Unusual Sobolev Chain rule

Question

Let $f \in W^{1, \infty}(\mathbb{R}^n)$ and $x, y \in \mathbb{R}^n \setminus \lbrace 0 \rbrace$. Then, is $g(t) := f(x+ty) \in W^{1, 1}([0, 1])$ and is $$ g'(t) = \nabla f(x+ty)^\top y? $$ The usual chain rule (as stated e.g. in Evans - Measure theory...) is not applicable, because in this case, the Sobolev function is on the outside. Thus any approximation arguments to prove weak differentiability will be very difficult. The only hope is probably to introduce some type of transformation...

Answer 1

As pointed out by Phoemuex, you have $g \in W^{1,\infty}([0,1])$. However, you will not get any sensible chain rule, since the points $\{ x + t y | t \in [0,1]\}$ have measure zero (if $n > 1$) and $\nabla f$ is only defined up to sets of measure zero. As an example, consider $n = 2$, $x = (0,0)$, $y = (1,0)$ and $$ f(x_1, x_2) = |x_2| + \text{something smooth}.$$