Suppose $f: [0,1] \rightarrow [0,1] $ and $f(x) \leq \int_0^x \sqrt{f(t)}dt$. Show that $f(x) \leq x^2$ for all $x \in [0,1]$.

I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $x\in [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<\epsilon< M_x.$ Choose $x_\epsilon\in [0,x]$ such that $f(x_\epsilon) > M_x-\epsilon.$ Then

$$M_x-\epsilon < f(x_\epsilon) \le \int_0^{x_\epsilon}\sqrt {f(t)}\,dt \le x\sqrt{M_x}.$$

Now let $\epsilon\to 0^+$ to see $M_x\le x\sqrt{M_x},$ which implies $\sqrt{M_x}\le x.$ Squaring, we see $f(x)\le M_x\le x^2,$ giving the result.

Tags:

Inequality

Integration

Calculus

Real Analysis

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