# Can the integral of a function be larger than function itself?

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 $[0,1]$?

• $f(t) = -c$ where $c > 0$ . Jan 7, 2017 at 17:08
• Note that this does not rule out Guru's answer - maybe you want to say non-constant instead? Jan 7, 2017 at 17:10
• sorry f(x) should be non negative Jan 7, 2017 at 17:13

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.

• I'm sure I'm being obtuse, but -- where does your first inequality come from? Jan 8, 2017 at 0:20
• @ruakh: The derivative is just $e^{-x}(f(x) - \int^x_0 f)$, which is non-negative everywhere by hypothesis. Jan 8, 2017 at 1:03
• @anomaly: Oh, I see now -- thanks! Jan 8, 2017 at 1:25

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

$$\int_0^{x_1} f(t)dt < f(x_1)$$

• 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 Jan 8, 2017 at 7:45