Function $f$ satisfying $f(x,y) = g(ax + by)$ and is continuous everywhere but not differentiable everywhere?

Question

I am looking for a function $f : \mathbb{R_{0}^2} \to \mathbb{R_0}$ (where $\mathbb{R}_0$ is the set of non-negative reals) such that $f(x,y) = g(ax + by)$ for some $a,b > 0$ and $g : \mathbb{R_0} \to \mathbb{R}_0$, $f$ is continuous everywhere but not differentiable everywhere?

You are free to choose any $a,b,g$ (they are not given). The construction I came up with is $f(x,y) = |x+y-1|$.

Is that correct? Can I get a better example that looks more natural? (What I mean by natural is, something that looks differentiable to an amateur eye, but is not.)

https://en.wikipedia.org/wiki/Weierstrass_function – Bob Dobbs Mar 27 '23 at 12:59 — Bob Dobbs, Mar 27 '23 at 12:59
I think your example is not differantiable on $x+y=1$. – Bob Dobbs Mar 27 '23 at 13:01 — Bob Dobbs, Mar 27 '23 at 13:01

score 1 · Accepted Answer · answered Mar 27 '23 at 13:03

If $f$ satisfies $f(x,y) = g(ax + by)$, then $f$ is differentiable/continuous iff $g : \mathbb R_{0} \to \mathbb R$ is (as long as you choose $a,b \neq 0$). So you are looking for a continuous, non-differentiable $g$. There are numerous examples, your function is one such example. A more natural function is of course a subjective question, but I don't see anything more suited.

Function $f$ satisfying $f(x,y) = g(ax + by)$ and is continuous everywhere but not differentiable everywhere?

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