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

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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 $b>0$:

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

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.

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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?

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  • $\begingroup$ 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
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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 $0.$

POSTSCRIPT:

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}

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  • $\begingroup$ 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
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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ $\ldots\,$\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

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