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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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Are we allowed to use paths with $\lambda'(0)=0$? Probably not, because smooth paths are normally understood to have nonvanishing velocity vector. –  user31373 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
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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. –  user31373 May 25 '12 at 23:04
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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 $t\mapsto (t,t^3)$. –  user31373 May 25 '12 at 23:18

1 Answer 1

up vote 1 down vote accepted

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. –  user31373 Jul 25 '12 at 17:38

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