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So far, I have shown that $f(x)=(x^{2} + 2x)W(x)$ is differentiable at 0, where the parameters of the Weierstrass function are $a = \frac{1}{2}$ and $b = 3$, and $f'(0)$ = 4. I am fairly certain that $f(x)$ is non-differentiable everywhere else on the domain $[-1, 1]$ because $x^{2} + 2x$ has no other roots in that interval, and when I calculated the derivative of $f$ at $0$, I noticed that the fact that $x = 0$ is a root of $x^{2} + 2x$ led to the simplification that allowed me to calculate the derivative, although I have not proven this statement yet. My question is, given the above information, can I say that $f(x)$ is non-monotonic for every subinterval in its domain that contains 0? If so, I would like to prove it, but I am not aware of how to approach the proof. This question is inspired by the following post.

Eliot
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  • $(x^{2} + 2x)W(x)$ is almost certainly non-monotonic in every interval (and $x^2 + 2x$ can almost certainly be replaced by any polynomial, probably by any locally Lipschitz function), and I suspect this can be deduced easily from known non-Hölder continuity properties of the Weierstrass function obtained by G. H. Hardy in 1916. See "but neither of them satisfies ..." at the end of 1.33 on p. 303 of Weierstrass’s non-differentiable function. (continued) – Dave L. Renfro Oct 28 '21 at 06:19
  • I'll be too busy the next few days to look at this sufficiently to write an appropriate answer, so if anyone else wishes to write an answer based on what I said about Hardy's result, that's fine with me. – Dave L. Renfro Oct 28 '21 at 06:20

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If $f$ were differentiable at some $x$ different from $0$,then $W$ would be differentiable at $x$ since it would be the quotient of two differentiable functions. Hence $f$ is only differentiable at the origin. Now if $f$ were monotone in some interval, then by the Lebesgue differentiation theorem, $f$ would be differentiable at a.e. Point in that interval, which is a contradiction. There is probably a more direct way to do this…

Gio67
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