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Please help me understand the following proof.

Q) Let $f$ be measurable and periodic with period $1$, that is, $f(t+1)=f(t)$. Suppose that there is a finite $c$ such that $$\int_0^1 | f(a+t) - f(b+t) | dt \le c$$ for all $a$ and $b$. Show that $f \in L(0, 1)$.

A)

  1. \begin{align} c &\ge \int_0^1 \int_0^1 |f(x+t) - f(-x+t)| ~dt~dx \\ &= \frac12 \int_0^1 \int_{-\xi}^\xi |f(\xi) - f(\eta)| ~d\eta~d\xi + \frac12 \int_0^1 \int_{\xi-1}^{1-\xi} |f(\xi)-f(\eta)| ~d\eta~d\xi \end{align}
  2. \begin{align} \int_0^2 \int_{-1}^1 |f(\xi)-f(\eta)|~d\eta~d\xi &\le 2c \\ \int_0^1 \int_0^1 |f(\xi)-f(\eta)|~d\eta~d\xi &\le\frac c2 \end{align}
  3. Thus, $|f(\xi)-f(\eta)|$ is integrable over the square $[0, 1]\times[0, 1]$.
  4. There is a theorem: for measurable $f$ on $(0, 1)$, if $f(x)-f(y)$ is integrable over $[0, 1]\times[0, 1]$, then $f\in L(0, 1)$.
  5. By this theorem, we can have $f$ is integrable over $(0, 1)$.

Can someone explane about the process $1$ and $2$? I do not understand them.

Danny_Kim
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1 Answers1

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The first inequality comes from the use of the assumption with $a=x$ and $b=-x$ followed by an integration over $x\in[0,1]$. Then the substitution $x+t=\xi$ and $-x+t=\eta$ is done.

Then in the second step, using periodicity and Chasles relation, we derive $$2c\geqslant \int_0^1\int_{-\xi}^{2-\xi}\left|f\left(\xi\right)-f\left(\eta\right)\right| \mathrm d\eta\mathrm d\xi.$$ Now, if $\xi$ belongs to $[0,1]$, then $ -\xi\leqslant 0$ and $1\leqslant 2-\xi$ so that the interval $[0,1]$ is contained in $[-\xi,2-\xi]$. This entails $$2c\geqslant \int_0^1\int_{0}^{1}\left|f\left(\xi\right)-f\left(\eta\right)\right| \mathrm d\eta\mathrm d\xi.$$

Davide Giraudo
  • 172,925