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For $n \geq 2$, let $f: \mathbb R^n \to \mathbb R$ be differentiable. Is it possible that $\nabla f$ is everywhere discontinuous?

I believe in dimension $1$, $\nabla f$ has to be continuous on a dense set by the Baire category theorem. The same argument shows that the partials $\partial_i f$ are continuous along lines on a dense set. However, this doesn't seem to apply directly to $\nabla f$ itself, which is a map to $\mathbb R^n$.

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    $\begingroup$ Does not Baire category argument prove that each partial derivative, being a pointwise limit of continuous functions, is continuous on the complement of a Baire category 1 set? $\endgroup$
    – Fedor Petrov
    Commented Jun 25 at 1:38
  • $\begingroup$ @FedorPetrov Please see my reply to Willie Wong’s answer. $\endgroup$
    – Nate River
    Commented Jun 25 at 4:12
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    $\begingroup$ I do not understand what is the essence of the discussion. Is not a gradient just a vector of partial derivatives, and a point where they all are continuous is a point where the gradient is continuous? $\endgroup$
    – Fedor Petrov
    Commented Jun 25 at 9:46
  • $\begingroup$ @FedorPetrov Uh, you are right, for some reason I thought that the Baire argument showed only continuity of the partials along lines, but the argument holds for functions on $\mathbb R^n$ too, so the conclusion does follow from that… $\endgroup$
    – Nate River
    Commented Jun 25 at 10:08
  • $\begingroup$ A good reference for properties of derivatives is Differentiation of Real Functions by Andrew M. Bruckner. Theorem 2.1 on page 46 says that $S$ is the set of discontinuities of a derivative if and only if $S$ is a dense $G_\delta$ set. (Admittedly this is a theorem about 1-dimensional functions.) $\endgroup$
    – Timothy Chow
    Commented Jun 25 at 12:25

1 Answer 1

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To spell out Fedor's comment:

  1. For each $i$, you have $\partial_i f(x) = \lim_{n\to\infty} n \left( f(x + n e_i) - f(x) \right)$ is the pointwise limit of continuous functions, and hence is Baire class 1.

  2. Denote by $C_i$ the set of points in $\mathbb{R}^n$ where $\partial_if$ is continuous, then Baire's theorem says that $C_i$ is comeagre.

  3. Since the dimension $n < \infty$, you have that $C := \cap_{i=1}^n C_i$ is also comeagre, and hence dense in $\mathbb{R}^n$ by the Baire Category Theorem.

  4. Finally we use the calculus results: (a) if a point $x_0\in \mathbb{R}^n$ is such that for each $i\in \{1, \ldots, n\}$, the partial $\partial_i f$ exists on an open neighborhood of $x_0$ and is continuous at $x_0$, then $f$ is strongly differentiable at $x_0$, in the sense of [1]. (b) if a function $f$ is differentiable on an open set and strongly differentiable at $x_0$, then $\nabla f$ is continuous at $x_0$.

    Putting things together we conclude that $\nabla f$ is continuous on $C$.

References:

[1] - Strong Derivatives and Inverse Mappings, Nijenhuis.

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  • $\begingroup$ Thanks for the answer! Concerning step 4, is it not required that the partials exist and are continuous in a neighbourhood of $x_0$? That’s the classical theorem from calculus I remember. Without the continuity in a neighbourhood, it seems like the rate of differentiability of the partials might be different and thus the function is not strongly differentiable. $\endgroup$
    – Nate River
    Commented Jun 25 at 4:11
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    $\begingroup$ Yes, it is unfortunate that there are two possible interpretations of the phrase "strongly differentiable" (one in opposition to "weakly differentiable" in PDEs and distribution theory, and one that is more restrictive than Frechet differentiability). To that: I am just the messenger, so don't shoot. $\endgroup$
    – Willie Wong
    Commented Jun 25 at 6:27
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    $\begingroup$ Strong differentiability at a point is what is required for inverse/implicit function theorems. // A nice interpretation: let $\daleth_x^sf(y) = \frac{1}{s} (f(x + sy) - f(x))$ be the "zooming-in" of the function $f$ centered at $x$ by the scale $s$. Frechet differentiability is equivalent to $\lim_{s\to 0} \daleth_x^s f$ converging uniformly on bounded sets. Strong differentiability is equivalent to convergence in the Lipschitz seminorm. This shows how strong differentiability is just a little bit better than Frechet. $\endgroup$
    – Willie Wong
    Commented Jun 25 at 12:18
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    $\begingroup$ @rfloc: mathoverflow.net/questions/473821/… (there's also a link in the answer that Nate added). $\endgroup$
    – Willie Wong
    Commented Jun 26 at 3:15
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    $\begingroup$ @NateRiver You can also check this Q&A on the notion of "strong differentiability" mathoverflow.net/questions/404397 $\endgroup$
    – Pietro Majer
    Commented Jun 26 at 5:31

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