I have a following problem:
Suppose f is a k-smooth function ($ \in C^k$) but not $(k+1)$-smooth ($f \notin C^{k+1}$) and $g(x)=ax+b, a\neq 0$).
Is it true, and if it is than how to prove it, that $f \circ g $ is also a k-smooth function ($ \in C^k$) but is not $k+1$-smooth ($ f \circ g \notin C^{k+1}$)? \
I can't see anywhere such questions. All I see is that the composition doesn't reduce smoothness.
Suppose $f$ is in $C^k$ but not in $C^{k+1}$. We want to show that $f \circ g$ is also in $C^k$ but not in $C^{k+1}$. Since $f \circ g(x) = f(ax+b)$, it is not difficult to see that $f \circ g$ belongs to $C^k$. The problem is to show that $f \circ g$ does not belong to $C^{k+1}$.
Suppose by way of contradiction it does belong to $C^{k+1}$. Let $h(x) = \frac{1}{a}x - \frac{b}{a}$. It is not difficult to see that $g(h(x)) = x$. Since $f \circ g$ belongs to $C^{k+1}$, so does $(f \circ g) \circ h$. But
$$ (f \circ g) \circ h(x) = f(g(h(x)) = f(x)$$ In other words, $(f \circ g) \circ h = f$. This is a contradiction, since we assumed that $f$ did not belong to $C^{k+1}$.