An integrable function $f$ on $\Bbb R$ satisfying $\lim_{h\to 0}\int_\Bbb R \frac{|f(x+h)-f(x)|}{h}dx=0$ must be constant

Question

Suppose $f:\Bbb R\to \Bbb C$ is an integrable function, i.e. $\int_\Bbb R|f|~dx<\infty$, satisfying $$ \displaystyle\lim_{h\to 0}\displaystyle\int_\Bbb R \dfrac{|f(x+h)-f(x)|}{h}dx=0.$$ I am trying to show that in this case that $f$ must be constant (of course in a.e. sense). Using Fatou's lemma, I could show that $\liminf_{h\to 0} (f(x+h)-f(x))/h=0$ for almost all $x\in \Bbb R$, but I can't see how to progress next. Can I get a hint?

Answer 1

Let $a<b$ and assume that $a$ and $b$ are Lebesgue points of $f$. Then $\int_a^{b} \frac {f(x+h)-f(x)} h dx \to 0$. This can be written as $\frac 1 h (\int_{a+h} ^{b+h} f(x)dx- \int_a^{b} f(x)dx) \to 0$ or $\frac 1 h(_b^{b+h}f(x)dx-\int_a^{a+h} f(x)dx )\to 0$. Since $a$ and $b$ are Lebesgue points this gives $f(a)=f(b)$. Hence $f$ is a constant almost everywhere. Note that $f$ is also integrable so $f=0$ almost everywhere!.