# Prove that $f'(x_o) =0$

Let $f$ be a function defined on an interval $I$ differentiable at a point $x_o$ in the interior of $I$.

Prove that if $\exists a>0$ $\ [x_o -a, x_o+a] \subset I$ and $\ \forall x \in [x_o -a, x_o+a] \ \ f(x) \leq f(x_o)$, then $f'(x_o)=0$.

I did it as follows:

Let b>0.

Since $f$ is differentiable at $x_o$, $$\exists a_o>0 \ \ \text{s.t} \ \ \forall x \in I \ \ \ \ \ 0<|x-x_o|<a_o \implies \left| \frac{f(x)-f(x_o)}{x-x_0} - f'(x_o)\right| <b$$ Let $x_1 \in (x_o,x_o+a) \forall x \in I; f(x_1) \leq f(x_o)$ $$\left| \frac{f(x_1)-f(x_o)}{x_1-x_0} - f'(x_o)\right| <b \\ -b < f'(x_o)-\frac{f(x_1)-f(x_o)}{x_1-x_0} <b \\ f'(x_o) < b+ \frac{f(x_1)-f(x_o)}{x_1-x_0} < b$$ $$f'(x_o) < b \tag{1}$$ Similarly Let $x_2 \in (x_o-a,x_o) \forall x \in I; f(x_2) \leq f(x_o)$

$$\left| \frac{f(x_2)-f(x_o)}{x_2-x_0} - f'(x_o)\right| <b \\ -b < \frac{f(x_2)-f(x_o)}{x_2-x_0} - f'(x_o) <b \\ -b< -b + \frac{f(x_2)-f(x_o)}{x_2-x_0} < f'(x_o)$$ $$-b<f'(x_o) \tag{2}$$

From $(1)$ and $(2)$, $$-b < f'(x_o) <b \\ |f'(x_o)|<b$$

I'm stuck here, how can I go to $f'(x_o)=0$ from here?

Any help?

• I think your notation might be confusing, when you use the differentiable property of $f$ you say there exists an $a$, and i'm guessing you don't mean the $a$ from before? Commented Aug 2, 2014 at 11:50
• Yes, don't reuse the $a$. Commented Aug 2, 2014 at 12:01
• Ok I'll edit it. Commented Aug 2, 2014 at 12:05
• It seems that you proved that $\left|f'\left(x_{0}\right)\right|<b$ for any $b>0$. This implies directly that $f'\left(x_{0}\right)=0$. Commented Aug 2, 2014 at 12:07
• Since $b>0$ is an arbitrary positive number, from last equation ($\left| {f({x_0})} \right|<b$) you conclude that $f'(x_0)=0$. Prove: if $f'(x_0)\ne 0$, then if we set $b=\left| {f({x_0})} \right|/2>0$, the last equation is not correct. So, $f'(x_0)$ must be zero. Commented Aug 2, 2014 at 12:08

You proved that $$\forall b>0, |f'(x_0)|<b$$ and this means that $f'(x_0)=0$. So your proof is already finish.

• So how can $|f'(x)|<b \implies f'(x)=0$ be true? I think I'm missing something because I don't get it. Commented Aug 2, 2014 at 12:08
• @S.Dan for every $b<0 |f'(x)|<b|$. letting $b \to 0$ the derivative is $0$ for the property of the norm Commented Aug 2, 2014 at 12:09
• You didn't prove that $|f'(x_0)|<b\implies f'(x_0)=0$ but $$\forall b>0, |f'(x_0)|<b.$$ In simple word, you proved that for all $b>0$ we have $|f'(x_0)|<b$. And this implies that $f'(x_0)=0$.
– idm
Commented Aug 2, 2014 at 12:10
• S.Dan you have proved it $\forall b>0$. This is very important to understand. when you use differentiability ALWAYS state this Commented Aug 2, 2014 at 12:10
• Oh I think I get it now, b can be any positive value, even something like 0.0000000....1 so if something is even smaller, it is fair to say that it's zero right? Commented Aug 2, 2014 at 12:12

It seems you're pretty much there. There are a couple of things you can do to make the argument clearer, both in the early line using the property of differentiable:

1.Differentiation talks about a limit, make sure you say $\forall b > 0$, as this is what gets you the final step.

2.You want to include the given property here, because it seems you have assumed this. What it should say is something like:

By differentiability at $x_0$ and the property given (about $f$ being max at $x_0$ ) we have:

$$\forall b> 0 \ \ \exists a > 0 \ \ \text{s.t} \ \ \forall x \in (x_0-a,x_0+a) \subset I \ \ \left| \frac{f(x)-f(x_0)}{x-x_0} - f'(x_0) \right| < b \ \ \text{and} \ \ f(x) \le f(x_0)$$

This is not an answer, but is too long for the comments.

Another approach is to show that if $f'(x_0) > 0$, where $x_0$ is in the interior of $I$, then for some $\delta>0$, then if $x \in (x_0-\delta,x_0)$ we have $f(x) <f(x_0)$ and for $x \in (x_0, x_0+\delta)$ we have $f(x) > f(x_0)$.

A corresponding result will hold for $f'(x_0)<0$, of course.

The proof is straightforward, since if $f'(x_0)>0$, we can find some $\delta>0$ such that ${ f(x)-f(x_0) \over x-x_0 } \ge {f'(x_0) \over 2}$ for all $x$ such that $|x-x_0|< \delta$. For $x>x_0$ this gives $f(x) \ge f(x_0) + {1 \over 2} (x-x_0) f'(x_0)$, and similarly for $x<x_0$ this gives $f(x) \le f(x_0) + {1 \over 2} (x-x_0) f'(x_0)$.

Hence if $f$ has a local maximum (or indeed a local minimum) at $x_0$, we must have $f'(x_0) = 0$ (otherwise we could find nearby points that violate the assumption).

A more elegant way of working is:

Define $g\left(x\right):=f\left(x+x_{0}\right)-f\left(x_{0}\right)$ and $J:=x_{0}+I$.

Then $g$ is a function defined on interval $J$ differentiable at $0\in\left[-a,a\right]\subset J$ with $g\left(0\right)=0$ and $g\left(x\right)\leq0$ for each $x\in\left[-a,a\right]$.

Since $g$ is differentiable at $0$ and $g\left(0\right)=0$ both limits $\lim_{x\rightarrow0+}\frac{g\left(x\right)}{x}$ and $\lim_{x\rightarrow0-}\frac{g\left(x\right)}{x}$ exist and coincide.

Here $\frac{g\left(x\right)}{x}\geq0$ for $x\in\left[-a,0\right)$ implying that $\lim_{x\rightarrow0-}\frac{g\left(x\right)}{x}\geq0$ and $\frac{g\left(x\right)}{x}\leq0$ for $x\in\left(0,a\right]$ implying that $\lim_{x\rightarrow0+}\frac{g\left(x\right)}{x}\leq0$.

So the limits can only coincide if both equal $0$.

This proves that $g'\left(0\right)=0$ or equivalently that $f'\left(x_{0}\right)=0$.