Differentiability of composition with every path implies differentiability

I’m struggling with this problem from my book Curso de Analise Vol. 2 by Elon Lages Lima.

Let $U\subseteq \mathbb{R}^n$ be an open set, $a\in U$ and $f:U\to \mathbb{R}$ with the following property: for every $v\in \mathbb{R}^n$ and every path and $g:(-\epsilon,\epsilon)\to U$ with $g(0)=a$ and $g'(0)=v$ the composite map $f\circ g:(-\epsilon,\epsilon)\to R$ satisfies $(f\circ g)'(0)=T(v)$ where $T:\mathbb{R}^n\to\mathbb{R}$ is a fixed linear transformation. Prove that $f$ is differentiable at $a$.

It’s clear that one must have $f'(a)=T$ so one has to prove
\tag 0\lim_{v\to 0}\frac{f(a+v)-f(a)-T(v)}{|v|}=0

I can’t think of any path other than $t\mapsto a+tv$ but then I’m not using the hypothesis to its full strength. I also tried a proof by negating that limit and trying to construct a path from there, but didn’t succeed.

How do I approach the problem? Any hints?


I’ll accept zhw.’s hint but write the full answer here because the question has received some attention.

Suppose the limit in $(0)$ doesnt exist, then there exists a sequence $v_k$ with $v_k\to 0$ and $v_k\neq 0$ such that

$$\tag 1\frac{|f(a+ v_k) – f(a) – Tv_k|}{|v_k|} \ge \epsilon$$

for all $k$. Write $v_k=r_ku_k$ where $r_k=|v_k|$ and $|u_k|=1$. Then some subsequence of $u_k,$ which I’ll still denote by $u_k,$ converges to some $u, |u|=1.$ Passing to a further subsequence if necessary, we can assume $r_1>r_2 >\cdots.$ Now this is the tricky part: how to define $g$. First we define $g$ in $[0,r_1]$. Set $g(0)=a$. For $t\in (0,r_1]$, since $r_k\to 0$ and $r_1>r_2>\cdots$ there is a $k$ such that $t\in[r_{k+1},r_k]$ and we set


It’s easy to see that $g$ is well defined and $g(r_k)=a+v_k$. Now if $t\in [r_{k+1},r_k]$


From which one can prove $g_+'(0)=u$. If we set $g(t)=-g(-t)$ for $t\in [-r_1,0]$ we get $g'(0)=u$.

We should have $(f\circ g)'(0)=Tu$, i.e.

$$\tag 2\lim_{t\to 0}\frac{f(g(t))-f(g(0))}{t}=Tu$$

Since $r_k\to 0$, $(2)$ implies that

$$\tag 3\lim_{k\to \infty}\frac{f(g(r_k))-f(g(0))}{r_k}=Tu$$

But $v_k/r_k=u_k\to u$ and $T$ is continuous so

$$\tag 4\lim_{k\to \infty}\frac{1}{r_k}Tv_k=\lim_{k\to \infty}T(\frac{v_k}{r_k})=Tu$$

$(3)$, $(4)$ and the fact that $g(r_k)=a+v_k$, $g(0)=a$ and $|v_k|=r_k$ imply that

$$\lim_{k\to \infty}\frac{f(a+ v_k) – f(a) – Tv_k}{|v_k|}=0$$

contradicting $(1)$.

Source : Link , Question Author : Zero , Answer Author : Zero

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