1

I'm trying to better understand how things get proved. I ran into a proof of one of the corollaries of Mean Value Theorem:

Suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f'(x)>0$ at each point $x\in (a,b)$, then $f$ is increasing on $[a,b]$. If $f'(x)<0$ at each point $x\in (a,b)$, then $f$ is decreasing on $[a,b]$.

The proof in the book uses Mean Value Theorem to show that $f(x_2)-f(x_1)>0$. But why did we need this theorem? Couldn't we repurpose our usual quotient of the difference limit? Since $f(x)$ is differentiable:

$$ \lim_{x_2 \to x_1^+}\frac{f(x_2)-f(x_1)}{x_2-x_1}>0 \implies f(x_2)-f(x_1)>0 $$

And since this is true for all $x_1\in(a,b)$, then it's true on the whole interval.

Stanislav Bashkyrtsev
  • 617
  • 4
  • 12
  • 9
    The statement on the RHS of your implication is false. Clearly, $\lim_{x_2 \rightarrow x_1^+} f(x_2) - f(x_1) = 0$ due to continuity. – Hyperbolic PDE friend May 21 '24 at 12:39
  • 1
    The fact that the limit if $f(x_2)-f(x_1)$ is positive doesn't mean that it is positive for all $x_2,$ just for $x_2$ close to $x_1.$ – Thomas Andrews May 21 '24 at 12:57
  • @ThomasAndrews Right, but you can extend that to everywhere using completeness. – Vivaan Daga May 21 '24 at 13:05
  • Proving that with completness sounds more complicated than via the mean value theorrem. Can you write that completeness proof? Waving hands is one thing, but the actual proof is going to be more work. A real analysis proof would use compactness of $[a,b],$ but compactness is not usually covered in calculus. – Thomas Andrews May 21 '24 at 13:10
  • @HyperbolicPDEfriend, that's interesting.. I think that in more than one proof I've seen reasoning about whether the result of the limit is positive or negative judging by the sign of the enumerator/denominator. E.g. to prove that $f(x)=0$ at the local extrema. I guess I should've simply written $f(x_2)-f(x_1)$ w/o the $lim$, right? – Stanislav Bashkyrtsev May 21 '24 at 13:36
  • @ThomasAndrews, got it, thanks! – Stanislav Bashkyrtsev May 21 '24 at 13:43
  • 1
    Without the $\lim$ indeed. First, you argue using continuity that $\tfrac{f(x_2)-f(x_1)}{x_2-x_1}>0$ in a neighborhood of $x_1$ and then the conclusion is immediate. – Hyperbolic PDE friend May 21 '24 at 13:47
  • @HyperbolicPDEfriend doesn't that require a compactness argument? – Steen82 May 21 '24 at 15:06
  • @Steen82 at least I can't think of one that I would need. Which one do you mean? – Hyperbolic PDE friend May 21 '24 at 15:31
  • @HyperbolicPDEfriend to prove it is increasing in the whole interval, not just in a neighborhood of each point. But maybe you have a different idea in mind. – Steen82 May 21 '24 at 15:36
  • @Steen82, logically in mind if $f(x_2) > f(x_1)$ and $f(x_3) > f(x_2)$, then $f(x_3) > f(x_1)$, and therefore this should cover the whole interval. But I guess I'm missing something deeper since apparently this argument doesn't count? :) – Stanislav Bashkyrtsev May 21 '24 at 16:00
  • @Stanislav Bashkyrtsev Zeno’s paradox could get in the way. Compactness is one way around that. – Steen82 May 21 '24 at 22:06
  • The result in question is not trivial and it requires completeness of reals in some form. Using mean value theorem is easier because it hides the completeness thing in mean value theorem itself. Another proof uses a combination of Rolle's and intermediate value theorem and then we have a direct proof via completeness. See https://math.stackexchange.com/a/960585/72031 for more details. – Paramanand Singh May 22 '24 at 18:04
  • The problem with your approach is that by assumption $x_1,x_2$ are fixed but arbitrary points of the interval and hence $x_2$ can't tend to $x_1$. – Paramanand Singh May 22 '24 at 18:08
  • See also https://math.stackexchange.com/a/2047743/72031 – Paramanand Singh May 22 '24 at 18:11

0 Answers0