Definition. Let us say that a function $f\colon \mathbb R^d\to \mathbb R$ is differentiable along hyperplanes in the point $0\in \mathbb R^d$, if $f\circ \varphi\colon \mathbb R^{d-1}\to \mathbb R$ is (totally) differentiable in $0\in \mathbb{R}^{d-1}$ for all linear maps $\varphi\colon \mathbb R^{d-1}\to \mathbb R^d$.

For $d=2$ there are some standard counterexamples to show that this notion is weaker than total differentiablity, e.g.: $$ f(x,y)=\frac{xy^2}{x^2+y^2} $$ This is of the form $f=p/q$ for homogeneous polynomials with $\deg p = 1+\deg q$. Differentiability along lines follows from the fact that homogeneous polyomials in one variable are automatically divisible, if the degree of the enumerator is larger than that of the denominator.

Question. Suppose $f\colon \mathbb R^3\to \mathbb R$ is differentiable along planes in the point $0$. Is $f$ then also totally differentiable in $0$?

If we restrict to homogeneous rational functions $f=p/q$ with $\deg p \ge 1+\deg q$ one can also ask for the stronger property that $q\circ \varphi$ shall divide $p\circ \varphi$ in the polynomial ring of $2$ variables and wonder whether this already implies that $q$ divides $p$. I've asked this algebraic question on math.stackexchange but I haven't received a satisfying answer (it seems tricky that we deal with not necessarily reducible polynomials over $\mathbb R$). Even if settled, this doesn't answer the full question asked here.

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    $\begingroup$ Nice question… and something I ran into too. I instead have differentiability on a dense set of hyperplanes (in the Grassmannian topology) and some additional regularity, and I want to conclude total differentiability. $\endgroup$
    – Nate River
    Commented Jun 11 at 9:36
  • $\begingroup$ Now you know your enemy. :) What kind of regularity do you impose? $\endgroup$
    – Jan Bohr
    Commented Jun 12 at 15:57
  • $\begingroup$ So far, continuous and differentiable a.e., though I hope more regularity will follow from my set up… $\endgroup$
    – Nate River
    Commented Jun 14 at 3:47

1 Answer 1


No, $f$ need not be continuous at $0$, here is a counterexample.

Let $(A_n)_n$ be a sequence of balls $A_n=B(p_n,r_n)$ with $0\not\in A_n$, $p_n\to0$, $r_n\to0$ and such that, for any $n_1<n_2<n_3\in\mathbb{N}$ and any vectors $v_i\in A_{n_i}$, the vectors $v_1,v_2,v_3$ are linearly independent (see a construction below).

And let $\phi_n$ be a smooth bump function supported inside $A_n$ with $\|\phi_n\|_\infty=1$. Then $f=\sum_n\phi_n$ is not continuous at $0$, but its restriction to any plane is a sum of two smooth bump functions, so it's smooth.

An example of a sequence of balls as above:

Let $p_n=\left(\frac{1}{10^n},\frac{1}{100^n},\frac{1}{1000^n}\right)$ and $r_n=\frac{1}{1000000^{n}}$, for $n=10,100,1000,\dots$

Now let $n_1<n_2<n_3\in\{10,100,1000,\dots\}$ and let $|\varepsilon_{i,j}|<\frac{1}{1000000^{n_i}}$, it will be enough to check that the following matrix is invertible:

$$\begin{pmatrix} \frac{1}{10^{n_1}}+\varepsilon_{1,1}&\frac{1}{100^{n_1}}+\varepsilon_{1,2}&\frac{1}{1000^{n_1}}+\varepsilon_{1,3}\\ \frac{1}{10^{n_2}}+\varepsilon_{2,1}&\frac{1}{100^{n_2}}+\varepsilon_{2,2}&\frac{1}{1000^{n_2}}+\varepsilon_{2,3}\\ \frac{1}{10^{n_3}}+\varepsilon_{3,1}&\frac{1}{100^{n_3}}+\varepsilon_{3,2}&\frac{1}{1000^{n_3}}+\varepsilon_{3,3}\\ \end{pmatrix}$$

This determinant of this matrix is a sum of $48$ terms, but the term $\frac{1}{10^{n_3}100^{n_2}1000^{n_1}}$ is bigger in absolute value than all other terms by a factor $>100$ (it's enough to check this for the other five terms not involving the $\varepsilon_{i,j}$), so the determinant is not $0$.

  • $\begingroup$ Great argument! You can also modify it to be continuous at $0$, by setting $f(x)=\sum_n a_n\phi_n(x)$ for a sequence $a_n\to 0$ and taking $\phi_n(p_n)=1$. Since $N(\epsilon)=\min\{n\in \mathbb N: A_n\cap B(0,\epsilon)\}\to \infty$, as $\epsilon\to 0$, we have $\sup_{|x|<\epsilon}|f(x)|\le a_{N(\epsilon)} \to 0$ as $\epsilon\to 0$, which gives continuity. If $f$ was differentiable at $0$, its differential would have to be zero and thus $b_n=f(p_n)/|p_n|\to 0$. But $b_n=a_n/|p_n|$ and we get a contradiction by choosing e.g. $a_n=|p_n|$. $\endgroup$
    – Jan Bohr
    Commented Jun 12 at 13:10
  • $\begingroup$ Yes, the construction provides a lot of flexibility, you can make it continuous if you want (I preferred to make it discontinuous as that made it, in my opinion, a stronger counterexample with the given hypotheses) $\endgroup$
    – Saúl RM
    Commented Jun 12 at 13:13
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    $\begingroup$ I was trying to see if the example survives strengthening the hypothesis of the question (i.e. differentiable along planes and continuous at zero). The next step would be to ask for a rational counterexample (which might not exist) -- but I'll post this as a separate question. $\endgroup$
    – Jan Bohr
    Commented Jun 12 at 13:18

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