Proving convergence of a sequence $\{x_{n}\}$ satisfying$|f( x_{n}) |=|f'( x_{n+1}) ||x_{n} -a|$,where $f$ is a function satisfying some conditions. Theorem: Let $\displaystyle f:[ a,b]\rightarrow \mathbb{R}$ be a function differentiable on $\displaystyle [ a,b]$ such that $\displaystyle f( a) =0$ and that there exists $\displaystyle A\in \mathbb{R}$ such that $\displaystyle |f'( x) |\leq A|f( x) |$ for all $\displaystyle x\in [ a,b]$, then $\displaystyle f\equiv 0$ on $\displaystyle [ a,b]$.
I tried to prove the above stated theorem and in the process I came across the question in title. Here's what I tried to prove the above stated theorem:
For $\displaystyle A=0,$ the result is true so let $\displaystyle A >0.\ $Suppose on the contrary that there exists $\displaystyle x_{0} \in ( a,b]$ such that $\displaystyle f( x_{0}) \neq 0$. By MVT, there exists an $\displaystyle x_{1} \in ( a,x_{0})$ such that $\displaystyle |f( x_{0}) -f( a) |=|f'( x_{1}) ||x_{0} -a|\Longrightarrow |f( x_{0}) |=|f'( x_{1}) ||x_{0} -a|$.
Similarly, there exists $\displaystyle x_{2} \in ( a,x_{1})$ such that $\displaystyle |f( x_{1}) |=|f'( x_{2}) ||x_{1} -a|$.  We continue like this and having found $\displaystyle x_{0} ,x_{1} ,x_{2} ,...,x_{n-1}$ we find $\displaystyle x_{n}$ by applying MVT on $\displaystyle [ a,x_{n-1}]$ so the construction of decreasing sequence $\displaystyle \{x_{n} \}$ can be done.
So we have a sequence $\displaystyle \{x_{n} \}$ such that $\displaystyle x_{n}  >x_{n+1}$ for all $\displaystyle n\in \mathbb{N} \cup \{0\}$ and $\displaystyle |f( x_{n}) |=|f'( x_{n+1}) ||x_{n} -a|$ for all $\displaystyle n\in \mathbb{N} \cup \{0\}$. It is clear that the sequence is bounded below by $\displaystyle a$ and hence it must converge. Intuitively, it seems to me that the sequence $\displaystyle \{x_{n} \}$ converges to $\displaystyle a$. But I am having difficulties proving it (please refer Note).
Assuming that $\displaystyle x_{n}\rightarrow a$, the theorem can be proved as follows:
We have:
$\displaystyle |f( x_{0}) |=|f'( x_{1}) ||x_{0} -a|\leq A|f( x_{1}) ||x_{0} -a|=A|f'( x_{2}) ||x_{0} -a||x_{1} -a|\leq A^{2} |f( x_{2}) ||x_{0} -a||x_{1} -a|$
Proceeding like this to get:
\begin{equation*}
|f( x_{0}) |\leq A^{n+1} |f( x_{n+1}) ||x_{0} -a||x_{1} -a|...|x_{n} -a|\leq A^{n+1} s|x_{0} -a||x_{1} -a|...|x_{n} -a|
\end{equation*}
where $\displaystyle s=\max |f|[ a,b]$, which is finite because $\displaystyle |f|$ is continuous on $\displaystyle [ a,b] .$
I take $\displaystyle n-$th root on both sides to get:
\begin{equation*}
|f( x_{0}) |^{\frac{1}{n}} \leq A^{\frac{1}{n} +1} s^{\frac{1}{n}}( |x_{0} -a||x_{1} -a|...|x_{n} -a|)^{\frac{1}{n}}
\end{equation*}
Now taking $\displaystyle n\rightarrow \infty $ on both sides to get $\displaystyle |f( x_{0}) |^{\frac{1}{n}}\rightarrow 0$ (please note that $\displaystyle ( |x_{0} -a||x_{1} -a|...|x_{n} -a|)^{\frac{1}{n}}\rightarrow 0$ because $\displaystyle x_{n}\rightarrow a$).
Now $\displaystyle \left( |f( x_{0}) |^{\frac{1}{n}}\rightarrow 0\right) \Longrightarrow |f( x_{0}) |=0$ , which is a contradiction.
