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If $f(x)$ is strictly convex, and

$$\lim_{x\to \infty}\left(f(x) - x - ue^{x}\right) = w$$

for some $u\ge 0$ and $w$ then what can be said about:

$$g(x) = ve^{-x} + f(x)$$

on $x\ge0$ where $v$ is some fixed real number. Can I say that it has exactly one minimum?

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    Do you mean to assume that $\lim_{x \to infty} \left( f(x) - x - ue^x\right) = 0$?2012-11-24
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    Yes, thanks for the correction.2012-11-24
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    If you mean that the limit is $0$ you should fix the question by replacing "limit = w" by "limit = $0$".2012-11-24
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    @coffeemath: I meant $w$, but I don't think it makes any difference?2012-11-24
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    Yes, no difference, it's just a constant and drops out of the derivative. In a comment above your "yes, thanks for the correction" led me to believe you meant limit = 0.2012-11-25

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