Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$. Is the set of points at which $f$ is differentiable a Borel set?

Question

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a function. Is the set of points at which $f$ is differentiable a Borel set?

The answer is "yes" for $n=1$, even for arbitrary $f$ (not assumed to be measurable).

But what happens for $n>1$? The proof for $n=1$ (refer Characterization of sets of differentiability) does not seem to generalize easily to higher dimensions.

(The motivation for asking this question came from the Rademacher's theorem, which states that any Lipschitz map $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is differentiable almost everywhere. I was wondering if the points where $f$ is not differentiable forms a Borel set. Therefore, if the solution to the original question seems obscure, please feel free to help me with partial results, especially for the Lipschitz case.)

Answer 1

Here is an adaptation of the answer you linked: Suppose $f$ is continuous and consider any norm $\Vert\cdot\Vert$ on $\mathbb{R}^n$ and the associated matricial/operator norm $\Vert\cdot\Vert$. Then $f$ is differentiable at $x$ iff $$\exists M>0:\forall\epsilon>0:\exists\delta>0:\exists A\in\mathbb{Q}_{n\times n}(\Vert A\Vert<M):\forall h(\Vert h\Vert<\delta):\Vert f(x+h)-f(x)-Ah\Vert\leq\epsilon\Vert h\Vert,$$ where we take all numbers and vectors rational, and $\mathbb{Q}_{n\times n}$ is the set of $n\times n$ rational matrices.

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Here's the idea: If $f$ is differentiable at $x$, then take rational $M>\Vert df(x)\Vert+1$. Given $\epsilon>0$, let $A\in\mathbb{Q}_{n\times n}$ such that $\Vert df(x)-A\Vert\leq\epsilon/2$ (matricial norm). Since $\lim\frac{f(x+h)-f(x)-df(x)h}{\Vert h\Vert}=0$ there exists $\delta>0$ such that if $\Vert h\Vert<\delta$ then $\Vert f(x+h)-f(x)-df(x)h\Vert\leq\epsilon\Vert h\Vert/2$, so $$\Vert f(x+h)-f(x)-Ah\Vert\leq \epsilon\Vert h\Vert/2+\Vert Ah-df(x)h\Vert\leq\epsilon\Vert h\Vert.$$

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In the other direction, suppose $x$ is such that the expression above is satisfied. Let $\epsilon_n\to 0^+$ and take $\delta_n$ and $A_n$ satisfy the expression above, with respect to $\epsilon_n$. Since $f$ is continuous, the inequality holds for all $h$ with $\Vert h\Vert\leq\delta$ (possibly non-rational). Since $\Vert A_n\Vert<M$ for all $n$ then we can pass to a subsequence and assume that the matrices $A_n$ converge to some matrix $A$.

If $\Vert h\Vert<\delta_n$, then \begin{align*} \Vert f(x+h)-f(x)-Ah\Vert&\leq\Vert f(x+h)-f(x)-A_nh\Vert+\Vert A_nh-Ah\Vert\\ &\leq(\epsilon_n+\Vert A_n-A\Vert)\Vert h\Vert. \end{align*} Since $A_n\to A$ and $\epsilon_n\to 0$ then $f$ is differentiable at $x$ with $df(x)=A$.

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Addendum: I just noticed we can ommit the "$\exists M>0$" part: Let $B$ be the matrix we obtain for $\epsilon=1$. Then for any $\epsilon$, take the matrix $A_\epsilon$ associated to it. For $\epsilon<1$ and every sufficiently small $h$ we have $\Vert Bh- A_\epsilon h\Vert\leq (\epsilon+1)\Vert h\Vert\leq 2\Vert h\Vert$, so $\Vert A_\epsilon\Vert\leq\Vert B\Vert+2$. This is enough for the arguments above.