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I'm a little rusty on calculus and would like to know how one goes about showing this without having to rely on differentiating the antiderivative twice.

Also, is there an extension to this in the multivariate case (i.e., what condition is needed to ensure that the antiderivative is concave).

Thanks!!!

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    exactly! I just need a proof.2012-07-20

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Suppose $f$ is a continuous decreasing (or more generally, nonincreasing) function on $(a,b)$ and $F$ is an antiderivative of $f$ there. For $0 < t < 1$ and $a < x < y < b$, we need to prove that $F(tx + (1-t) y) \ge t F(x) + (1-t) F(y)$. Rearrange this as $ t \left(F(m) - F(x)\right) \ge (1-t) \left( F(y) - F(m) \right)$, where $m = tx + (1-t) y$. Now by the Mean Value Theorem, $F(m) - F(x) = (m-x) f(u)$ for some $u$ between $x$ and $m$, while $F(y) - F(m) = (y-m) f(v)$ for some $v$ between $m$ and $y$. Note that $u \le v$ so $f(u) \ge f(v)$, and $m - x = (1-t)(y-x)$ while $y-m = t (y-x)$. So indeed $t \left(F(m) - F(x) \right) = t(1-t) (y-x) f(u) \ge t(1-t)(y-x) f(v) = (1-t) \left(F(y) - F(m)\right)$

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    I see. Suppose, I define $F$ by $F(x_1,\cdots, x_n)=\int_{-\infty}^x_1\cdots \int_{-\infty}^x_n f(y_1,\cdots,y_n) dy_1\cdots dy_n$. What conditions of f>0 does one need to guarantee that $F$ is concave on that region. Thanks2012-07-20