is easy because $f_{\mu=0}(x)\to-\infty$ as $x\to-\infty$ and $f_{\mu=0}(x)\to\infty$ for $x\to\infty$. By multiplying $2^{\mu x}$ we get the original $f$, which has the same root as $f_{\mu=0}$ has. $\endgroup$ – byk7 Mar 13 at 12:43

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