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The (simplest form of the) Rademacher theorem reads as follows:

Any Lipschitz continuous function $f: \mathbb{R} \to \mathbb{R}$ is Lebesgue-almost everywhere differentiable.

In other words: If the finite difference $\Delta_h^1[f](x) := f(x+h)-f(x)$ satisfies $$|\Delta_h^1[f](x)| \leq C |h|, \qquad x,h \in \mathbb{R} \tag{1}$$ for some absolute constant $C>0$, then $f$ is almost everywhere differentiable.

Question: Is there a generalization for derivatives of second order?

More precisely, if we replace the finite difference $\Delta_h^1$ by a second-order difference, for instance

$$\Delta_h^2[f](x) := f(x+h)-2f(x)+f(x-h), \tag{2}$$

then does

$$|\Delta_h^1[f](x)| \leq C_1|h| \qquad \quad |\Delta_h^2[f](x)| \leq C_2 |h|^2, \qquad x,h \in \mathbb{R} \tag{3}$$

imply that $f$ is almost everywhere twice differentiable? If not, then what additional information on the regularity gives the estimate $(3)$ compared to the almost everyhwere differentiability which follows from the weaker assumption $(1)$?

The obvious idea would be to try to apply the Rademacher theorem twice (first to $f$ and then to its derivative $f'$), but unfortunately the estimate

$$|\Delta_h^1[f'](x)| \leq C |h|$$

will, in general, only hold for Lebesgue almost every $x,h$, and therefore it is not possible to apply the Rademacher theorem directly to $f'$.

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I believe the answer is yes. Writing $D$ for the derivative in the sense of distributions, and $f'$ for the pointwise derivative:

It's easy to see that $\frac1{h^2}\Delta_h^2 f\to D^2f$ in the sense of distributions (or $2D^2f$ or $\frac 12D^2f$ or whatever it is). So the hypothesis implies $$D^2f\in L^\infty.$$This implies that $Df$ is a Lipshitz function. In particular, $Df$ is continuous, which implies that $f$ is differentiable everywhere and $f'=Df$. Since $Df$ is Lipschitz this implies that $f'$ is differentiable almost everywhere.

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  • $\begingroup$ Thanks a lot for your answer! Right now I don't see any flaw in your reasoning (which is great :)), but I will think it over carefully tomorrow with a fresh mind. $\endgroup$ – saz May 27 '18 at 18:36
  • $\begingroup$ @saz Yeah, I said I believe the answer's yes - there are a lot of details there... $\endgroup$ – David C. Ullrich May 27 '18 at 19:14

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