It's straight forward to prove by contradiction that for differentiable function if $f(a)=\max(f)$, then $f'(a)=0$.


Proof: Without loss of generality, assume to the converse that $f'(a)>0$

Now by definition of derivative and limit for all $x$, and for all $\varepsilon$, there exists $b>0$:


Since the function is differentiable and $f(a)$ is the maximum, we can select $x$ such that: $$0\geq f(a-x)-f(a)>-b/2$$ and $x>0$.

Then select $\varepsilon=b/2x$. Now:

$$\frac{f(a-x)-f(a)+b}{x}<b/2x$$ $$\frac{f(a-x)-f(a)}{x}<-b/2x$$ $$f(a-x)-f(a)<-b/2$$

A contradiction.


However, is the result provable by a similar epsilon-delta direct proof? I tried to find a direct proof but could not, I wonder if there is a reason (can it be done?). Would the following work?

Direct proof:

Without loss of generality prove the statement for one sided limit. So by the definition of limit and derivative we must prove, for all $\varepsilon$ exists $\delta$, such that: $$0<x<\delta \rightarrow \left|\frac{f(a-x)-f(a)} x \right|<\varepsilon $$

Now $f(a-x)-f(x)\leq 0$, and we can assume $x>0$:

$$-\varepsilon<\frac{f(a-x)-f(a)}{x}<\varepsilon $$

If $f(a-x)-f(a)=0$ in the immediate neighborhood the case is already proven. As such it can be assumed $f(a-x)-f(x)< 0$. Also, assume $x<1$:

$$-\frac{\varepsilon}{f(a-x)-f(a)}>\frac{1}{x}>\frac{\varepsilon}{f(a-x)-f(a)} $$

$$\frac{f(a-x)-f(a)}{\varepsilon}<x<-\frac{f(a-x)-f(a)}{\varepsilon} $$

So take $0<x<\delta=\min(-\frac{f(a-x)-f(a)}{\varepsilon},1) $

$$\left|\frac{f(a-x)-f(a)} x \right|=-\frac{f(a-x)-f(a)}{x}<\varepsilon$$

I am not sure that this is correct, especially the bolded part seems dodgy. Thoughts?

  • $\begingroup$ Where you have $\dfrac{f(a-x) -f(x)-b} x,$ you need $\dfrac{f(z+x) - f(x)} x - b.$ The reason for that should quickly become apparent if you think it through. $\qquad$ $\endgroup$ – Michael Hardy Jun 12 '18 at 17:07

You need to say that for every value of $\varepsilon>0,$ there exists $\delta>0$ such that if $-\delta<x<\delta$ and $x\ne 0,$ then $$ -\varepsilon<\frac{f(a+x) - f(a)} x -b <\varepsilon. $$ If $b\ne0,$ just pick $\varepsilon= |b|/2.$ That way, everything between $b\pm\varepsilon$ differs from $0.$


If $f$ has a local maximum at $a,$ then for $x$ in some open neighborhood of $a,$ we have $f(a+x) \le f(a).$ Therefore $$ \frac{f(x+a) - f(x)} x \begin{cases} >0 & \text{if } x<0, \\ <0 & \text{if } x>0. \end{cases} $$ Therefore \begin{align} & \lim\limits_{x\,\uparrow\,0} \dfrac{f(x+a) - f(x)} x \ge 0 \\[10pt] & \lim\limits_{x\,\downarrow\,0} \dfrac{f(x+a) - f(x)} x \le 0. \end{align}

  • $\begingroup$ Thanks, this is a very efficient way to to complete the indirect proof. But what about the direct proof? (Also, is there some way to put the indirect proof behind a spoiler, it was meant just as reference, maybe it would be better to just link some other proof...). $\endgroup$ – Dole Jun 12 '18 at 19:15
  • $\begingroup$ @Dole : One way to do a direct proof is to say that if a local maximum occurs at $a,$ then $f(a+x) \le f(a)$ for all $x$ close enough to $a,$ and then observe that that means $$ \frac{f(x+a) - f(x)} x \begin{cases} >0 & \text{if } x<0, \\ <0 & \text{if } x>0. \end{cases} $$ From that it follows that if that quotient has a limit as $x\to0,$ then it can only be $0. \qquad$ $\endgroup$ – Michael Hardy Jun 12 '18 at 19:37
  • $\begingroup$ Although is that not a proof by contradiction again? $\endgroup$ – Dole Jun 12 '18 at 19:41
  • $\begingroup$ @Dole : How so? A proof by contradiction would assume that either the limit does not exist or it is not $0.$ Nothing in that argument makes such an assumption. The things asserted to be positive or negative here are NOT limits as $x$ approches anything. $\endgroup$ – Michael Hardy Jun 12 '18 at 19:47
  • $\begingroup$ I think I got it now. I was referring to the part "then it can only be 0",which hints at assuming the converse and deriving a contradiction. $\endgroup$ – Dole Jun 12 '18 at 19:49

what you wrote in the first line is not correct: Take the function $f(x)=x$ for $ x \in [0,1]$, then $\max(f)=1=f(1)$, but the differential $f'$ is vanishing nowhere. You probably mean are local extrema?

  • 1
    $\begingroup$ I changed $max(f)$ in this answer to $\max(f),$ coded as \max(f). The backslash does not only prevent italicization but also provides proper spacing in things like $\max f$ (so you don't see $\text{max}f,$ without proper spacing) and the spacing is context-dependent, so there's more space to the right of $\max$ in $\max f$ than in $\max(f),$ and also in a displayed, rather than inline, context, when you write $$ \max_{x\,\in\,S} f(x) $$ then the subscript is directly below $\max. \qquad$ $\endgroup$ – Michael Hardy Jun 12 '18 at 17:17
  • $\begingroup$ oh thank you! @MichaelHardy by the way, how do I write a proper subscript that is actually under the expression? $\endgroup$ – Simonsays Jun 12 '18 at 17:27
  • 1
    $\begingroup$ If the expression is an "operator" like $\lim$ or $\max$ or $\sum,$ it will be directly under the expression in a displayed context but not in an inline context, thus: $$ \lim_{x\,\to\,0}, \qquad \max_{x\,\in\,S}, \qquad \sum_{x=0}^n $$ but $ \lim_{x\,\to\,0}, \qquad \max_{x\,\in\,S}, \qquad \sum_{x=0}^n. $ But in an inline context you can write \max\limits_{x\,\to\0} and you get $\max\limits_{x\,\to\,0};$ thus using \limits changes that. In a displayed context you see this: $$ \int_a^b, $$ coded as \int_a^b, but $$\int\limits_a^b$$ is coded as$\,\ldots\qquad$ $\endgroup$ – Michael Hardy Jun 12 '18 at 17:40
  • 1
    $\begingroup$ $\ldots\,$\int\limits_a^b. You can make something get typeset according the conventions used with "operators" by writing \operatorname{abcdxyz}, thus: $$ \operatorname{abcdxyz} f(x). $$ If you use an asterisk, thus \operatorname*{abcdxyz}, then subscripts will behave as with $\lim$ and $\max,$ thus: $$ \operatorname*{abcdxyz}_{pqr} f(x) $$ is coded as \operatorname*{abcdxyz}_pqr}f(x). Finally, in other cases you can use \underset, thus $$\underset{pq} {ABCD} $$ is coded as \underset{pq} {ABCD}. $\qquad$ $\endgroup$ – Michael Hardy Jun 12 '18 at 17:43
  • $\begingroup$ @MichaelHardy thank you very much for that detailed answer! $\endgroup$ – Simonsays Jun 12 '18 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.