Lebesgue integral of difference function

Question

I'm working on the following problem: Given $f$ integrable over $\mathbb{R}$ with $f$ absolutely continuous and $$\lim\limits_{h \to 0}\int_{-\infty}^{\infty} \frac{|f(x+h)-f(x)|}{|h|}=0$$ Then $f$ is constant.

I know I want to use the fact that since $f$ is absolutely continuous I can write $f$ as an indefinite integral. I also know the result would then follow if I can show that $f'(x)=0$, but I'm having trouble working with the integral to show my derivative is zero.

Answer 1

Idea. The claim will follow if one can show that

$$ \lim_{h \to 0} \int_{-\infty}^{\infty} \left| \frac{f(x+h) - f(x)}{h}\right| \, \mathrm{d}x = \int_{-\infty}^{\infty} \lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h}\right| \, \mathrm{d}x, $$

since Lebesgue's Differentiation Theorem tells that $\lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h}\right| = \left| f'(x) \right|$ for a.e. $x$. We will indeed show a version of this claim to solve the problem.

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Proof. Perhaps this problem is easily solved by invoking the following variant of Dominated Convergence Theorem:

Theorem. Let $f_n, g_n, g \in L^1(\mu)$ satisfy the following conditions:

1. $|f_n| \leq g_n$ for all $n \geq 1$,
2. $f_n$ converges pointwise $\mu$-a.e., and $g_n \to g$ pointwise $\mu$-a.e.
3. $\int g_n \, \mathrm{d}\mu \to \int g \, \mathrm{d}\mu $

Then $$ \lim_{n\to\infty} \int f_n \, \mathrm{d}\mu = \int \lim_{n\to\infty} f_n \, \mathrm{d}\mu. $$

As in the case of the original DCT, it is proved in a routine way, i.e., by invoking Fatou's Lemma. The proof is included at the end of this answer.

Returning to the problem, for any $h > 0$ and $x$, we have

$$ \left| \frac{f(x+h) - f(x)}{h} \right| = \left| \frac{1}{h} \int_{x}^{x+h} f'(t) \, \mathrm{d}t \right| \leq \frac{1}{h} \int_{x}^{x+h} \left| f'(t) \right| \, \mathrm{d}t. $$

Moreover, Lebesgue's Differentiation Theorem tells that $ \frac{1}{h} \int_{x}^{x+h} \left| f'(t) \right| \, \mathrm{d}t \to |f'(x)|$ for a.e. $x$ and Fubini-Tonelli Theorem tells that

\begin{align*} \int_{-\infty}^{\infty} \frac{1}{h} \int_{x}^{x+h} \left| f'(t) \right| \, \mathrm{d}t \mathrm{d}x = \int_{-\infty}^{\infty} \frac{1}{h} \int_{t-h}^{t} \left| f'(t) \right| \, \mathrm{d}x \mathrm{d}t = \int_{-\infty}^{\infty} \left| f'(t) \right| \, \mathrm{d}t. \end{align*}

So, if $(h_n)$ is any sequence of positive real numbers converging to $0$, then the above theorem and Lebesgue's Differentiation Theorem together show that

$$ \lim_{n \to \infty} \int_{-\infty}^{\infty} \left| \frac{f(x+h_n) - f(x)}{h_n}\right| \, \mathrm{d}x = \int_{-\infty}^{\infty} \lim_{n \to \infty} \left| \frac{f(x+h_n) - f(x)}{h_n}\right| \, \mathrm{d}x = \int_{-\infty}^{\infty} \left| f'(x) \right| \, \mathrm{d}x. $$

Now the assumption tells that this is zero, and so, $f' = 0$ a.e. and the desired conclusion follows. $\square$

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Addendum - Proof of Theorem. Write $f = \lim_{n\to\infty} f_n$. Then Fatou's lemma tells that

\begin{align*} \int (g - f) \, \mathrm{d}\mu &= \int \liminf_{n\to\infty} (g_n - f_n) \, \mathrm{d}\mu \leq \liminf_{n\to\infty} \int (g_n - f_n) \, \mathrm{d}\mu \\ &= \int g \, \mathrm{d}\mu - \limsup_{n\to\infty} \int f_n \, \mathrm{d}\mu \end{align*}

and similarly

\begin{align*} \int (g + f) \, \mathrm{d}\mu &= \int \liminf_{n\to\infty} (g_n + f_n) \, \mathrm{d}\mu \leq \liminf_{n\to\infty} \int (g_n + f_n) \, \mathrm{d}\mu \\ &= \int g \, \mathrm{d}\mu + \liminf_{n\to\infty} \int f_n \, \mathrm{d}\mu. \end{align*}

Therefore we have

$$ \int f \, \mathrm{d}\mu \leq \liminf_{n\to\infty} \int f \, \mathrm{d}\mu \leq \limsup_{n\to\infty} \int f_n \, \mathrm{d}\mu \leq \int f \, \mathrm{d}\mu, $$

proving the desired claim. $\square$