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 $[0,1]$?

Solution 1:

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.

Solution 2:

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

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