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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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    Differentiating an antiderivative is not exactly difficult!2012-07-20
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    But differentiating it twice is, if the original function is not differentiable.2012-07-20
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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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    Thanks, RObert! I wonder if there is a natural extension to the multivariate case.2012-07-20
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    A continuously differentiable function $F$ of several variables is concave if for all vectors $\bf a$ and $\bf u$ the directional derivative $D_{\bf u} F({\bf a}+t {\bf u})$ of $F$ in the direction $\bf u$ at ${\bf a}+t{\bf u}$ is nonincreasing in $t$.2012-07-20
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    $F$ in the multivariate case is not an "antiderivative", but rather a scalar potential for a conservative vector field.2012-07-20
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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