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According to wikipedia, "let $f$ denote a real-valued, continuous and strictly increasing function on [0, c] with c > 0 and f(0) = 0. Let $f^{−1}$ denote the inverse function of $f$. Then, for all a ∈ [0, c] and b ∈ [0, f(c)],"

$$ab \le \int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx$$

I am wondering: does the inequality reverse if $f$ is decreasing, and $f(a) =0$ ? $($and, specifically with $a = b$, if that is necessary$)$

EDIT: I am pretty sure it neither holds nor reverses, as I can find examples of both. However, I don't understand why it doesn't hold. Consider $a=b=1$.

If I take an increasing function $f$, where $f(0) =0$ and $f(1) = 1$, everything is fine. But if I redefine $f$ as $f_{new}(x) = f(1-x)$ then the function is decreasing and the inequality falls apart.. even though the integral of $f_{new}$ equals that of the old, and I would think the same must be true for the inverses.

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  • $\begingroup$ Or, actually, now I think perhaps the inequality stays the same? I suppose more specifically, I am asking: Why does f need to be increasing? $\endgroup$
    – Angada
    Sep 24, 2011 at 19:54
  • $\begingroup$ Can you tell us exactly what your hypothesis is? You might want to consider $g(x)=f(a)-f(x)$ $\endgroup$
    – Henry
    Sep 24, 2011 at 19:58
  • $\begingroup$ First, why not try a few special cases? Second, if you think about the "chop the rectangle in half with some surplus" proof of Young's inequality, you should be able to figure out a version for decreasing functions (but you'll have to mess around with the integration limits, etc. a bit). Or just perform the transformation suggested by Henry (I think he means $f(0) - f(x)$). $\endgroup$
    – user83827
    Sep 24, 2011 at 20:04

1 Answer 1

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If $f$ is strictly decreasing on $[0,c]$, then $-f$ is strictly increasing on $[0,c]$. Then, noting that the inverse of $-f(x)$ is $f^{-1}(-x)$ we get that for all $a \in [0,c]$ and $b \in [0,-f(c)]$

$ab \leq -\int_0^a f(x) dx + \int_0^b f^{-1}(-x) dx$

(some of the intervals in the theorem doesn't make sense unless we agree that $(a,b)=(b,a)$ when $b < a$)

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