Differentiability in composition of function with a differentiable path.

Question

Let $f:U\to \Bbb R^n$ be Lipschitz in the open $U\subset \Bbb R^m$. For $a\in U$, suppose that, for all $v\in \Bbb R^m$, there exists the directional derivative $\dfrac{\partial f}{\partial v}(a)$ and it depends linearly on $v$. Prove that, for all path $g:(-\epsilon,\epsilon)\to U$, with $g(0)=a$, differentiable in $t=0$, there exists $(f\circ g)'(0)$. Conclude that $f$ is differentiable in the point $a$.

I did not have a good idea for this question. I know that, since $\dfrac{\partial f}{\partial v}(a)$ is linear, $\dfrac{\partial f}{\partial v}(a) = Tv$ where $T$ a linear transformation.But, I don't know how to use that. Any hint?

Answer 1

First part: Let $g'(0)=v.$ Then

$$g(t) = g(0) + tv +o(t).$$

Because $f$ is Lipschitz near $g(0),$

$$f(g(t)) = f(g(0) + tv + o(t)) = f(g(0) + tv) + o(t).$$

Thus

$$\frac{f(g(t)) - f(g(0))}{t} = \frac{f(g(0)) + tv)+o(t)-f(g(0))}{t}$$ $$ = \frac{f(g(0)) + tv)-f(g(0))}{t} +o(1) \to Df_v(g(0)).$$

Thus $(f\circ g)'(0) = Df_v(g(0)).$

Note we did not use the linearity of $v\to D_vf(a)$ for this part.