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It is well-known that every monotone function $f : \mathbb{R} \to \mathbb{R}$ is differentiable almost everywhere (with respect to Lebesgue measure). It is also known if $E$ has measure $0$, then there exists a continuous, monotone function that is differentiable at no point of $E$.

The proofs of these results, at least those I have seen, are a bit too technical for first-year calculus students to digest. On the other hand, I'm willing to settle for a much weaker result:

Every monotone function is differentiable at some point.

Is there an elementary way avoiding all measure theory, and preferably also avoiding Baire's theorem or other topological concepts that won't be familar to most calculus students showing this?

Edit: to clarify, I'd like to avoid integrals too. See the motivating example further down. If we have the Riemann integral at our disposal, there are much simpler ways to define $x^y$ which gives differentiability with less effort.

If you need to assume continuity to simplify the proof, that's ok.

(One possible) motivation

Let's try to define exponentiation $x^y$ for $x >0$, $y \in \mathbb{R}$.If $y$ is a positive integer, of course $$ x^n = \underbrace{x \cdot x \cdots x}_{n~\text{times}}. $$ Assuming we have dealt with $q$:th roots of real numbers, the extension to rational exponents is also straight-forward: $$ x^{p/q} = \big(\sqrt[q]{x}\big)^p $$ Finally, it's a little tedious, but not too bad to extend first to negative rational numbers $x^{-r} = 1/x^r$ and finally to real exponents by "continuity". Doing all this will give (for a fixed $x$) a continuous monotone function $f(y) = x^y$ satisfying the functional equation $$f(y_1+y_2) = f(y_1)f(y_2).$$ How do we prove that this function $f$ is differentiable? (See Show $\lim\limits_{h\to 0} \frac{(a^h-1)}{h}$ exists without l'Hôpital or even referencing $e$ or natural log for an expanded version of this question.) Among the answers is a clever way to do it using convexity, but I'm still curious if it's possible to give an elementary solution by just exploiting monotonicity.

If we can show that $f$ is differentiable at a single point, then the functional equation implies diffentiability everwhere.

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  • $\begingroup$ Is there any proof avoiding measure theory but using Baire's category theorem? Riesz's proof is elementary, but certainly not for those who is taking 1st year calculus course. $\endgroup$
    – Yai0Phah
    Nov 11, 2015 at 10:49
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    $\begingroup$ I believe it was not known whether monotone functions had to be differentiable anywhere until Lebesgue showed they're differentiable almost everywhere, but I'm not sure right now and don't have time to look into this. But if I'm correct about this historical issue, then I suspect there isn't a known easier way to prove this than by proving the stronger Lebesgue result (otherwise someone would likely have thought of it in the 1880s or 1890s). Reminds me of squaring the circle. We only need that $\pi$ is not ruler-and-compass constructable, but the only proofs known show $\pi$ is transcendental. $\endgroup$
    – Dave L. Renfro
    Nov 18, 2015 at 21:31
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    $\begingroup$ I would suspect that there isn't an "elementary" way of showing this, even with some more advanced topological language. Sets of measure zero can still be "large" things topologically. To demonstrate the difficulty, you might consider $$E=\cap_{m=1}^\infty\cup_{n=1}^\infty(r_n-2^{-n-m}, r_n+2^{-n-m}).$$ where $\{r_n\}$ is an enumeration of the rationals. Demonstrating a monotone function is differentiable somewhere doesn't seem to be easier than determining whether or not $E$ is not all of $\mathbb{R}$. We seem to only know by measure theory. $\endgroup$
    – user123641
    Nov 21, 2017 at 22:25

1 Answer 1

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Only a partial answer, for your functional equation $f(y_1+y_2)=f(y_1)f(y_2)$ assuming that $f$ is continuous (and not the zero function): Put $\displaystyle F(x)=\int_0 ^x f(t)dt$. Then $$F(x+y)-F(x)=\int_x^{x+y}f(t)dt=\int_0^y f(t+x)dt=f(x)F(y)$$ Now if you fix a $y$ such that $F(y)$ is not $0$, you have that $f(x)$ is differentiable, as $F$ is differentiable.

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  • $\begingroup$ That's fine for the motivating example (I should probably have said that I prefer solutions not relying on integrals either. Once we have integrals, it's a lot easier to define $\ln x$ as $\int_1^x \frac{dt}{t}$, get differentiabilty for free and go on to exponential functions from there.) $\endgroup$
    – mrf
    Nov 11, 2015 at 10:08

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