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GRE 9768 #64

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From here

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How do you prove that? I tried this approach based on a question and solution in the Princeton GRE exam

The question:

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The solution:

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(Btw, it's supposed to be $f'(y)$ not $f'(x)$?)

Approach:

$$0 \le \lim_{x \to y} \frac{|f(x) - f(y)|}{|x-y|} \le \lim_{x \to y} \frac{E|x-y|}{|x-y|} = E$$

Thus $|f'(y)| = \lim_{x \to y} \frac{|f(x) - f(y)|}{|x-y|}$ exists, namely, it is some constant between $0$ and $E$ and hence $f'(y)$ exists?

If the above can work, why only almost everywhere? Where is compactness used?

Otherwise, how can I approach this please?

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    $\begingroup$ By the way, the given answer to the second problem is wrong in two ways. First, the answer is (E), since condition I vacuously implies $f(x)$ is constant (anything follows from a falsehood). Second, the proof that $f'(x)=0$, as written, starts by assuming that $f'(x)$ exists at all (though you can modify the proof to not make this assumption). $\endgroup$
    – Eric Wofsey
    Oct 25, 2016 at 5:47
  • $\begingroup$ @EricWofsey I knew it was vacuous! Also how to modify please? It just looks like we have to remove the $f'(x)$ (supposed to be $f'(y)$?) there then show that the nonnegative limit is zero and then since it exists, it is equal to $f'(x)$ for all x (supposed to be $f'(y)$ for all y?) ? $\endgroup$
    – BCLC
    Oct 25, 2016 at 5:51
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    $\begingroup$ Oh, I didn't even notice that the variables $x$ and $y$ were swapped, that's a third error. Yeah, your fix is right--you just prove directly that the limit defining $f'(y)$ exists and is $0$ because $\left|\frac{f(x)-f(y)}{x-y}\right|\leq|x-y|$ and so whenever $|x-y|<\epsilon$ the difference quotient is within $\epsilon$ of $0$. $\endgroup$
    – Eric Wofsey
    Oct 25, 2016 at 5:54
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    $\begingroup$ Re problem 64, the easiest way to see that III fails is to look at $f(x) = \sqrt x,$ or something similar. It's difficult to see why the solution author would suggest groping through all of that as the first order of business. $\endgroup$
    – zhw.
    Oct 25, 2016 at 16:28

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Your approach doesn't work, all that you get from the inequalities you wrote is that the difference quotients are bounded, that doesn't imply that the limit exists.

The a.e. differentiability of Lipschitz continuous functions - defined on (open, or otherwise nice enough to talk about differentiability) subsets of $\mathbb{R}^n$ - is Rademacher's theorem, which is not trivial to prove.

For $n = 1$, we have a less mighty hammer available. Lipschitz continuous functions defined on an interval $I \subset \mathbb{R}$ are absolutely continuous, and absolutely continuous functions are almost everywhere differentiable (and they are the integral of their a.e. defined derivative). Not that this hammer is trivial either.

Compactness of the domain plays no role whatsoever for either argument.

As has been mentioned in the comments, an easier example to show that III doesn't need to hold in question 64 would be $f(x) = x^{\alpha}$ for some $\alpha \in (0,1)$.

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  • $\begingroup$ Thanks Daniel Fischer♦! ^-^ $\endgroup$
    – BCLC
    Oct 23, 2018 at 3:25

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