2
$\begingroup$

I'm currently working through chapter 5 of baby Rudin and came across

Theorem 5.8 Let $f$ be defined on $[a,b]$; if $f$ has a local maximum at a point $x\in(a,b)$, and if $f'(x)$ exists, then $f'(x)=0$.

Now, this seemed like a theorem that shouldn't be too hard to prove myself so I made an attempt before moving on:

Proof: Consider $\delta>0$ such that $d(x,q)<\delta \implies f(q)\leq f(x)$. Suppose $f'$ exists, then $\lim_{t\rightarrow x} \frac{f(t)-f(x)}{t-x}$ exists. Let $d(t,x)<\delta$, then $f(t)-f(x)$ is nonpositive. Suppose that $f(t)-f(x)<0$, then $$\lim_{t\rightarrow x^+} \frac{f(t)-f(x)}{t-x}\neq\lim_{t\rightarrow x^-} \frac{f(t)-f(x)}{t-x},$$ which contradicts the existence of $f'$. Therefore $f'=0$

But I'm not entirely sure that I'm allowed to do everything I do. Specifically, I'm not certain whether the use of unequal right and left limit is allowed in the way I do it here.

$\endgroup$
  • $\begingroup$ When you assume that $f(t) - f(x) < 0$, are you assuming that for all $t$ near $x$, for at least one $t$, or something else? $\endgroup$ – user49640 Jun 12 '17 at 10:36
  • 2
    $\begingroup$ You find a contradiction and then you conclude that $f'=0$. I reckon you mean $f'(x)=0$. But such conclusion can only be made if the contradiction is based on the hypothese $f'(x)\neq0$. In the proof you do not make that hypothese. $\endgroup$ – drhab Jun 12 '17 at 10:45
  • $\begingroup$ @user49640 For all, based on the definition of a local maximum. $\endgroup$ – Mitchell Faas Jun 12 '17 at 11:11
  • 1
    $\begingroup$ @MitchellFaas That's not really the negation of what you're trying to prove. In fact, if $f(t) = -t^2$, then it will certainly be the case that $f(t) - f(0)$ is negative for $t \ne 0$. Since you know that $f'(x)$ exists, the negation of what you're trying to prove would be to assume $f'(x) \ne 0$. $f'(x)$ is defined as a certain limit, but a nonzero function can have a zero limit. This is the case when you write down the limit defining $f'(0)$ in the example I gave. $\endgroup$ – user49640 Jun 12 '17 at 11:13
1
$\begingroup$

Here is the proof of Fermat’s Theorem.

Paul's Online Math Notes

http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeAppsProofs.aspx

$\endgroup$
  • 1
    $\begingroup$ This might be suitable to place as a comment. Definitely not as an answer. $\endgroup$ – drhab Jun 12 '17 at 10:40
  • 1
    $\begingroup$ Sorry I don't have 50 reputation to comment. $\endgroup$ – Ray Cheng Jun 12 '17 at 10:44
  • $\begingroup$ Rudin's proof is much neater. This was just me trying to prove it before actually reading the given proof. $\endgroup$ – Mitchell Faas Jun 12 '17 at 11:27
  • $\begingroup$ @Mitchell Faas Actually I'd say that the proof given above is more explicit than Rudin's (having just looked at both). I like Rudin's book quite a lot myself, but it's definitely important to consult other sources - like the one Ray gave - if something he does has you confused. Rudin will sometimes condense non-trivial steps in ways that the reader is expected to unpack, and you can miss important details if you aren't careful. $\endgroup$ – Chris Jun 12 '17 at 17:08

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.