$A=\{x∈[0,1] : \text{for all } y∈[0,x] \text{ we have }f(y)≤1/2\}=[0,1]$, where $f(x) \leq \int_{0}^{x} f(y)^2 dy$.

Question

Let $f:[0,1] \to [0, \infty)$ is a non-negative continuous function so that $f(0)=0$ and for all $x \in [0,1]$ we have $$f(x) \leq \int_{0}^{x} f(y)^2 dy$$
Now consider the set $$A=\{x∈[0,1] : \text{for all } y∈[0,x] \text{ we have }f(y)≤1/2\}$$ Prove that $A=[0,1]$.
Since $f$ is bounded of $[0,x]$, I think $f$ may be $0$. But I am not able to do this. Please help me to solve this.

Answer 1

Clearly $A$ is non-empty since $f(0) = 0$ and thus by continuity there is $\delta > 0$ so that $f(y) \le 1/2$ for all $y \in [0,\delta]$.

Take $x \in A$. Then $f(y) \le 1/2$ for all $y \in [0,x]$. But then $$f(x) \le \int^x_0 f(y)^2 dy \le \int^x_0 \frac 1 4 dy = \frac x 4 \le \frac 1 4.$$ Thus again by contiuity, there is $\delta > 0$ so that $f(y) \le 1/2$ for all $y \in [0,x+\delta]$. This shows that $(x-\delta, x+\delta) \subset A$, and thus $A$ is an open set, since for any element of $A$, we can find a ball surrounding that element that remains in $A$.

Conversely, if $x \not\in A$, then there is $y \in [0,x]$ such that $f(y) > 1/2$. But then by continuity, $f(y-\epsilon) > 1/2$ for some small $\epsilon > 0$ and this shows that $x-\epsilon \not \in A$, and thus $(x-\epsilon,x+\epsilon) \subset A^c$. Similar to above, this shows that $A^c$ is open, and so $A$ is closed.

Since $[0,1]$ is connected, we conclude that $A = [0,1]$, since this is the only open and closed subset of $[0,1]$.

Answer 2

Just a comment that is too long to post it as a comment.

$$f(x)\le\int_0^xf(y)^2dy\le x\sup_{0\le t \le 1}f(t)^2$$

Since the supremum of $f$ is reached at some point $x=x_M$ (Weierstrass' theorem), $$M=f(x_M)\le x_Mf(x_M)^2$$ If $M=f(x_M)=0$, the statement is obvious. Otherwise, $$M\ge Mx_M\ge 1$$

The other answer shows that $M\le\frac12$. Thus, as you guessed, $f\equiv 0$.

Answer 3

A bit of a different approach I believe also to be correct: since $f(x)\leq\int_0^xf(y)^2dy$ for all $x\in[0,1]$, we can say that $f$ is bounded from above (i.e. $f(x)\leq\bar{f}(x)$) on $[0,1]$ by a function $\bar{f}$ defined by \begin{align} \bar{f}(x)& = \int_0^x\bar{f}(y)^2dy,\quad\text{for all }x\in[0,1],\\ \bar{f}(0) &= 0. \end{align} This is an ODE for $\bar{f}$ with solution $\bar{f}(x)=(-x+c)^{-1}$. Using $\bar{f}(0)=0$, we find that $$ \bar{f}(0)=\frac{1}{-0+c}=0\implies c\rightarrow\infty. $$ Since $x\in[0,1], \bar{f}(x)=\lim_{c\rightarrow\infty}(-x+c)^{-1}=0$ for all $x$. Thus, $f(x)\leq0\leq1/2$ for all $x\in[0,1]$ and $A=[0,1]$.