Note: Let $\displaystyle x_{n}\rightarrow l$ then from the iterative definition of our $x_{n}$, i.e. $\displaystyle |f( x_{n}) |=|f'( x_{n+1}) ||x_{n} -a|$ and noting that $|f|$ is continuous,
we should get $\displaystyle |f( l) |\leq A|f( l) ||l-a|$, which is where I got stuck as it does not give me $\displaystyle l=a$.
Please help me to prove that $\displaystyle x_{n}\rightarrow a$. Thanks.
 A: The case $A=0$ is clear, so assume $A>0$. Assume first that $A(b-a)<1$. Then if $f(x)\neq 0$, there must exist $x_0\in (0,b]$ such that $f(x_0)\neq 0.$ By the MVT, one has $$|f(x_0)|=|f(x_0)-f(a)|=|f'(x_1)||x_0-a|\leq A|f(x_1)||x_0-a|,~x_1\in (a,x_0).$$ Similarly, one has $$|f(x_1)|\leq A|f(x_2)|x_1-a|,~x_2\in (a,x_1),$$
$$\cdots$$
$$|f(x_{n-1})|\leq A|f(x_n)|x_{n-1}-a|,~x_2\in (a,x_{n-1}).$$ Note that $f(x_i)\neq 0,~{\rm for ~}i=0,\cdots,n.$ Multiplying the above inequalities and cancel common factors on both sides, one has $$|f(x_0)|\leq A^n|f(x_n)||x_0-a||x_1-a|\cdots |x_{n-1}-a|$$ $$\Rightarrow |f(x_0)|\leq A^nM|b-a|^n,\qquad (1)$$ where $M:=\max_{x\in [a,b]}|f(x)|.$ Statement (1) is absurd, since $|f(x_0)|>0$, but $A^n|b-a|^n=(A(b-a))^n\rightarrow 0,$ as $n\rightarrow \infty.$ This proves the case when $A(b-a)<1$.
To prove the general case, let $m=\lceil A(b-a)\rceil+1.$ Then $\Delta:=\frac{b-a}m$ satisfies $A\Delta=\frac{A(b-a)}m<1.$ Make a partition of $[a,b]$ with $x_0=a,x_i=x_0+i\Delta,i=1,\cdots,m.$ Then the interval $I_i=[x_{i-1},x_i]$ satisfies $A(x_i-x_{i-1})<1$ for $1\leq i\leq m.$ Applying the above argument $m$ times starting with $I_1$, the general case is proven. QED
Remark. The result is a consequence of Grönwall's inequality: Write the original assumption as $$|f'(x)|\leq A|f(x)|~{\rm on~}[a,b]\qquad (2)$$ with $f(a)=0$. One needs to prove that $f(x)\equiv 0.$ If the result is not true, then multiplying $f$ by $-1$ if necessary, one may assume that there exists $b'\in (a,b]$ such that $f(b')>0.$ Let $$a':=\inf\{t~|~f(x)>0~{\rm on~}(t,b']\}.$$ Then by continuity of $f$, one has $$f(a')=0,~{\rm and~}f(x)>0~{\rm on~}(a',b'].$$ Then (2) implies that $$f'\leq Af~{\rm on~}(a',b'),$$ so by Grönwall's inequality, one has $$f(x)\leq f(a'){\rm exp}\left(\int_{a'}^x A~ds\right)=0~{\rm on~}[a',b'],$$ a contradiction.
[Edit] (In response to a question in the comments) Without looking into the local behavior of $f(x)$ near $a$, one cannot rule out the possibility that $\lim x_n=c>a$. One example was suggested in my previous comment: Namely take a piecewise defined function $f(x)$ such that $f(x)=0$ on $[a,c]$ and $f(x)=g(x)$ on $[c,x_0],$ where $g$ is concave up with $g(c)=g'(c)=0.$ Another example can be constructed as follows: Let $g$ defined on $[a,c]$ satisfy $g(a)=0,g(c)>0,$ and $g'(c)=\frac{g(c)}{c-a}.$ Then define $f$ by $$f(x)=\left\{\begin{array}{cc}g(x),& a\leq x\leq c\\
                         g(c)+\frac{g(c)}{c-a}(x-c),&c\leq x\leq x_0.\end{array}\right.$$ In the first example, there exists a sequence $x_n$ satisfying the MVT requirement such that $\lim x_n=c,\lim f'(x_n)=0$ and $f(c)=0.$ In the second example, there exists a sequence $x_n$ satisfying the MVT requirement such that $\lim x_n=c,\lim f'(x_n)=\frac{g(c)}{c-a}>0,$ and $f(c)=g(c)>0.$
