# the differentiable function in real-analysis

Could someone help me to figure it out? I totally have no idea about that.I'd appreciate it if someone could help me.

Assume f is diﬀerentiable at some point r. Prove that if f(r) > 0, then there is a δ > 0 such that f(r) < f(x) for all x such that r < x < r + δ and f(r) > f(x) for all x such that r − δ < x < r.

• Note that this question has nothing to do with functional analysis, I suggested to remove that tag. Commented Mar 31, 2017 at 19:12

I think you mean $f'(r)>0$. In this case, for $x$ sufficiently close to $r$, with $x>r$, you have $\frac{f(x)-f(r)}{x-r} > \frac{f'(r)}{2}$, so $f(x)>f(r)+(x-r)\frac{f'(r)}{2}>f(r)$. The case $x<r$ is analogous.
Recall the definition of derivative. $$f'(r)=\lim_{h\rightarrow 0}\frac{f(r+h)-f(r)}{h}$$ So in particular we have, $$f'(r)=\lim_{h\rightarrow 0+}\frac{f(r+h)-f(r)}{h} \ \ \ \ \ \text{and} \ \ \ \ f'(r)=\lim_{h\rightarrow 0-}\frac{f(r+h)-f(r)}{h}$$ So, $\exists \ \delta_1>0$ such that if $0< h< \delta_1$ then $\frac{f(r+h)-f(r)}{h}>0$ which implies that $f(r+h)>f(r)$ since $h>0$.
Similarly, chose a $\delta_2>0$ such that if $0<|h|<\delta_2$ then $\frac{f(r+h)-f(r)}{h}>0$ which implies that $f(r+h)<f(r)$ since $h<0$.
Now chose $\delta = \min \{\delta_1,\delta_2\}$.
• @YUANFENG, If you chose $\delta$ as suggested then for we will simultaneously have both the above inequalities for $|h|<\delta$. In particular, if $h<\delta$ then $f(r-h)<f(r)<f(r+h)$. Commented Apr 1, 2017 at 20:38