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

I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?

I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ 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 $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then

$$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$

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

Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.