Question: Let $f:[0,1] \to \mathbb{R}$ be a real valued continuous function which is differentiable on $(0,1)$, and satisfies $f(0)=0$. Suppose there exists a $c \in (0,1)$ such that $|f'(x)| \leq c|f(x)|$ for all $x \in (0,1)$. Show that $f(x)=0$.
Solution attempt:
$f$ being continuous at $x=0$, for a given $\epsilon >0$ $\exists$ a $\delta>0$, such that $|f(x)|< \epsilon$ whenever $x \in [0, \delta) \cap [0,1]$.
Now, consider $|f'(h)|=|\lim_{k \to 0}\frac{f(h+k)-f(h)}{k} |\leq |cf(h)| \implies \lim_{ \ k \to 0} |f(2h)| \leq |f(h)|[1+ c|k|] $ Being continuous at $x=0$, $\lim_{h \to 0} f(h) = f(0) = 0 \implies f(2h) = 0$ [by applying $|f(2h)| \leq |f(h)|(c|k|+1) \ $].
In this manner, let us partition the interval $[0,1)$ into $n$ subintervals of length $h$ each. As the length $h \to 0$, $n \to \infty$, and recursively, we get $|f(rh)| \leq |f(h)|(c|k|+1)^r \ $. Hence, finally for all $r$, we get $f(rh)=0$. By continuity, we can safely say, for all $x$ in those respective subintervals, $f(x)=0$. Again, by continuity, we have $\lim_{ x \to 1} f(x) = 0$.
Hence, $f(x)=0$ for all $x$ in $[0,1]$.
Let $M$ be the sup norm of $f$. Let us prove by induction that $|f(x)| \leq \frac{c^nMx^n}{n!}$. If $n=0$, this is obvious.
Let $n \geq 0$ be such that $|f(x)| \leq \frac{c^nMx^n}{n!}$ for each $x$.
Then $|f’(x)| \leq \frac{c^{n+1}Mx^n}{n!}$ for each $x$.
Since $f(0)=0$, integration yields the desired inequality.
Now, the $RHS$ goes to $0$ as $n$ goes to $\infty$, thus $f=0$ (note that $c<1$ is unimportant).