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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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    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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Let $f(x)=x+e^x+0$ (taking $w=0$ from your statement) and $v=1$ so that $g(x)=e^{-x}+x+e^x.$ Then $g(x)$ has exactly one minimum, but not on $x \ge 0$, and numerically the min of $g(x)$ is at the point $(-0.482,1.754)$. Replacing the $0$ by an arbitrary $w$ just shifts this example up or down, with the same negative $x$ value at which the minimum occurs.

Now for another example let $v=-1$ (and $w=0$ again) so that this time $g(x)=-e^{-x}+x+e^x.$ This function has no minimum at all, not even a local minimum.