In a comment on this question, Tom Goodwillie proposed a notion of higher differentiability that I elaborate to something like the following:

Let $f:\mathbb{R}^n \to \mathbb{R}$. Let's say that $f$ is strongly twice differentiable at $x$ if there is a bilinear map $d^2 f_x:\mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}$ such that

$$ f(x+v+w) - f(x+v) - f(x+w) + f(x) = d^2f_x(v,w) + E(v,w){|v|}{|w|} $$

where $\lim_{v,w\to 0} E(v,w) = 0$. He pointed out that such a $d^2 f_x$, if it exists, is symmetric, since the LHS above is symmetric in $v$ and $w$. On the other hand, the usual proof of equality of mixed partials essentially shows that if $f$ is $C^2$ in a neighborhood of $x$, then it is strongly twice differentiable in this sense.

  1. Does strong twice differentiability at $x$ imply that $f$ is differentiable at $x$?
  2. If $f$ is differentiable in a neighborhood $U$ of $x$, does strong twice differentiability at $x$ imply that $f$ is twice differentiable in the usual sense, i.e. that $d f:U \to L(\mathbb{R}^n,\mathbb{R})$ is differentiable at $x$? Do we need $f$ to be strongly twice differentiable on a whole neighborhood $U$?
  3. Can you give any reference for a definition such as this?
  • 2
    $\begingroup$ For 1, no: $f$ might be linear over $\mathbb Q$ but not over $\mathbb R$. $\endgroup$ – Tom Goodwillie May 10 '14 at 2:02
  • $\begingroup$ Are you implicitly assuming continuity? $\endgroup$ – François G. Dorais May 10 '14 at 2:24
  • 1
    $\begingroup$ Francois' comment reminds me of how, in topology, to define $n$-connected, you first require $k$-connected for all $k < n$. That is, a simply connected space must first be connected, and so on. Most likely whatever hopes you have for this shortcut to twice-differentiability will have this particular problem. $\endgroup$ – Ryan Reich May 10 '14 at 5:18
  • 1
    $\begingroup$ You may also be interested in Tom's answer to my question here: mathoverflow.net/questions/80724/… $\endgroup$ – Steven Gubkin May 10 '14 at 17:44
  • $\begingroup$ @RyanReich It is possible to define $n$-connected without defining $k$-connected first: just ask that the $n$-truncation be contractible. I think there may be a similar solution here. $\endgroup$ – Mike Shulman May 11 '14 at 18:33
  1. No. Think of a $\mathbb Q$-linear map $f:\mathbb R\to \mathbb Q$.

  2. Yes. For small nonzero $v$ and $w$ write $$ E(v,w)=\frac{f(a+v+w)-f(a+v)-f(a+w)+f(a)-b(v,w)}{|v||w|}. $$

By assumption you have bilinear $b$ such that the limit of $E(v,w)$ as $(v,w)\to (0,0)$ is zero. If you also assume that $f$ is differentiable at $a+v$ for all small enough $v$ then for small nonzero $v$ you can take the limit of $E(v,w)$ as $w\to 0$ and get $$ \frac{df_{a+v}(\frac{w}{|w|})-df_{a}(\frac{w}{|w|})-b(v,\frac{w}{|w|})}{|v|}. $$ The limit of this as $v\to 0$ is zero. Therefore the derivative of $x\mapsto df_x$ at $x=a$ exists and corresponds to $b$.

Note that this implies that you get continuity of $df$ at a point if you have existence of $df$ in a neighborhood of the point and existence of $d^2f$ (in the sense we are exploring) at the point.

| cite | improve this answer | |
  • $\begingroup$ Really, really dumb question: what's a nontrivial linear map from $\mathbb{R}$ to $\mathbb{Q}$? $\endgroup$ – Deane Yang May 10 '14 at 19:41
  • $\begingroup$ I meant $\mathbb Q$-linear, sorry. I'll edit. But don't ask me for an example. $\endgroup$ – Tom Goodwillie May 10 '14 at 20:09
  • $\begingroup$ I understood it as $\mathbb{Q}$-linear, but does a nontrivial example exist? I'll settle for a proof that such an example exists. $\endgroup$ – Deane Yang May 11 '14 at 1:59
  • $\begingroup$ Do you believe that every $\mathbb Q$ vector space has a basis? $\endgroup$ – Tom Goodwillie May 11 '14 at 2:02
  • $\begingroup$ Oy. Stuff like this gives me a headache. It reminds me why analysts and geometers avoid studying differentiable functions per se. $\endgroup$ – Deane Yang May 11 '14 at 2:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.