# Antiderivative of a function that is decreasing is concave.

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! –  André Nicolas Jul 20 '12 at 18:07
But differentiating it twice is, if the original function is not differentiable. –  Robert Israel Jul 20 '12 at 18:13
exactly! I just need a proof. –  Stuck_pls_help Jul 20 '12 at 18:24

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 $$t \left(F(m) - F(x) \right) = t(1-t) (y-x) f(u) \ge t(1-t)(y-x) f(v) = (1-t) \left(F(y) - F(m)\right)$$
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 $t$. –  Robert Israel Jul 20 '12 at 22:28
$F$ in the multivariate case is not an "antiderivative", but rather a scalar potential for a conservative vector field. –  Robert Israel Jul 20 '12 at 22:38
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. Thanks –  Stuck_pls_help Jul 20 '12 at 22:45