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Let $f: [0, \infty) \to \mathbb{R}$ be a continuous function. Prove that $f$ is increasing if and only if: $$\int_a^b f(x) dx \leq bf(b) - af(a), \, \forall \, \, 0 \leq a \leq b.$$

I have no difficulties in proving that if $f$ is increasing then the inequality holds. But I haven't figured out yet how to prove it the other way around, that is knowing the inequality and proving that $f$ is increasing.

Thank you!

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3 Answers 3

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Prove that the integral equation implies that $f$ is increasing via contraposition, viz. prove that if $f$ is not increasing then there exist $a_0$ and $b_0$ such that $$\int_{a_0}^{b_0} f(x) dx > b_0f(b_0) - a_0f(a_0).$$

Indeed, if $f$ is differentiable, not increasing and not everywhere constant, then there exists an interval $[a_0, b_0]\in\mathbb{R^+}$ such for all $x\in [a_0, b_0]$ we have that $$f(a_0)\geq f(x) \geq f(b_0)\mbox{ and } f(a_0)> f(b_0).$$ Therefore, we have that $$\int_{a_0}^{b_0} f(x) dx \geq f(b_0)(b_0 - a_0)= f(b_0)b_0 - f(b_0)a_0 > f(b_0)b_0 -f(a_0)a_0.$$

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    $\begingroup$ Are you sure about the statement you make about the existence of such a non-empty interval $[a_0,b_0]$? See e.g. this for the existence of a continuous function nowhere monotone. (I.e., such that there is no non-empty open interval where the function is non-increasing or non-decreasing.) $\endgroup$
    – Clement C.
    Jan 19, 2017 at 19:17
  • $\begingroup$ Clement, you're right. I proved it assuming differentiability of $f$, which is not stated in the question. $\endgroup$
    – mathisfun
    Jan 19, 2017 at 19:36
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    $\begingroup$ @ClementC.: Unless I am mistaken, you don't need $f$ to be differentiable. If $f(a) > f(b)$ then the minimum of $f$ in $[a, b]$ is attained at some point $b_0$, and the maximum of $f$ in $[a, b_0]$ is attained at some point $a_0$. $\endgroup$
    – Martin R
    Jan 19, 2017 at 20:04
  • $\begingroup$ @MartinR I am not saying one needs differentiability of $f$ to establish the statement made in the OP's question (or even to "fix" this answer). My comment was pointing out that the argument of this particular answer made a logical jump that needed further elaboration, or at least was not as straightforward (to me) as the answer suggested. $\endgroup$
    – Clement C.
    Jan 19, 2017 at 20:09
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HINTS:

Show that, WLOG, we can assume $a = 0$. Work with

$$\int_0^{b} f(x)dx \leq bf(b)$$

It should make the result more clear.

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  • $\begingroup$ and for $\int_a$.? $\endgroup$
    – Nosrati
    Jan 19, 2017 at 16:08
  • $\begingroup$ @MyGlasses what do you mean? $\endgroup$
    – RGS
    Jan 19, 2017 at 16:10
  • $\begingroup$ How add $a$ in proof? $\endgroup$
    – Nosrati
    Jan 19, 2017 at 16:10
  • $\begingroup$ @MyGlasses that is the point: we can show that we can assume $a = 0$ and what I did was substitute $a = 0$ into the expression $\endgroup$
    – RGS
    Jan 19, 2017 at 16:12
  • $\begingroup$ If someone could please point me what is wrong with my answer, I would be glad! That way I can improve my work. Simply downvoting won't help me nor the OP $\endgroup$
    – RGS
    Jan 19, 2017 at 16:18
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Note that if $f(a) > f(b) $ then by continuity we can choose $c\in(a, b] $ such that $f(x) > f(c) $ for all $x\in [a, c) $ and hence $$\int_{a} ^{c} f(t) \, dt>(c-a) f(c) $$ And given condition on $f$ implies that integral above is not greater than $cf(c) - af(a) $. It now follows that $f(a) <f(c) $ which is contrary to $f(a) >f(c) $. Thus we must have $f(a) \leq f(b) $.

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