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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.
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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 $$(\varphi \circ F)(x) = \int_0^xg(t)dt$$

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