Are there any non zero(i.e not $f \equiv0 $) and non-negative continuous function $f$ defined on $[0,1]$, which satisfies $\int_0^x{f(t)}dt \geq f(x)~ \forall x$ in $[0,1]$?
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4$\begingroup$ $f(t) = -c$ where $c > 0$ . $\endgroup$– BlencerJan 7, 2017 at 17:08
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$\begingroup$ Note that this does not rule out Guru's answer - maybe you want to say non-constant instead? $\endgroup$– Noah SchweberJan 7, 2017 at 17:10
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$\begingroup$ sorry f(x) should be non negative $\endgroup$– SaravananJan 7, 2017 at 17:13
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2 Answers
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Use the integrating factor $e^{-x}$ to find $$ \frac{d}{dx}\left(e^{-x}\int_0^xf(s)ds\right)\le 0 $$ so that after integration $$ e^{-x}\int_0^xf(s)ds-e^0·0\le 0 $$ and thus $$ f(x)\le\int_0^xf(s)ds\le 0 $$
As $f$ is supposed to be non-negative, the only solution is the zero function.
See also Grönwall lemma.
answered Jan 7, 2017 at 17:18
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No. Since $f$ is continuous on a compact set it is bounded and achieves both its minimum and maximum, $m=f(x_0)$ and $M=f(x_1)$. Since $f$ is non-constant these are distinct. Pick any $c\in(m,M)$. Without loss of generality assume $x_0<x_1$ By continuity there is some interval $(a,b)\in[0,1]$ containing $x_0$ on which $f<c$. Thus
$$\int_a^b f(t)dt \le \int_a^b cdt = c(b-a) < M(b-a)$$
Note the strict inequality. Then
$$\int_0^{x_1} f(t)dt$$
$$=\int_0^a f(t)dt + \int_a^b f(t)dt + \int_b^{x_1} f(t)dt$$
$$\le M(a-0) + \int_a^b f(t)dt + M(x_1-b)$$
$$< M(a-0) + M(b-a) + M(x_1-b)$$
$$= Mx_1\le M=f(x_1)$$
Thus
$$\int_0^{x_1} f(t)dt < f(x_1)$$
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$\begingroup$ I got it, but after looking your answer I got that just have to apply mean value theorem on any smaller sub interval [0,b] It will be an two line argument $\int^{x_1}_0{f(x)}dx=f(c)(x_1-0)\geq{f(x_1)}$ the equality case only f is 0 $\endgroup$ Jan 8, 2017 at 7:45