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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 $\lim_{x \to infty} \left( f(x) - x - ue^x\right) = 0$? –  Hans Engler Nov 24 '12 at 16:06
    
Yes, thanks for the correction. –  Neil G Nov 24 '12 at 16:06
    
If you mean that the limit is $0$ you should fix the question by replacing "limit = w" by "limit = $0$". –  coffeemath Nov 24 '12 at 22:12
    
@coffeemath: I meant $w$, but I don't think it makes any difference? –  Neil G Nov 24 '12 at 22:58
    
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. –  coffeemath Nov 25 '12 at 10:39
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1 Answer

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.

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