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If you know the solution for this exercise, I would appreciate a HINT:

Let $f:U\longrightarrow\mathbb{R}$ a function defined in an open subset $U$ of $\mathbb{R}^m$. Given $p\in U$, suppose that, for every path $\lambda:(-\epsilon,\epsilon)\longrightarrow U$, with $\lambda(0)=p$, that has a velocity vector $v=\lambda '(0)$ at $t=0$, the composed path $f\circ\lambda:(-\epsilon,\epsilon)\longrightarrow\mathbb{R}$ also has a velocity vector $(f\circ\lambda)'(0)=Tv$, where $T:\mathbb{R}^n\longrightarrow\mathbb{R}$ is linear. Prove that, under these conditions, $f$ is differentiable at $p$.

[ NOTE: I've been thinking about it for a while now. In doing so, I came up with this other question (poorly formulated, but please see my comments on the second answer): Always a differentiable path through a convergent sequence of points in $\mathbb{R}^n$? ]

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Weltschmerz
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  • Are we allowed to use paths with $\lambda'(0)=0$? Probably not, because smooth paths are normally understood to have nonvanishing velocity vector. –  May 25 '12 at 16:00
  • @Leonid I don't know. You're probably right that we aren't, but what could be said if we are? – Weltschmerz May 25 '12 at 16:02
  • You were on the right path :), but one missing ingredient is this: by passing to a subsequence, you can make sure that the vectors $x_n/|x_n|$ have a limit as $n\to \infty$. –  May 25 '12 at 16:23
  • Do you need some additional continuity hypothesis above? It would seem to me that you could define $f$ to be $0$ on the set $\cup_{t \in (0,2\pi]} [0, \sqrt{t}(\cos t,\sin t)]$, and $1$ outside. Then with $T=0$, it satisfies all the hypotheses, but fails to be differentiable at $0$? (By $[0,x]$ I mean the ray extending from $0$ to $x$.) – copper.hat May 25 '12 at 16:53
  • This is a related concept. The present exercise is asking to prove that, in the finite-dimensional case, a Gâteaux differentiable function whose differential is linear (hence also continuous) is Fréchet differentiable. – Giuseppe Negro May 25 '12 at 18:01
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    I suspect the claim to be false as it stands. As a counterexample I'd take a function $f \colon \mathbb{R}^2\to \mathbb{R}$ given in polar coordinates by

    $$f\left(re^{i \theta}\right)=r^2 u(\theta)$$

    for an unbounded $u\colon [0, 2\pi]\to \mathbb{R}$ such that $u(0)=u(2\pi)$ (in particular, $u$ can not be continuous). This $f$ has got a zero directional derivative along every direction, thus (to be checked!) satisfies the hypothesis with $T \equiv 0$, but it is not differentiable at the origin since it is not even continuous.

     – Giuseppe Negro May 25 '12 at 18:53
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    @GiuseppeNegro The assumption in the present problem is stronger than Gateaux differentiability, because it involves derivatives along paths more general than straight lines. –  May 25 '12 at 23:04
  • @copper.hat I don't see how your example works. It seems to have f(0,0)=1 but f(x,0)=0 for all x>0, so the x-derivative does not exist. –  May 25 '12 at 23:05
  • @LeonidKovalev: Let $S$ be the set I defined. The intersection of $S$ with every line through $0$ has a small 'open segment' (in the 'line subspace')containing $0$, so $f$ is $0$ there (this includes $0$) and the Gateaux differential is defined there (and hence $0$). Look at the '$t$' index, it ranges through $(0, 2 \pi]$. – copper.hat May 25 '12 at 23:11
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    @copper.hat Sorry, I misread your example at first. But it seems that your function is not differentiable along the path $t\mapsto (t,t^3)$. –  May 25 '12 at 23:18
  • @LeonidKovalev: Excellent point, my 'counterexample' is not a counterexample. – copper.hat May 25 '12 at 23:21
  • I am guessing its true if you assume $f$ to be continuous near $p$. – copper.hat May 26 '12 at 00:10
  • I'm sure it's true as stated, and a proof (by contradiction) can be obtained by extracting a sequence of "bad" points that approach $p$ from a certain direction (as I mentioned in my older comment). –  May 26 '12 at 00:36
  • @Leonid :) That's what my idea was about and my reason for the other post. Why can you make sure that in general a subsequence exists such that $x_n/|x_n|$ has a limit ? – Weltschmerz May 26 '12 at 00:57
  • Because the unit sphere is a compact set. –  May 26 '12 at 01:10
  • @Leonid: Oh, yes. Alright, thank you very much. I hope to post the solution in the course of this weekend. – Weltschmerz May 26 '12 at 01:13
  • @LeonidKovalev: (Regarding Gateaux differentiation). I see my mistake, thank you. I thought that differentiability along every path was equivalent to differentiability along every straight direction but evidently I was wrong. Indeed, the example posted by copper.hat is differentiable along every straight direction but not along every smooth path, as you pointed out. – Giuseppe Negro May 26 '12 at 15:36
  • @Leonid: I'm still missing something. I seem to think that there needs to be a differentiable path through all the points $v_k$ such that $v_k/|v_k|$ approaches $v$, and I need that somehow it is parametrized such that $t_k=|v_k|$. Why can this always be done? – Weltschmerz May 30 '12 at 16:13

1 Answers1

2

The chain of comments became too long, so I'm switching to the answer box.

  1. We may assume that $T=0$ by subtracting $Tx$ from our function.
  2. Suppose $f$ is not differentiable. Pick a sequence $v_k\to 0$ such that $|f(v_k)|\ge \epsilon |v_k|$.
  3. Passing to a subsequence, make sure that $v_k/|v_k|\to u$, and also that $|v_{k+1}|\le \frac{1}{2}|v_k|$. (We don't want the sequence to jump back and forth.)
  4. Connect the points by line segments. Parametrize this piecewise-linear curve $\lambda$ by arclength, which is finite.
  5. The length of $\lambda$ between $0$ and $v_k$ is bounded by $4|v_k|$ or some such multiple.
  6. $\lambda$ has a one-sided derivative when it reaches $0$. Extend it to get two-sided derivative at that point (apparently, this is all the statement requires; the entire path need not be smooth)
  7. By assumption, $|f(\lambda(t))|/|t|\to 0$ as $t\to\infty$. This contradicts 2&5.
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    My answer was somewhat sloppy (which I figure is OK since OP only wanted a hint). To really make this work, use a stronger bound $|v_{k+1}|\le 2^{-k}|v_k|$ in #3 and estimate the length by $(1+o(1))|v_k|$ in #5. And of course, $t\to 0$ in #7. –  Jul 25 '12 at 17:38