Good evening everybody.

The following question came to my mind while reading an old American Math. Monthly paper entitled "Double integrals as initial value problems".

Let $f:\mathbb R^2\to\mathbb R$ be a function. Assume that $f$ has the following property: for any $(x,y)\in\mathbb R^2$, it holds that $$\frac{f(x+h,y+h)-f(x+h,y)-f(x,y+h)+f(x,y)}{h^2}\to 0\qquad\hbox{as $h\to 0$}\tag{$*$} $$ Does it follow that $f$ has the form $f(x,y)=u(x)+v(y)$, where $u$ and $v$ are $1$-variable functions?

Note that the answer is "Yes" if $f$ is known to be twice differentiable, since in this cas $(*)$ means that $\frac{\partial^2f}{\partial x\partial y}\equiv 0$. The point is that no regularity assumption is a priori made on $f$.

Thanks in advance for any answer.

  • 2
    $\begingroup$ Your title does not really make sense... $\endgroup$
    – amsmath
    Commented Sep 7, 2018 at 15:41
  • $\begingroup$ Hopefully, the question does... $\endgroup$
    – Etienne
    Commented Sep 7, 2018 at 15:53
  • 1
    $\begingroup$ The question does already. But something that doesn't exist cannot be zero because zero exists. $\endgroup$
    – amsmath
    Commented Sep 7, 2018 at 15:56
  • $\begingroup$ What about taking this just as a joke (possibly not a good one)? $\endgroup$
    – Etienne
    Commented Sep 7, 2018 at 19:44
  • $\begingroup$ Etienne It wasn't a joke. I like your question (upvoted it), but your title still is nonsense. $\endgroup$
    – amsmath
    Commented Sep 8, 2018 at 14:10


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