Given $f$ is differentiable in a neighborhood of 0. And given the following equality holds:

Question

Given $f$ is differentiable in a neighborhood of 0. And given the following equality holds: $|f(\frac{x}{4})|\le \frac{1}{7}|f(x)|$. Prove $f'(0)=0$. I just need a hint from where to start, I tried using every single theorem or criterion I've been taught in real analysis and I can't quite get to a result. I thought about using the reverse triangle inequality and then the mean value between $x,\frac{x}{4}$ but it's not getting me anywhere useful.

Answer 1

Suppose $f$ is differentiable on $(-2\delta, 2\delta)$ for some $\delta > 0$. Observe that $$ \left| f \left( \frac{\delta}{4^n} \right) \right| \leq \frac{1}{7} \left| f \left( \frac{\delta}{4^{n-1}} \right) \right| \leq \cdots \leq \frac{1}{7^n} \left| f \left( \delta \right) \right| $$ for each positive integer $n$. It follows from this that $$ \frac{4^n}{\delta} \left| f \left( \frac{\delta}{4^n} \right) \right| \leq \frac{1}{\delta} \left( \frac{4}{7} \right)^n \left| f \left( \delta \right) \right|. $$ The squeeze theorem now gives us $$ \lim_{n \to \infty} \frac{4^n}{\delta} f \left( \frac{\delta}{4^n} \right) = 0. $$ Plugging $x = 0$ into the inequality $ \left| f \left( \tfrac{x}{4} \right) \right| \leq \tfrac{1}{7} \left| f(x) \right| $ shows that $f(0) = 0$. Since $\lim_{n \to \infty} \tfrac{\delta}{4^n} = 0 $ and $f'(0)$ exists, we have $$ f'(0) = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{n \to \infty} \frac{f \left( \frac{\delta}{4^n} \right)}{\frac{\delta}{4^n}} = \lim_{n \to \infty} \frac{4^n}{\delta} f \left( \frac{\delta}{4^n} \right) = 0. $$