A sufficient condition for differentiability of a function of several variables in terms of differentiability along paths.

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 $\mathbb{R}^n$? ]

Solutions Collecting From Web of "A sufficient condition for differentiability of a function of several variables in terms of differentiability along paths."

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.