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Let,$f:[0,1]\to [0,\infty)$ be a continuous function such that $\int_0^x f(t)dt \geq f(x)$. Then how many such function exists?

I think $e^x$ is such a function. Again, from the given condition $f(0) = 0$. But $e^x$ does not assume value $0$ at any point in $[0,1]$.

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    $\begingroup$ If there is a nonzero one there are infinitely many of them. $\endgroup$ – user99914 Jun 6 '18 at 4:46
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    $\begingroup$ $\int_0^x e^x=e^x-1$, so it isn't this kind of function. $\endgroup$ – Sonal_sqrt Jun 6 '18 at 4:49
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    $\begingroup$ Why do you say "$f(0) = 0$" is a given condition? Also, why are you concerned that $\mathrm{e}^x$ does not take the value $0$ in $[0,1]$? (Is the function required to be surjective?) $\endgroup$ – Eric Towers Jun 6 '18 at 9:52
  • $\begingroup$ In addition to the answers below, you can also just iterate the thing like in the proof of gronwall. $\endgroup$ – shalop Jun 6 '18 at 14:25
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Let $g(x) = \int_{0}^x f(t)dt$. Then, by FTC, $g'(x) = f(x)$, and $g(x) \geq g'(x)$ on the given interval, with $g(0) = 0$.

Rearranging, $g(x) - g'(x) \geq 0$ on the interval.

At this stage, we have a common trick.

Multiplying by $e^{-x}$, one sees that $e^{-x}g(x) - e^{-x}g'(x) \geq 0$. But the derivative of $e^{-x} g(x)$ is $e^{-x}g'(x) - e^{-x}g(x) = e^{-x}(g'(x) - g(x))$.

Consequently, $(e^{-x}g(x))' \leq 0$ on the interval. Therefore, $e^{-x}g(x)$ is decreasing on the given interval.

But $g(0) = 0$, and $g \geq 0$, since it is defined as the integral of a non-negative function. This forces $e^{-x} g(x) = 0$ on the interval, and therefore $g \equiv 0$ and $f \equiv 0$.

So there is only one such function, the zero function.

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Here is a (hopefully) simpler argument than the one given by астон. Just intergrate both sides from 0 to 1. If you interchange the integrals on the left you get $\int_0^{1} (1-x)f(x)dx \geq \int_0^{1} f(x)dx$. Since $(1-x) \leq 1$ we get $(1-x)f(x)=f(x)$ for all $x$ which implies $f \equiv 0$.

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  • $\begingroup$ I'm not sure I follow how you're interchanging the integrals on the LHS. Can you explain? $\endgroup$ – C. Helling Jun 6 '18 at 16:13
  • $\begingroup$ $\int_0^{1} \int_0^{x} f(t)\, dt \, dx= \int_0^{1} \int_t^{1} f(t)\, dx \, dt=\int_0^{1} (1-t) f(t) \, dt$ $\endgroup$ – Kavi Rama Murthy Jun 7 '18 at 0:02
  • $\begingroup$ And you can exchange the values of the integral from 0 to x to the values t to 1 because the function f is defined on the unit interval [0, 1]? $\endgroup$ – C. Helling Jun 7 '18 at 14:59
  • $\begingroup$ Yes, you are right. $\endgroup$ – Kavi Rama Murthy Jun 7 '18 at 23:40
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You can also prove it using the mean value theorem for integrals:

Given any $x \in (0,1]$, there exists a $y \in [0,x]$ with $\int_0^x f(t)dt = xf(y)$. With the given conditions on $f$, this implies that for all $x \in (0,1]$ there exists a $y \in [0,x]$ with $xf(y) \geq f(x)$.

Pick $x$ so that it maximizes $f(x)$ on $[0,1]$. If $x = 0$ then the fact that $f$ is nonnegative implies that $f$ is constantly 0. On the other hand, if $x > 0$ we have that $f(y) \geq f(x)/x$. This would contradict the maximality of $f(x)$ unless $x = 1$. But then $f(y) \geq f(1)/1$, with $f(1)$ maximal, which implies that $f(y) = f(1)$.

Putting this together, we have that $\int_0^1 f(t)dt = f(1)$, with $f(1)$ maximal. The only way that the integral of a continuous function on $[0,1]$ can equal the maximum value of that function on $[0,1]$ is if the function is constant. Hence $f(x)$ is constant. But since the given conditions imply that $f(0) = 0$, we must have that $f$ is constantly $0$ in this case as well.

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