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$f: [0,1] \to [0, \infty )$ is continuous such that $\int_{0}^{x} f(t) dt \geq f(x)$ for all $x \in [0,1]$.

How many such functions are possible?

Edit: So far by taking limit of both sides as $x$ goes to $0$, I got that $f(0) \leq 0$. Also as $f(0) \geq 0$, $f(0)=0$.

I don't know how to proceed from here.

If you are downvoting this question, kindly let me know the reason, if that's not too much trouble.

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    $\begingroup$ Although you've got a answer already still you may have a look at Gronwall inequality! $\endgroup$ – Arpit Kansal Dec 10 '16 at 16:44
  • $\begingroup$ Can I ask where you found that question? $\endgroup$ – rhaldryn Nov 6 '18 at 6:25
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We claim that $f = 0$ is the only function satisfying the constraints. To this suppose that $a = \max_{x \in [0,1]} f(x) > 0$, and let $x_a \in [0,1]$ be a value with $f(x_a) = a$. Now, we have that $$ a = f(x_a) \leq \int_0^{x_a} f(t) \,\mathrm dt \leq \int_0^{x_a} a \,\mathrm dt = ax_a. $$ As a result, it follows that $x_a = 1$ and the inequalities are in fact equalities. Hence, by linearity of the integral we obtain $$ \int_0^1 a - f(t) \,\mathrm dt = 0, $$ and we know that $a - f(x)$ is a non-negative continuous function, so $a - f(x) = 0$. But this contradicts $f(0) = 0$, which concludes the proof.

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  • $\begingroup$ I was thinking in this direction. I wasn't sure how to prove though. Thanks a lot! $\endgroup$ – Prince Kumar Dec 10 '16 at 15:41
  • $\begingroup$ How do you justify the $a=$ in the first line? I don't think you need it. Just say because $x_a \le 1$ the right side is $\lt a = f(x_a)$ where the strict inequality comes from continuity at $0$ so there is some interval near zero where $f(x) \lt a$. This violates the stated requirement. $\endgroup$ – Ross Millikan Dec 10 '16 at 15:49
  • $\begingroup$ @RossMillikan, you're right, my bad, I meant $a \leq$. Edited now. $\endgroup$ – Mees de Vries Dec 10 '16 at 15:52
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Let $x\in[0,1]$ be a global maximum of $f$, i.e. $f(x) = \|f\|_\infty$. Then we have $$0\leq f(x) \leq \int_0^x f(t)dt \leq \int_0^x f(x)dt = xf(x) \leq f(x).$$ But this yields $xf(x) = f(x)$ and $f(x) = \int_0^xf(t) dt$. If not $f(x)=0$, then $x=1$ but then the second identity becomes $\|f\|_\infty = \int_0^1 f(t)dt$ which yields $f(t) = \|f\|_\infty$ for each $t\in[0,1]$ because of continuity, so $f$ is constant. But then, as you concluded, since $f(0) = 0$, we have $f\equiv 0$.

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  • $\begingroup$ I edited my answer. Is it more clear now? $\endgroup$ – Tim B. Dec 10 '16 at 15:40
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    $\begingroup$ Yes, that's good. $\endgroup$ – Arthur Dec 10 '16 at 15:41
  • $\begingroup$ @TimB. Could you explain how "$f(t) = \|f\|_\infty$ for each $t\in[0,1]$ because of continuity", assuming $\|f\|_\infty=sup f(x) , x\in [0,1]$. I am not familiar with that notation. $\endgroup$ – rhaldryn Nov 6 '18 at 7:19

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