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Let $p$ be an increasing, continuous function on the interval $[t_0.t_1]$. I wish to show the inequality $$ \int_{t_0}^{t_1}p(t)\exp (t_0-t)dt\leq \int_{t_0}^{t_1}p(t)\exp(t-t_1) $$ I have checked this inequality in the simple case that $p(t)=\sqrt{t},p(t)=t$ and $p(t)=t^2$, and it should be a simple inequality to proof but somehow it is giving me more struggles than it should.

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    $\begingroup$ The conclusion is wrong in general. As a hint towards understanding when it holds, recall that $\exp(a+b) = \exp(a) \exp(b),$ and $\int a f(t) \mathrm{d}t = a \int f(t) \mathrm{d}t$ for a constant $a$. $\endgroup$ Jun 17, 2022 at 19:58

2 Answers 2

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As the function $\exp(.)$ is increasing, we have $\exp(t_0)\leq \exp(t_1)$. If we suppose that $\int_{t_0}^{t_1}p(t)\exp (-t)dt \geq 0$, we will get \begin{align*} \exp(t_0)\int_{t_0}^{t_1}p(t)\exp (-t)dt &\leq \exp(t_1)\int_{t_0}^{t_1}p(t)\exp (-t)dt\,\\ \int_{t_0}^{t_1}p(t)\exp (t_0-t)dt &\leq \int_{t_0}^{t_1}p(t)\exp(t_1-t) \end{align*}

So your result will be true if $p$ is positive. But if $p$ is negative, your result will be false.

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  • $\begingroup$ Sorry there was a typo in the original question $\endgroup$
    – James
    Jun 18, 2022 at 6:13
  • $\begingroup$ You should accept the answer if it is the answer of your question. $\endgroup$ Jun 18, 2022 at 11:28
  • $\begingroup$ But it is not the answer, since $t$ and $t_1$ are different $\endgroup$
    – James
    Jun 18, 2022 at 12:51
  • $\begingroup$ Aren't you integrating over $t$? I mean it is an integrale $dt$ right? $\endgroup$ Jun 18, 2022 at 13:18
  • $\begingroup$ It's about $t-t_1$ instead of $t_1-t$ $\endgroup$
    – James
    Jun 18, 2022 at 13:33
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For the amended question, we can show the following generalisation:

Let $p$ and $q$ be increasing on $[a,b].$ Then $$ \int_a^b p(t) q(a+b-t) \mathrm{d}t\le \int_a^b p(t) q(t) \mathrm{d}t.$$

Proof. By rearranging the inequality, it suffices to argue that $$ \int_a^b (q(t) - q(a+b - t)) p(t) \mathrm{d}t \overset?\ge 0.$$

Now, to add some symmetry to our expressions, define $\mu = (a+b)/2, \delta = (b-a)/2,$ and make the substitution $t = z + \mu.$ We end up with the goal $$ \int_{-\delta}^\delta (q(\mu + z) - q(\mu - z)) p(z + \mu) \mathrm{d}z \overset?\ge 0,$$ where $p, q$ are increasing with $z$ in its range.

But observe by substituting $z \mapsto -z$ that $$ \int_{-\delta}^0 (q(\mu + z) - q(\mu - z)) p(z + \mu) \mathrm{d}z = \int_0^\delta - (q(\mu + z) - q(\mu - z)) p(-z + \mu) \mathrm{d}z,$$ and so it suffices to argue that $$ \int_0^\delta (p (\mu + z) - p(\mu - z)) (q ( \mu + z) - q(\mu-z)) \mathrm{d}z\overset?\ge 0.$$ But this is obvious: since both $p$ and $q$ are increasing, and since $z \ge 0,$ the integrand above is positive. QED


Thanks to @William M. for helpful comments. They also point out that the statement holds if $p$ and $q$ are both decreasing as well, which follows since if $p$ and $q$ were decreasing, then $-p$ and $-q$ would be increasing, and $p q = -p \cdot -q$

Further, if one of $p$ and $q$ is increasing and the other decreasing, then applying the statement to one of $(p,-q)$ or $(-p,q)$ yields that $\int_a^b p(t) q(a+b - t) \ge \int_a^b p(t) q(t).$

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  • $\begingroup$ The main trick really is only the first step. Once we do that and end up with $\int_a^b (p(t) - p(a+b - t)) q(t),$ it should be clear that when the difference is positive, the $q(t)$ term is big, and when it is negative, the $q(t)$ term is small, so overall the integral has to be positive. The remainder of the argument is just making this point clearer. $\endgroup$ Jun 20, 2022 at 14:41
  • $\begingroup$ It seems everything cheks when $p$ and $q$ are both the same type of monote functions. $\endgroup$
    – William M.
    Jun 20, 2022 at 15:09
  • $\begingroup$ @WilliamM. Yeah good spot, although one could argue that the statement already includes this (since if both $p$ and $q$ were decreasing, then $-p$ and $-q$ would increase, and have the same product). $\endgroup$ Jun 20, 2022 at 15:17
  • $\begingroup$ Also - the statement is very reminiscent of the (extreme cases of the) rearrangement inequality. In fact, one should be able to directly derive the claim using this and the definition of Riemann integrals. $\endgroup$ Jun 20, 2022 at 15:19
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    $\begingroup$ (That's exactly what I was thinking. Also, the problem is indeed symmetric in $p$ and $q,$ so your first 4 lines are redudant. Nice solution. +1.) $\endgroup$
    – William M.
    Jun 20, 2022 at 15:20

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