$\int_{0}^{x} f(t) dt \ge {f(x)}$ holds or not on $[0,1]$

Solution 1:

Your reasoning has a big hole - the condition is supposed to hold for all $x$, and you talk about just one $x$.

Say $f\le c$. Then the inequality shows that $f(x) \le cx$. Now integrate $cx$ and you see that $f(x)\le c x^2/2$. And then $f(x)\le cx^3/6$. By induction $f(x)\le cx^n/n!$. Let $n\to\infty$ and you see $f(x)=0$. So there's exactly one such function.

Solution 2:

Let us modify the problem slightly.

Assume that the function $$ f:[0,a]\to[0,\infty) $$ is continuous and that $$ \int_{0}^{x}f(t)\, dt \ge f(x) \quad \text{for all } x\in[0,a]. $$ Then $f(x) = 0$ for all $x\in[0,a]$.

Proof. We can rewrite the inequality as $$ \dfrac{d}{dx}\left(e^{-x}\cdot\int_{0}^{x}f(t)\, dt\right) \le 0 \quad \text{for all } x\in[0,a]. $$ Consequently the function $$ x\mapsto e^{-x}\cdot\int_{0}^{x}f(t)\, dt $$ is decreasing and it takes its maximum at $x=0$. Thus $$ e^{-x}\cdot\int_{0}^{x}f(t)\, dt \le 0 \Leftrightarrow \int_{0}^{x}f(t)\, dt \le 0. $$ However $f$ is non-negative and continuous. We conclude that $f(x) = 0$ for all $x\in[0,a]$.

Solution 3:

Let $f$ be such a function. If $0<x< 1$ then by the MWT, there exists $\xi\in (0,x)$ with $$\int_0^xf(t)\,\mathrm dt=xf(\xi)\stackrel{(1)}\le x\int_0^\xi f(t)\,\mathrm dt\stackrel{(2)}\le \int_0^\xi f(t)\,\mathrm dt\stackrel{(3)}\le \int_0^xf(t)\,\mathrm dt.$$ We conclude that $(1)$, $(2)$, and $(3)$ are in fact equalities. The second means that $\int_0^\xi f(t)\,\mathrm dt=0$, the third that $\int_\xi^x f(t)\,\mathrm dt=0$, so that in fact $\int_0^x f(t)\,\mathrm dt=0$. As this holds for all $x\in(0,1)$ and $f$ is continuous, we conclude $f(x)=0$ for all $x\in[0,1]$. So, $C$ - final answer.