Continuity of linear functionals on $L^1([a,b])$ with the $L^1$ functions composed with $C^1$ functions I've been thinking about the following idea involving continuity of linear functionals on $L^1([a,b])$. Let $f\in C^1([a,b])$ be fixed. Define $\alpha:=\min f$ and $\beta:=\max f$. Consider the space $L^1([\alpha,\beta])$.
Define the linear functional $F:L^1([\alpha,\beta]) \to \mathbb{R}$ by
$$
F(g)=\int_a^bg(f(x))\,dx.
$$
Under what conditions is $F$ bounded? For example, an easy case is if $f$ is strictly increasing with a derivative bounded below by a positive number. Say $f'>c>0$. Then using substitution, we may write
$$
F(g)=\int_{f(a)}^{f(b)}\frac{1}{f'(f^{-1}(u))}g(u)\,du=\int_{\alpha}^{\beta}\frac{1}{f'(f^{-1}(u))}g(u)\,du
$$
so
$|F(g)|\leq \frac1c\|g\|_1$. Obviously, we recover the case where $f(x)=x$, where $f'(x)=1>0$.
What I have in mind is more along the lines of $f$ having a mix of regions of increasing/decreasing so that $f'(x)=0$ occurs. Certainly, we can apply the idea above to the strictly increasing/decreasing regions by splitting up the integral into a sum of integrals. But the issue is always that $\frac{1}{f'(f^{-1}(u))}$ blows up around the critical point so it's hard (if not impossible) to formulate a bound in terms of $\|g\|_1$.
Does anyone have any thoughts on this idea?
 A: Defining  a new measure $\mu $ on $[\alpha , \beta ]$ by
$$
  \mu (E) = \lambda (f^{-1}(E)),
  $$
for every Borel measurable set $E\subseteq [\alpha , \beta ]$,
one has that
$$
  F(g) = \int_a^b g(f(x))\, dx =   \int_\alpha ^\beta  g(y)\, d\mu (y).
  $$
Regarding boundedness of $F$,
the first important point  is whether or not $\mu $ is absolutely continuous w.r.t. Lebesgue measure (here denoted by $\lambda $), which is to say
that
$$
  \lambda (E)=0 \Rightarrow \mu (E)=0.
  \tag {1}
  $$
If this fails for some $E$, then $F$ is not even well defined on $L^1([\alpha ,\beta ])$ because if $g$ is  the characteristic
function of $E$, then $g=0$ as an element of $L^1([\alpha ,\beta ])$, but
$$
  F(g)=\mu (E)\neq 0.
  $$
This problematic situation occurs, for instance, if $f$ is constant on some nontrivial interval.
On the other hand, if (1) holds, then the Radon-Nykodym Theorem asserts that there exists some measurable function $\varphi $
on $[\alpha ,\beta ]$, such that
$$
  \int_\alpha ^\beta  g(y)\, d\mu (y) =   \int_\alpha ^\beta  g(y)\varphi (y)\, d\lambda (y),
  $$
and then $F$ defines a bounded linear functional iff $\varphi $ is in $L^\infty ([\alpha ,\beta ])$. In this case one also has
$$
\|F\| = \|\varphi\|_{\infty}.
$$

EDIT:  (1) Notice that the above holds for every measurable function $f:[a,b]\to[\alpha,\beta]$.
(2) In the special case,
already noted by the OP, in which $f$ is of class $C^1$ and  $f'$ is never zero (and hence also bounded away from zero because $[a,
b]$ is compact), then $\mu \ll\lambda $, with bounded Radon-Nykodym derivative
$$
  \varphi (u) = {d\mu  \over d\lambda }(u) = {1\over f'(f^{-1}(u))},
  \tag {2}
  $$
so  it is clear that $F$ is a bounded linear functional.
(3) When $f$ is of class $C^1$,  the above is in fact the only case leading to a bounded $F$!  This is because if
$f'(t)=0$  for a single point $t$ in $[a, b]$,  and assuming of course that $\mu \ll\lambda $, then the Radon-Nykodym derivative
will still be given by (2) but it will be unbounded on any neighborhood of $f(t)$ (notice that $\varphi$ will not be defined at $f(t)$, but that is OK since $\varphi$ is only supposed to be defined a.e.).
