Non-continuous higher differentiability, II

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?
• For 1, no: $f$ might be linear over $\mathbb Q$ but not over $\mathbb R$. – Tom Goodwillie May 10 '14 at 2:02
• Are you implicitly assuming continuity? – François G. Dorais May 10 '14 at 2:24
• 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. – Ryan Reich May 10 '14 at 5:18
• You may also be interested in Tom's answer to my question here: mathoverflow.net/questions/80724/… – Steven Gubkin May 10 '14 at 17:44
• @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. – 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.

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