Let $f\in C^\infty(\mathbb R;\mathbb R)$ and let us define $g(x)=\vert f(x)\vert$. It is easy to verify that $g$ is locally Lipschitz-continuous function, but I would like to find an example of a smooth function $f$ where $g''$ is a distribution with positive order, that is a distribution which not a Radon measure.
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1$\begingroup$ It is not "easy to verify", because it's wrong in general. $f=g=x\mapsto x^2$ is not Lipschitz $\endgroup$– Johannes HahnCommented May 26, 2021 at 15:33
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1$\begingroup$ Do you mean locally Lipschitz? Or was $f$ supposed to be compactly supported or something? $\endgroup$– Nate EldredgeCommented May 26, 2021 at 15:38
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$\begingroup$ I meant locally Lipschitz-continuous. $\endgroup$– BazinCommented May 26, 2021 at 15:38
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1$\begingroup$ @bathalf15320 No, as every measure, $2\delta$ has order $0$. $\endgroup$– Jochen WengenrothCommented May 26, 2021 at 19:54
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1$\begingroup$ @Christian Remling: I want to find a $C^\infty$ function $f$ such that, with $g=\vert f\vert$, the distribution second derivative of $g$ is not a Radon measure. $\endgroup$– BazinCommented May 27, 2021 at 10:27
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2 Answers
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No, this is not possible. Clearly, the singular part of $g''$ results from points where $f(x)=0$ and $f'(x)\neq 0$. A given interval $[a,b]$ can contain at most countably many such points. Denote the set of these points by $X$. We need to show that $\sum_{x\in X} |f'(x)|$ is finite. Take any finite subset of $X$, ordered such that $a\le x_1<...<x_N\le b$. By Rolle's theorem, there is a zero of $f'$ between $x_i$ and $x_{i+1}$, hence $$|f'(x_{i+1})|\le \int_{x_i}^{x_{i+1}} |f''(x)|\,dx.$$ It follows that $$\sum_{i=1}^N |f'(x_i)|\le |f'(a)|+\int_a^b |f''(x)|\,dx.$$
answered May 27, 2021 at 18:59
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$\begingroup$ Can you please explain the first claim in more detail. In particular, what is the precise meaning of "results from"? (If you are claiming that $g''$ is a function on some open set, what is that set?) $\endgroup$ Commented May 27, 2021 at 21:53
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$\begingroup$ $g''$ is a function on any interval where $f\neq 0$. Because on any such interval $g$ is either $f$ or $-f$. $\endgroup$ Commented May 27, 2021 at 22:21
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$\begingroup$ @MichaelRenardy: Yes, thanks, I figured it out in the meantime myself. (What confused me was the case of a more complicated zero set, say a Cantor set.) Maybe a more explicit version of the start of your argument would be to write $(g'',\varphi)=\int g\varphi'' = \int_{f\not= 0}g\varphi''$, and then we integrate by parts twice on each component of the open set $\{ f\not= 0\}$. $\endgroup$ Commented May 28, 2021 at 12:39
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This is not an answer but rather a perhaps non-obvious example. A natural try is a smooth function which changes signs very often like $f(x)=h(x)\sin(\pi/x)$ with, e.g., $h(x)=\exp(-1/x)$ and $h(x)=0$ for $x<0$ . The idea is that the second derivative would contain (in some sense) a sequence $c_n \delta_{1/n}$ which might prevent order zero.
However, for a regular distribution of sign changes in points $x_n\to 0$ this will not work, because by partial integration $$ \langle g'',\varphi\rangle =\int \varphi''(x) |f(x)| dx =\sum_{n=1}^\infty (-1)^n \int_{x_n}^{x_{n+1}} \varphi''(x) f(x)dx$$ $$ = \sum_{n=1}^\infty (-1)^n \left(\left[\varphi'(x)f(x)-\varphi(x)f'(x)\right]_{x_n}^{x_{n+1}}+\int_{x_n}^{x_{n+1}} \varphi(x) f''(x)dx\right).$$ Since $x_n$ are zeros of $f$, the critical terms involving $\varphi'$ disappear and one can estimate $\langle g'',\varphi\rangle$ by the sup-norm of the test function $\varphi$, i.e., $g''$ has order $0$.
I don't know whether this phenomenon can be made a proof because the set of sign changes of a smooth function can probably be very wild.
answered May 27, 2021 at 5:16
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$\begingroup$ Thanks for your example: you are certainly right that oscillatory behaviour should be the key of this question, but I guess that for your function the series of $c_n$ is absolutely convergent (as $e^{-n}$). $\endgroup$– BazinCommented May 27, 2021 at 10:32
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$\begingroup$ Yes, indeed. This is another reason that the particular $f$ can't be an example. $\endgroup$ Commented May 27, 2021 at 11:18