# 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 $x \in [0,1]$.

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?

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

• 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 $[0,1]$ – Thinking Nov 30 '18 at 20:54
• @Thinking I don't see the problem. Where exactly do you think my proof breaks down? – zhw. Nov 30 '18 at 21:04
• You are right, I completely missounderstood the proof. It's actually clever. (+1) – Thinking Nov 30 '18 at 21:08
• @Thinking I've edited to remove the continuity hypothesis. – zhw. Nov 30 '18 at 21:24
• @Thinking Yes of course. Otherwise $M_x-\epsilon$ is an upper bound. – 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, $$f(x)\le a_nx^{b_n}\to \frac 14x^2$$.