Linear dilatation and differentiability of homeomorphisms of \(\mathbb R^n\) (Q2845870)

In this article, the author proves the following two results.NEWLINENEWLINESuppose that \(d\geq 2\). Let \(h: [0,\infty)\to [0,\infty)\) be a homeomorphism satisfying NEWLINE\[CARRIAGE_RETURNNEWLINE\frac{h(t)}{t^{d-1}}\to 0\; \text{as \(t\to 0^+\)}. CARRIAGE_RETURNNEWLINE\]NEWLINEThen there exists a homeomorphism \(f: [0,1]^d\to f([0,1]^d)\subset {\mathbb R}^d\) and a set \(S\subset [0,1]^d\) such thatNEWLINE\begin{itemize}NEWLINE\item \({\mathcal H}^h(S)=0\)NEWLINE\item \(h_f(x)=1\) for all \(x\in [0,1]^d\setminus S\),NEWLINE\item \(H_f(x)=\infty\) for all \(x\in [0,1]^d\),NEWLINE\item \(f\) is nowhere differentiable on \([0,1]^d\),NEWLINE\end{itemize}NEWLINEwhere NEWLINE\[CARRIAGE_RETURNNEWLINEH_f(x):=\limsup_{r\to 0} \frac{\max\{|f(x)-f(x)|\; |x-y|\leq r\}}{\min\{|f(x)-f(x)|\; |x-y|\geq r\}}, CARRIAGE_RETURNNEWLINE\]NEWLINE\[CARRIAGE_RETURNNEWLINEh_f(x):=\liminf_{r\to 0} \frac{\max\{|f(x)-f(x)|\; |x-y|\leq r\}}{\min\{|f(x)-f(x)|\; |x-y|\geq r\}}. CARRIAGE_RETURNNEWLINE\]NEWLINENEWLINEThe second result is as follows.NEWLINENEWLINESuppose that \(d\geq 1\). Let \(h: [0,\infty)\to [0,\infty)\) be a homeomorphism satisfying NEWLINE\[CARRIAGE_RETURNNEWLINE\frac{h(t)}{t^{d-1}}\to 0\; \text{as \(t\to 0^+\)}. CARRIAGE_RETURNNEWLINE\]NEWLINEThen there exists a continuous function \(f: [0,d]\to {\mathbb R}\) and a set \(S\subset [0,1]^d\) with \({\mathcal{H}}^h(S)=0\) such that NEWLINE\begin{itemize}NEWLINE\item \(\liminf_{r\to 0} \frac{\max\{|f(x)-f(x)|\; |x-y|\leq r\}}{r}=0\) for all \(x\in [0,1]^d\setminus S\),NEWLINE\item \(\limsup_{r\to 0} \frac{\max\{|f(x)-f(x)|\; |x-y|\leq r\}}{r}=\infty\) for all \(x\in [0,1]^d\).NEWLINE\end{itemize}NEWLINEThe first result shows that in the following statement, which can be found in [\textit{S. Kallunki} and \textit{O. Martio}, Proc. Am. Math. Soc. 130, No. 4, 1073--1078 (2002; Zbl 0990.30014)], one cannot replace \(H_f\) by \(h_f\):NEWLINENEWLINESuppose that a homeomorphism \(f: \Omega \to {\mathbb R}^n\) satisfies \(H_f<\infty\) a.e. in \(\Omega\). Then \(f\) is a.e. differentiable.NEWLINENEWLINEFurther the result shows that the following statement from [\textit{S. Kallunki} and \textit{P. Koskela}, Am. J. Math. 122, No. 4, 735--743 (2000; Zbl 0954.30008)] is quite sharp with respect to the size of the exceptional set.NEWLINENEWLINESuppose that \(S\) is a subset of a domain \(\Omega\) with \(\sigma\)-finite \((d-1)\)-dimensional measure, \(f: \Omega\to f(\Omega)\) is a homeomorphism, \(h_f(x)<\infty\) for all \(x\in \Omega\setminus S\) and there is a \(K<\infty\) such that \(h_f(x)\leq K\) a.e. on \(\Omega\). Then \(f\) is quasiconformal on \(\Omega\).NEWLINENEWLINEIn [Proc. Am. Math. Soc. 134, No. 9, 2667--2675 (2006; Zbl 1100.26007)], \textit{Z. Balogh} and \textit{M. Csörnyei} constructed for every \(d\geq 1\) a nowhere differentiable, continuous function \(f: [0,1]^d\to {\mathbb R}\) such that \(l_f(x)=0\) for a.e. \(x\in [0,1]^d\). The second result of the author strengthens the result by Balogh and Csörnyei and shows that in the following result by Balogh and Csörnyei in the above cited paper, \(E\) can have Hausdorff dimension at most \(d-1\).NEWLINENEWLINEAssume that \(l_f(x)<\infty\) on \(\Omega\setminus E\), where \(\Omega\subset {\mathbb R}^d\) is a domain, \(f: \Omega\to {\mathbb R}\) is continuous, and \(E\) has \(\sigma\)-finite \((d-1)\)-dimensional measure. Assume that \(l_f\in L^p_{\text{loc}}(\Omega)\) where \(p>d\) if \(d>1\) and \(p=1\) if \(d=1\). Then \(f\) is differentiable a.e. on \(\Omega\).NEWLINENEWLINEThe author proves the theorems by first detailing the construction required in the second result in the case \(d=1\). The general statements then follow easily.