Hello, I am having trouble understanding the above argument from Lang's Calculus of Several Variables, pp.64-65. I understand the author is not being completely rigorous in this book.
Nonetheless, when we replace $h$ by $-h$, shouldn't we replace $f(x+h)$ with $f(x-h)$. I'm just very unclear about what's going on here. If someone could give some color, I'd be more than happy.
Edit: I think that I should also add how the author continues - with the converse - assuming the existence of such a function $g(h)$ and proving differentiability. I think it clarifies.
Let $g$ be as above for $h \neq 0$ and define $g(0) = 0$. Then $g$ is continuous at zero and $g(0) = 0$.
Now define $\tilde{g}(h) = \operatorname{sgn} (h) g(h)$, then $hg(h) = |h| \tilde{g}(h)$.
Since $f(x+h) = f(x) + f'(x)h + h g(h)$ we have $f(x+h) = f(x) + f'(x)h + |h| \tilde{g}(h)$.
Note that it is the fact that $g(0) = 0$ that lets us do this and keep $\tilde g$ continuous.