Does there exist a function $\mathbb{R}^2 \rightarrow \mathbb{R}$ whose directional derivatives are infinity in almost every direction, at all points?

Question

For $\mathbf{x}\in \mathbb{R}^2$ and $\mathbf{r}\in S^1$ (here $S^1=\{\mathbf{r}\in \mathbb{R}^2: \|\mathbf{r}\|=1\}$), let us write $f'(\mathbf{x};\mathbf{r})=\infty$ if $\liminf\limits_{t\rightarrow 0}\frac{f(\mathbf{x}+t\mathbf{r})-f(\mathbf{x})}{t}=\infty$.

My question is the following: does there exist a (not necessarily measurable) function $f:\mathbb{R}^2\rightarrow \mathbb{R}$, such that for all $\mathbf{x}\in \mathbb{R}^2$, the set $$\{\mathbf{r}\in S^1: f'(\mathbf{x};\mathbf{r})=\infty\text{ or }f'(\mathbf{x};-\mathbf{r})=\infty\}$$ has full measure as a subset of $S^1$? Here $S^1$ is equipped with the "arc length" measure.

Can a function $f:\mathbb{R} \rightarrow \mathbb{R}$ be "infinitely steep" on a set with non-zero Lebesgue outer measure? tells us that on any line parallell to $\mathbf{r}$, the set of points $\mathbf{x}$ for which $f'(\mathbf{x};\mathbf{r})=\infty\text{ or }f'(\mathbf{x};-\mathbf{r})=\infty$, is a null set. One is then tempted to argue that by Tonelli's, the set

$$A=\{(\mathbf{r}, \mathbf{x})\in S^1\times \mathbb{R}^2: f'(\mathbf{x};\mathbf{r})=\infty\text{ or }f'(\mathbf{x};-\mathbf{r})=\infty\}$$ is a null set. On the other hand, Tonelli's seems to imply that when $\{\mathbf{r}\in S^1: f'(\mathbf{x};\mathbf{r})=\infty\text{ or }f'(\mathbf{x};-\mathbf{r})=\infty\}$ has full measure for all $\mathbf{x}\in \mathbb{R}^2$, then $A$ has full measure, so there appears to be a contradiction. However, none of these applications of Tonelli are valid, since Tonelli only works when one integrates over a measurable set and a set may be non-measurable even if all of its sections are null sets. Indeed, there are non-measurable subsets of $\mathbb{R}^2$ for which all sections are singletons (see https://mathoverflow.net/questions/89375/sections-measure-zero-imply-set-is-measure-zero).

All of this is to say, I cannot straightforwardly deduce from the "$1$-dimensional case", i.e. the first link, that that there cannot exist a function with the properties as in my question. My hope is that I have not overlooked some obvious argument here and that either the answer to my question is negative for some "deeper" reason, or, even better, that the answer is affirmative!

Answer 1

No, there is no such function $f.$ The main idea behind this argument is to make use of intersections between different directions.

Restrict $f$ to the open unit circle $B.$ Define sets of directions \begin{align*} S_0&:=\{(x,y):0<x<y<4x/3\}\\ S_1&:=\{(-x,y):(x,y)\in D_0\}\\ S_2&:=\{(-x,-y):(x,y)\in D_0\}\\ S_3&:=\{(x,-y):(x,y)\in D_0\}\\ \end{align*}

Define $E_i$ to be the set of $x\in B$ such that there exist $i\in\{0,1,2,3\}$ and $r\in S_i$ and $r'\in S_{i+1\pmod 4}$ such that $f'(x,r)=\infty$ and $f'(x,r')=-\infty.$ So points in $E_0$ have some $+\infty$ derivative "up and to the right" and some $-\infty$ derivative "up and to the left". Considering the possible cases we have $B=\bigcup_{i=0}^3 E_i.$ Therefore there exists $i$ such that $E_i$ has outer measure at least $1/4.$ Composing $f$ with a rotation if necessary, assume $i=0$ works.

Pick $K$ such that the set $E_0\cap \{x:|f(x)|\leq K\}$ has outer measure $>1/5.$ This is justified by countable subadditivity. Define $$A_{j,\epsilon}=\{x:(\exists r\in S_0)(\forall t)0<t<\epsilon\implies (f(x+tr)-f(x))/t>j\\ \wedge(\exists r\in S_1)(\forall t)0<t<\epsilon\implies (f(x+tr)-f(x))/t<-j\}.$$ Pick $\epsilon_1>0$ such that $E_0\cap \{x:|f(x)|\leq K\}\cap A_{1,\epsilon_1}$ has outer measure $>1/5.$ Recursively, for each $k=2,3,\dots$ pick $\epsilon_k>0$ such that $Z_k:=E_0\cap \{x:|f(x)|\leq K\}\cap \bigcap_{j=1}^k A_{j,\epsilon_j}$ has outer measure $>1/5.$ Let $\overline{Z_k}$ denote the closure of $Z_k$ and define $C=\bigcap_{k=1}^\infty \overline{Z_k}.$ By Fubini's theorem, there exists $y$ such that $C_y:=\{x:(x,y)\in C\}$ has measure at least $1/10.$ Define $g:C_y\to\mathbb R$ by $g(x)=\limsup_{(x',y')\to (x,y)\\(x',y')\in Z_k\\k\to\infty}f(x,y).$

Lemma. If $x,x'\in C_y$ and $0<x'-x<\tfrac 3 5\epsilon_k$ then $(g(x')-g(x))/(x'-x)\geq k.$

Proof. Let $\delta>0$ be arbitrary. Pick points $z,z'\in Z_k$ within distance $\delta$ of $(x,y)$ and $(x',y)$ respectively and such that $|f(z)-g(x)|<\delta$ and $|f(z')-g(x')|<\delta.$ By definition of $A_{k,\epsilon_k}$ there are directions $r\in S_0$ and $r'\in S_1$ such that $(f(z+tr)-f(z))/t>k$ and $(f(z'+t'r')-f(z'))/t'<-k$ as long as $0<t,t'<\epsilon_k.$ There are unique $t,t'$ with $z+tr=z'+t'r'.$ If I've got the geometry correct, the condition $0<x'-x<\tfrac 3 5\epsilon_k$ gives $0<t,t'<\epsilon_k$ for $\delta$ sufficiently small. Hence $f(z')-f(z)>(t+t')k>(x'-x)k.$ Taking $\delta\to 0$ gives the result. $\square$

So we have a function $g:C_y\to\mathbb R$ with a kind of infinite derivative at each point of $C.$ Extend $g$ to $g':\mathbb R\to\mathbb R$ by linear interpolation in each connected component of $\mathbb R\setminus C.$ The lower Dini derivatives at each point $x$ are not $-\infty$ - either there is a sequence $x_n\in C$ converging to $x$ from below, so the Lemma gives $(g(x_n)-g(x))/(x_n-x)\to+\infty,$ or there is no such sequence and $g'$ is defined to be an affine function in some interval $(x-\epsilon,x).$ And the upper Dini derivatives are $+\infty$ at each point of $C_y$ that is a two-sided limit. This contradicts the Denjoy-Young-Saks theorem, which says that a.e. point with a $+\infty$ upper left Dini derivative must have a $-\infty$ lower right Dini derivative.