Exercise 1.3.21 (Absolute summability is a special case of absolute integrability)

Question

Let $(c_n)_{n\in\mathbb{Z}}$ be a doubly infinite sequence of complex numbers, and let
$$f:\mathbb{R}	o\mathbb{C}, \qquad x \mapsto c_{\lfloor x floor} = \sum_{n\in\mathbb{Z}} c_n \cdot 1_{[n,n+1)}(x)$$
Show that $f$ is absolutely integrable if and only if the series $\sum_{n\in\mathbb{Z}} c_n$ are absolutely convergent, in which case we have
$$\int_{\mathbb{R}^d}f = \sum_{n\in\mathbb{Z}} c_n$$

Elu Thingol · Accepted Answer

We immediate that $f$ is the limit of simple functions $f_k = \sum_{n=-k}^{k} c_n\cdot 1_{[n,n+1)}(x)$, and thus Lebesgue measurable. We try to demonstrate tha both absolute integral and absolute sum are in fact equal.
\begin{align*}
\int_{\mathbb{R}^d} |f| &= \int_{\mathbb{R}^d} \left| \sum_{n\in\mathbb{Z}} c_n \cdot 1_{[n,n+1)} ight|
\end{align*}
The deciding step which serves to prove both directions of the equivalence is improving to inequality $|\sum_{n\in\mathbb{Z}} c_n| \leq \sum_{n\in\mathbb{Z}} |c_n|$ from Analysis I, Proposition 7.2.9 to an equality in our case.
This is possible since for all $x\in\mathbb{R}$ we have
$|\sum_{n\in\mathbb{Z}} c_n \cdot 1_{[n,n+1)}(x)| = |c_{\lfloor x floor}| = \sum_{n\in\mathbb{Z}} |c_n|  \cdot 1_{[n,n+1)}(x)$.
    \begin{align*}
      &=  \int_{\mathbb{R}^d} \sum_{n\in\mathbb{Z}} |c_n| \cdot 1_{[n,n+1)}
    \end{align*}
    The term inside the integral look "almost" simple, yet it sums up infinitely many values. This is, however, easy to overcome by approximating this integral using vertical truncation from \href{./exercise_1-3-10}{Exercise 1.3.10} (ix):
    \begin{align*}
      &= \lim_{k	o\infty}  \int_{\mathbb{R}^d}\left[ \sum_{n\in\mathbb{Z}} |c_n| \cdot 1_{[n,n+1)}ight] \cdot 1_{[-k,k]}\
        &= \lim_{k	o\infty}  \int_{\mathbb{R}^d}\sum_{n=-k}^{k} |c_n| \cdot 1_{[n,n+1)}
    \end{align*}
    This is a simple function; by \href{./exercise_1-3-10}{Exercise 1.3.10}(i) we have:
    \begin{align*}
      &= \lim_{k	o\infty}  \sum_{n=-k}^{k} |c_n| \cdot m({[n,n+1)}) \ &= \lim_{k	o\infty}  \sum_{n=-k}^{k} |c_n| = \sum_{n\in\mathbb{Z}} |c_n|
    \end{align*}
    Therefore, both absolute integrals are equal and are bound to be finite if one of them is finite. \par

At last, we demonstrate the the (conditional) integrals are equal as well when the absolute integrals are finite:
    \begin{align*}
      \int_{\mathbb{R}^d} f &= \int_{\mathbb{R}^d} \Re f + i\int_{\mathbb{R}^d}\Im f = \int_{\mathbb{R}^d} (\Re f)_+ - \int_{\mathbb{R}^d}(\Re f)_- + i\int_{\mathbb{R}^d}(\Im f)_+ - i\int_{\mathbb{R}^d} (\Im f)_-\
      &= \int_{\mathbb{R}^d} \Re_+\left[\sum_{n\in\mathbb{Z}} c_n  1_{[n,n+1)}ight] - \int_{\mathbb{R}^d}\Re_-\left[\sum_{n\in\mathbb{Z}} c_n  1_{[n,n+1)}ight]
          + i\int_{\mathbb{R}^d}\Im_+\left[\sum_{n\in\mathbb{Z}} c_n  1_{[n,n+1)}ight] - i\int_{\mathbb{R}^d}  \Im_-\left[\sum_{n\in\mathbb{Z}} c_n  1_{[n,n+1)}ight]\
              &\stackrel{??}{=} \int_{\mathbb{R}^d} \sum_{n\in\mathbb{Z}} \Re_+(c_n)  1_{[n,n+1)}
                - \int_{\mathbb{R}^d}\sum_{n\in\mathbb{Z}} \Re_-(c_n)  1_{[n,n+1)}
                  + i\int_{\mathbb{R}^d}\sum_{n\in\mathbb{Z}} \Im_+(c_n)  1_{[n,n+1)}
                    - i\int_{\mathbb{R}^d} \sum_{n\in\mathbb{Z}} \Im_-(c_n)  1_{[n,n+1)}
    \end{align*}
    We have arrived at unsignden Lebesgue measurable functions. Using the same trick as before with the vertical truncation we get
    \begin{align*}
      &{=} \sum_{n\in\mathbb{Z}} \Re(c_n)_+  m({[n,n+1)}) - \sum_{n\in\mathbb{Z}} \Re(c_n)_-  m({[n,n+1)}) + i\sum_{n\in\mathbb{Z}} \Im(c_n)_+ m({[n,n+1)})
            - i \sum_{n\in\mathbb{Z}} \Im(c_n)_-  m({[n,n+1)})\
              &{=} \sum_{n\in\mathbb{Z}} \Re(c_n)_+  - \Re(c_n)_-   +i\Im(c_n)_+ - i \Im(c_n)_- =   \sum_{n\in\mathbb{Z}} c_n
    \end{align*}

Exercise 1.3.21 (Absolute summability is a special case of absolute integrability)

Answers

Comments

Exercise 1.3.21 (Absolute summability is a special case of absolute integrability)

Answers

Comments

Add answer