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?

  • $\begingroup$ This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $\alpha = 1/2$. It follows that $f(x) \le x^2/4$. $\endgroup$ – Martin R Nov 30 '18 at 20:31
  • $\begingroup$ @MartinR $f$ is not necessarily conitnuous here $\endgroup$ – Thinking Nov 30 '18 at 20:38
  • $\begingroup$ @Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that. $\endgroup$ – Martin R Nov 30 '18 at 20:43
  • $\begingroup$ @MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix. $\endgroup$ – Thinking Nov 30 '18 at 20:46

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.

  • $\begingroup$ There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$ $\endgroup$ – Thinking Nov 30 '18 at 20:54
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    $\begingroup$ @Thinking I don't see the problem. Where exactly do you think my proof breaks down? $\endgroup$ – zhw. Nov 30 '18 at 21:04
  • $\begingroup$ You are right, I completely missounderstood the proof. It's actually clever. (+1) $\endgroup$ – Thinking Nov 30 '18 at 21:08
  • $\begingroup$ @Thinking I've edited to remove the continuity hypothesis. $\endgroup$ – zhw. Nov 30 '18 at 21:24
  • $\begingroup$ @Thinking Yes of course. Otherwise $M_x-\epsilon$ is an upper bound. $\endgroup$ – zhw. Nov 30 '18 at 21:52

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 $b_{n+1}-\frac 12b_n=1\Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, 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 $a_1=\frac 23\le 1 =a_0$, 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 $b_n\to 2$, and therefore $a=1/4$. Finally, $f(x)\le a_nx^{b_n}\to \frac 14x^2$.


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