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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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    "strictly increasing" means what? $x_1 < x_1 \Longrightarrow h(x_1) < h(x_2)$ ... if that is the definition, then the result could fail.2011-09-23
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    If $f(x) = x$ and $g(x) = -x$ then the first inequality holds. If we take $h(x) = x^2$, then certainly we will have equality in the second case.2011-09-23
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    @Jonas, but $h(x)=x^2$ is not strictly increasing in the relevant domain.2011-09-23
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    @GEdgar: I am not aware of any other definition of strictly increasing, so, yes, I think that it what I mean.2011-09-23
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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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    Good. Point. But, would it hold if all functions involved are monotonic increasing?2011-09-23
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    Makes no difference. You could put the high part of $f$ in [1,2] instead of [0,1], and add $\epsilon x$ to both $f$ and $g$ to make them monotonic. If $\epsilon$ is chosen small enough, it won't change the inequalities.2011-09-23
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    Ok, so if they're all continuous and monotonic, would that be enough?2011-09-23
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    No. Bridge the discontinuity in $f$ by replacing the values for $(1-\epsilon, 1+\epsilon)$ with a linear segment that joins the existing $f(1-\epsilon)$ and $f(1+\epsilon)$. Again, if $\epsilon$ is small enough, it won't affect the inequalities.2011-09-23
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    Ok... one more try. So, if all functions are monotonic, continuous, and smooth, then the inequality is preserved?2011-09-23
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    No. Use $f(x)=\frac{3}{1+e^{K(1-x)}}$ for large enough $K$.2011-09-23
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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.