# Show that there exists $g\in L^1([0,1])$ such that $\varphi(F[x])=\int_0^x g(t)dt$ for all $x\in [0,1]$

I would be glad if someone could help me to solve the following exercise.

Suppose $$f\in L^1([0,1])$$. For $$x\in[0,1]$$ let $$F(x)=\int_0^x f(t)dt.$$ Let $$\varphi:\Bbb R\to \Bbb R$$ be a Lipschitz function with $$\varphi(0)=0$$. Show that there exists $$g\in L^1([0,1])$$ such that for every $$x\in [0,1]$$ we have $$\varphi(F[x])=\int_0^x g(t)dt$$.

$$Observations.$$

Suppose such a function $$g$$ exists.

1. If $$\varphi$$ is differentiable on $$\Bbb R$$, then for any $$x\in [0,1]$$ we have $$\varphi(F(x))=\int_0^xg(t)dt\implies\varphi'(F(x))F'(x)=\frac{d}{dx}\int_0^xg(t)dt=g(x).$$ So we must have $$g(x)=\varphi'(F(x))f(x)$$ for all $$x\in [0,1]$$. Conversely, $$\int_0^x g(t)dt=\int_0^x\varphi '(F(t))f(t)dt=\int_0^x\varphi '(F(t))F'(t)dt=\int_0^x(\varphi\circ F)'(t)dt=(\varphi\circ F)(x) \qquad \text{for all x\in [0,1].}$$
2. Since $$\varphi$$ is Lipschitz, there exists $$C\geq 0$$ such that $$|\varphi(x)-\varphi(y)|\leq C|x-y|$$ for all $$x,y\in \Bbb R$$. In particular, $$|\varphi(x)|\leq C|x|$$ for all $$x\in \Bbb R$$, since $$\varphi(0)=0$$. Again, in particular, $$|\varphi(F(x))|\leq C|F(x)|\qquad \text{for all x\in[0,1].}$$ So, for any $$x\in [0,1]$$, $$|\int_0^xg(t)dt|\leq C|\int_0^xf(t)dt|$$ implies that $$\frac{d}{dx}\left|\int_0^xg(t)dt\right|\leq C \frac{d}{dx}\left|\int_0^xf(t)dt\right|$$ Since $$|g(x)|=|\frac{d}{dx}\int_0^xg(t)dt|=\frac{d}{dx}\left|\int_0^xg(t)dt\right|$$ and $$|f(x)|=|\frac{d}{dx}\int_0^xf(t)dt|=\frac{d}{dx}\left|\int_0^xf(t)dt\right|$$, we must have $$|g(x)|\leq C|f(x)| \qquad \text{for all x\in[0,1],}$$ if such a function $$g$$ exists.

Theorem: $$f : [a,b] \rightarrow \Bbb R$$ is absolutely continuous iff
• $$\exists f'$$ a.e. and $$f' \in L^1([a,b])$$
• $$f(x) = f(a) + \int_a^xf'(t)dt$$ for every $$x \in [a,b]$$
Let us prove that $$\varphi \circ F$$ is absolutely continuous on $$[0,1]$$, we know that $$\forall \epsilon > 0 \: \exists \delta(\epsilon)>0 \: \text{s.t.}\: \sum_i (b_i - a_i) < \delta \Rightarrow \sum_i |F(b_i) - F(a_i)| < \epsilon$$ where $$\{(a_i,b_i)\}_{i \in \Bbb N}$$ is a countable collection of disjoint open intervals in $$[0,1]$$ (in fact by Lebesgue's differentiation theorem $$\exists F'(x) = f(x)$$ for a.e. $$x \in [0,1]$$, $$F' \in L^1([0,1])$$ and $$F(x) = \int_0^x F’(t)dt$$ so $$F$$ is absolutely continuous)
Observe that for every $$i$$ $$|\varphi \circ F(b_i) - \varphi \circ F(a_i)| \leq C |F(b_i) - F(a_i)|$$ but now is easy to conclude that $$\varphi \circ F$$ is absolutely continuous on $$[0,1]$$ $$\sum_i |\varphi \circ F(b_i) - \varphi \circ F(a_i)| \leq C \sum_i |F(b_i) - F(a_i)| < C \epsilon$$ so $$\exists (\varphi \circ F)'(x) =: g(x)$$ for a.e. $$x \in [a,b]$$, $$g \in L^1([0,1])$$ and $$(\varphi \circ F)(x) = \int_0^xg(t)dt$$