If $\left| f'(x) \right| \leq A |f(x)|^\beta $ then f is a constant function Problem Let $f(x)$ be a differentiable function on $[a,b]$ satisfying $f(a)=0$. If there exist $A \ge 0$ and $\beta \ge 1$ such that the inequality
$$\left| f'(x) \right| \leq A \left| f(x) \right|^\beta $$
holds for all $x$ in $[a,b]$, then $f(x) = 0$ for any x in that interval.
My attempted solution
We shall prove by contradiction. Suppose that for some $x$ in $[a,b]$ we have $f(x) \neq 0$.
Put $S:= \left\lbrace x \right. $ such that $a \le x \le b $ and $f(x) \neq 0 \left. \right\rbrace$ and $c:=\inf S$ (the infimum value of set $S$). If $c=b$ then $ f(x) = 0$   $\forall  x < b$ and the continuity of $f$ implies that $f(b) = 0$, too (contradiction). Therefore $c \neq b$. If $c=a$, then $f(c) = 0$ by the hypothesis. Otherwise, when $a < c < b$ we have $f(x) = 0$ $\forall x < c$ and again, by continuity, $f(c) = 0$. Thus in all cases considered, we have $f(c) = 0$ and $a \leq c < b$
As $f$ is continuous at $c$ and $f(c) = 0$, one can choose $d>c$ and close enough to $c$ such that $|f(x)| \le 1$ and $$\tag{$\star$} A(x-c) \le \frac12\quad\text{for all }x\text{ with }c \le x \le d.$$ Since $f$ is continuous on $[c,d]$, it has maximum and minimum values on this closed interval, leading to the existence of $c \le t \le d$ such that $|f(t)| = \max_{c \le x \le d} |f(x)|$. As already noted, $f(c) = 0$ and $c$ is the infimum of the set of numbers whose image under $f$ is nonzero. Therefore, for any $\epsilon>0$, there is some $c<m<c+\epsilon$ satisfying $f(m) \neq 0$. This observation, together with the definition of $t$, gives us $$\tag{$\star\!\star$}f(t) \neq 0,$$ $t \neq c$ and so $c<t$. Applying the Lagrange theorem gives:
$$ |f(t)| = |f(t) - f(c)| = |t-c||f'(u)| \leq|t-c|A|f(u)|^\beta $$ where $c<u<t\le d$. By $(\star)$:
$$ |t-c|A|f(u)|^\beta \le \frac12 |f(u)|^\beta \leq \frac12 |f(u)| $$ 
Combining the two inequalities and noting that $|f(t)| \ge |f(u)|$ (because of the definition of $t$), we have $f(u) = f(t) = 0$ and arrived at a contradiction with $(\star\star)$.
Question
I hope someone can verify if my solution is correct. Of course, other ideas, comments or solutions are welcome.
I like to post problems and my solutions to the forum because I think it's beneficial to the community, and for learners like me. First, I can hardly know if there's flaw in my own argument. Second, I may get new insights/solutions for my problem.
Thank you.
 A: Let $B = \{x\in [a,b]: |f(x)|=0\}$. By continuity, this set is closed. Since $f(0) = 0$, the set is non-empty. We show that it is also open.
Let $x\in B$, i.e. $f(x) =0$. By continuity, there exists $0<r<(2A)^{-1}$, such that $|f(y)|<\frac 12$ for all $y\in B_r(x)$. Fix such a $y\in B_r(x)$. We want to show that $f(y) =0$.
By the mean value inequality, we have $$|f(y)| = |f(y) - f(x)| \le |f'(\xi_0)| |y-x| \le |f'(\xi_0)|r$$
for some $\xi_0\in B_r(x)$ lying on the segment between $x$ and $y$. By assumption, we have $|f'(\xi_0)|\le A |f(\xi_0)|^\beta$, hence $$|f(y)|\le |f'(\xi_0)|r \le A|f(\xi_0)|^\beta (2A)^{-1} \le \frac 12|f(\xi_0)|^\beta.$$ By the same argument applied to $\xi_0\in B_r(x)$ instead of $y$, we see that there exists $\xi_1\in B_r(x)$, such that $|f(\xi_0)|\le \frac 12|f(\xi_1)|^\beta$. Iterating yields a sequence $\xi_n \in B_r(x)$, such that $|f(\xi_{n})|\le \frac 12|f(\xi_{n+1})|^\beta$ for all $n$. Note that by our choice of $r$, we have $|f(\xi_{n})|\le \frac 12$ for all $n$. We conclude that
$$|f(y)|\le \frac 12 |f(\xi_0)|^\beta \le \frac 1{2^2}|f(\xi_1)|^{\beta^2} \le \frac{1}{2^3}|f(\xi_2)|^{\beta^3}\le \dots \le \frac{1}{2^n}|f(\xi_{n-1})|^{\beta^n}\le \frac 1{2^{n+\beta^n}},$$
for all $n\in \mathbb N$. Letting $n\to \infty$ shows that $|f(y)| = 0$ (using also $\beta \ge 1$). This implies that $B_r(x)\subset B$, hence $B$ is open.
Since $[a,b]$ is connected, it follows that $B$ must be all of $[a,b]$, hence $f= 0$.
A: ( Edit.  The ODE analysis also explains why $\beta \geq 1$ is necessary. The condition for $|f'(x)| \leq A|H(f(x))|$ to have the property that $f$ does not change sign, when $H$ is a function such that $H(x)=0$ only at $x=0$, is that $\frac{1}{H(x)}$ an has a non-integrable singularity at $0$. Knowing that this is the answer, there is no loss in considering singularities that are powers of $f$, which is what most cases would look like and how they would be recognized.)
The differential equation $f'(x) = A(x) f(x)^\beta$ can be solved explicitly, by separation of variables.  The calculation shows that $f(a)=0$ is necessary, or we could write down nowhere zero solutions, by solving the equation.  This bears further analysis, because superficially the problem looks like a Lipschitz condition with $|f(x) - f(y)| \leq |x-y|^\beta$ which (as is very well known and often solved as an exercise) implies constant $f$ when $\beta > 1$.
The DE for constant $\beta > 1$, writing $y$ for $f(x)$, is
$$ d(\frac{1}{(\beta - 1)y^{\beta - 1}}) = A(x) dx$$
The coordinates that trivialize the equation are therefore $Y = \frac{1}{(\beta - 1)y^{\beta - 1}} $ and $X = \int A(x) dx$, in which the solutions are $Y = X + c$ for constant $c$.  This brings out the idea that $A$, which appears at first to be a harmless constant removable by a linear change of variable, has a meaningful part in the problem, as velocity of the independent variable. 
Under this coordinate change, $f(a)=0$ becomes $f(a)=\infty$.  On a maximal interval where $f$ is nonzero, there cannot be any solutions that hit infinity at the boundary in finite "time" (here $X$ is time, so we mean a finite change in $\int A(x) dx$) while satisfying $Y - X = $ constant.  
So the meaning of the problem seems to be: no singularity can be reached in finite time for this ODE.    
A: Supposing that $f(x)=0$ only for $a$ and that $f(x)>0$ for the rest $x$ in $[a,b]$.
Then $\lim _ { x \to a ^ + } \frac{f(x) - f(a)}{x-a}$ will be greater or equal to zero. If greater then $f'(a)=m >0$ and that's contradicting with the given inequality.Therefore $f(x)=0$ for all $x$ in $[a,b]$. If equal, then I'm sorry but I need to give it some more thought.
In the same manner, you can prove for $f(x)<0$.
I just posted this insufficient answer because I thought it would be simpler to understand.
