Is ‎$‎f‎$‎ ‎monotone ‎when ‎$‎f‎$ ‎is ‎concave?‎

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‎Let ‎$‎f:[1, +‎\infty‎)‎‎‎\rightarrow‎‎\mathbb{R}$ ‎be a‎ ‎concave ‎function. Suppose‎ $‎F:[1, +‎\infty‎)‎‎\rightarrow‎‎\mathbb{R}‎$ is a primitive function of ‎$‎f‎$‎. My ‎questions ‎are‎:‎

‎‎ ‎(a) ‎‎What other condition is required to ‎monotone ‎‎$‎f‎$‎‎?‎

‎ ‎ ‎(b) Is $‎‎‎F‎$‎ concave‏?

‎‎‎‎I‎ ‎know ‎that,‎ ‎suppose $f$ is differentiable on $(a‎, ‎b)$‎. ‎Then $f$ is concave if and only if $f^\prime$ is decreasing. Also, ‎‎$‎f‎$ ‎is ‎concave, ‎then ‎‎for any ‎$‎x‎$‎ and ‎$‎y‎$‎ in the interval and for any ‎$‎\alpha \in [0,1]‎$‎‎‎ ‎ ‎‎$‎f(‎\alpha x ‎‎+ ‎(‎1 - ‎\alpha‎)y) ‎\geq ‎‎\alpha ‎f(x) +‎ ‎‎‎(‎1 - ‎\alpha‎)f(y)‎$‎.

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For a), you can use something along those lines:

$$\lim_{x \to -\infty} f^\prime(x) < 0$$

Having that and concavity $f^{\prime\prime}(x) < 0$, you can easily conclude that $f^\prime(x) < 0$, so f is a decreasing concave function.

There's also the symmetric case, which yields an increasing concave function: $$\lim_{x \to \infty} f^\prime(x) > 0$$

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For (a), I don't think a simple answer exists. Concavity and monotonicity are fairly unconnected properties. You can have concave monotone, concave non-monotone, non-concave monotone or non-concave non-monotone functions.

For (b), the answer is a simple no. For example, $f(x)=-3(x-5)^2$ is a concave function, however its integral, $F(x)=-(x-5)^3$, is not a concave function.