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If $\int_a^b f(x) \,dx > \int_a^b g(x) \,dx$

and there is a function $h(x)$ that is strictly increasing with $x$, does that imply that $\int_a^b h(f(x)) \,dx > \int_a^b h(g(x)) \,dx$ ?

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    @Jonas: At least the relevant domain must include anything that $f$ and $g$ produces between $a$ and $b$, such as the negative values of $g(x)$ in your example.2011-09-23

2 Answers 2

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No. Consider $a=0$, $b=2$, $f(x) = \begin{cases}3 & x<1 \\ 0 & x \ge 1\end{cases}$ $g(x) = 1$ $h(x) = \sqrt{x}$

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    @Angada: The problem here has nothing to do with the niceness of the functions. It's that $f$ is smaller than $g$ for the larger values of $x$, and so on those values the application of $h$ increases the integral by more. You can achieve this with almost any level of "niceness" you desire.2011-09-23
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For any continuous real-valued function $f$ on $[a,b]$, let $V(f) = \int_a^b f(x)\, dx$. Just for convenience, I'll take $a=-1/2$ and $b = 1/2$. Suppose for some continuous function $h$ on $\mathbb R$ and all polynomials $f$ and $g$ of degree $\le 1$ with real coefficients, $V(f) < V(g)$ implies $V(h(f)) < V(h(g))$. Then $V(f) = V(g)$ implies $V(h(f)) = V(h(g))$. Now for $f(x) = c x + d$ and $g(x) = d$ we have $V(f)=d$, so $V(h(g)) = V(h(g))$, i.e. $\int_{-1/2}^{1/2} h(cx+d)\ dx = \int_{-1/2}^{1/2} h(d)\ dx = h(d)$. Using the change of variables $t = cx+d$, this becomes $\int_{d-c/2}^{d+c/2} h(t)\ dt = c h(d)$. Take the derivative with respect to $c$ to get $\frac{h(d+c/2) + h(d - c/2)}{2} = h(d)$. It is well-known that for continuous (or even measurable) $h$ this implies $h$ is an affine function.