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Let $$f\colon [0,1]\rightarrow [0, \infty )$$ be continuous. Suppose that

$$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for all }x\in[0,1].$$

Then

$A.$ No such function exists.

$B.$ There are infinitely many such functions.

$C.$ There is only one such function

$D.$ There are exactly two functions.

Now I was thinking, since $\int_{0}^{x} f(t) dt$ is the area under the graph of $f(t)$ from $0$ to $x$ so it can be easily made greater than the value of $f(t)$ at one point, namely $x$. So there will be infinitely many such functions satisfying this condition.

Is my reasoning correct or not?

Thanks.

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    $\begingroup$ Well, this is not a reasoning... $\endgroup$
    – ajotatxe
    Sep 5, 2015 at 16:46
  • $\begingroup$ If $f(t)\not\equiv 0$ satisfy the inequality then so do the function $\lambda f(t)$ for any $\lambda>0$. This rules out D. $\endgroup$
    – Winther
    Sep 5, 2015 at 16:52
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    $\begingroup$ @Winther Unless $f \equiv 0$, so that $\lambda f = f$ for every $\lambda>0$. $\endgroup$
    – Siminore
    Sep 5, 2015 at 16:52
  • $\begingroup$ It is known that the derivative of a function cannot be bounded by function itself. $\endgroup$
    – Harish
    Sep 5, 2015 at 18:02
  • $\begingroup$ OP: What about correcting the mess there? $\endgroup$
    – Did
    Sep 6, 2015 at 0:30

3 Answers 3

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Your reasoning has a big hole - the condition is supposed to hold for all $x$, and you talk about just one $x$.

Say $f\le c$. Then the inequality shows that $f(x) \le cx$. Now integrate $cx$ and you see that $f(x)\le c x^2/2$. And then $f(x)\le cx^3/6$. By induction $f(x)\le cx^n/n!$. Let $n\to\infty$ and you see $f(x)=0$. So there's exactly one such function.

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Let us modify the problem slightly.

Assume that the function $$ f:[0,a]\to[0,\infty) $$ is continuous and that $$ \int_{0}^{x}f(t)\, dt \ge f(x) \quad \text{for all } x\in[0,a]. $$ Then $f(x) = 0$ for all $x\in[0,a]$.

Proof. We can rewrite the inequality as $$ \dfrac{d}{dx}\left(e^{-x}\cdot\int_{0}^{x}f(t)\, dt\right) \le 0 \quad \text{for all } x\in[0,a]. $$ Consequently the function $$ x\mapsto e^{-x}\cdot\int_{0}^{x}f(t)\, dt $$ is decreasing and it takes its maximum at $x=0$. Thus $$ e^{-x}\cdot\int_{0}^{x}f(t)\, dt \le 0 \Leftrightarrow \int_{0}^{x}f(t)\, dt \le 0. $$ However $f$ is non-negative and continuous. We conclude that $f(x) = 0$ for all $x\in[0,a]$.

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    $\begingroup$ This is the nicest approach. $\endgroup$ Sep 6, 2015 at 15:38
  • $\begingroup$ Yes, it's certainly the nicest approach. Seems curious that $e^x$ is important here, while the terms in the power series for $e^x$ come up in the solution I gave - could be that in some sense they're really the same...(?) $\endgroup$ Sep 6, 2015 at 15:53
  • $\begingroup$ The function $e^{-x}$ is an integrating factor. The given inequality could be changed to $\displaystyle \int_{0}^{x}f(t)\, dt \ge g(x)f(x) \quad \text{for all } x\in[0,a], $ where $g$ is a positive, continuous function. As before $f(x) = 0$ for all $x\in[0,a]$ will be the only solution. $\endgroup$
    – JanG
    Sep 6, 2015 at 17:35
  • $\begingroup$ @DavidC.Ullrich, I forgot to address my comment above to you. $\endgroup$
    – JanG
    Sep 8, 2015 at 9:57
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Let $f$ be such a function. If $0<x< 1$ then by the MWT, there exists $\xi\in (0,x)$ with $$\int_0^xf(t)\,\mathrm dt=xf(\xi)\stackrel{(1)}\le x\int_0^\xi f(t)\,\mathrm dt\stackrel{(2)}\le \int_0^\xi f(t)\,\mathrm dt\stackrel{(3)}\le \int_0^xf(t)\,\mathrm dt.$$ We conclude that $(1)$, $(2)$, and $(3)$ are in fact equalities. The second means that $\int_0^\xi f(t)\,\mathrm dt=0$, the third that $\int_\xi^x f(t)\,\mathrm dt=0$, so that in fact $\int_0^x f(t)\,\mathrm dt=0$. As this holds for all $x\in(0,1)$ and $f$ is continuous, we conclude $f(x)=0$ for all $x\in[0,1]$. So, $C$ - final answer.

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    $\begingroup$ Of course unless I'm missing something this depends on the fact that $x\le 1$; mine would work just as well for $f:[0,2]\to[0,\infty)$. $\endgroup$ Sep 5, 2015 at 17:05
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    $\begingroup$ @DavidC.Ullrich In fact yours would work imediately for any $[0,n]\to[0,\infty)$ and then per union even for $[0,\infty)\to[0,\infty)$. -Then again, any method that shows that $f$ must be $\equiv 0$ on some $[0,\epsilon]$ can be adjusted to work for $[0,\infty)\to[0,\infty)$: Assume $f$ is nonzero and let $a=\inf\{\,x\in[0,\infty): f(x)\ne 0\,\}$. Then apply the argument to $[a,a+\epsilon)$ ... $\endgroup$ Sep 5, 2015 at 17:09
  • $\begingroup$ I don't understand your second equality. Should it be inequality (where one uses the property of $f$ given in the problem)? $\endgroup$
    – mickep
    Sep 5, 2015 at 17:53
  • $\begingroup$ i think same as mikep $\endgroup$
    – R.N
    Sep 5, 2015 at 18:25
  • $\begingroup$ @mickep Ack, I missed that - but it turns out to be an equality in the end as well by the squeezing ... $\endgroup$ Sep 6, 2015 at 9:41

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