Lemma to the Riesz-Fischer Theorem (1 ≤ p < ∞)

Lemma to the Riesz-Fischer Theorem (1 ≤ p < ∞)

 Lemma 1: Let $(X, \mathfrak T, \mu)$ be a measure space and let $1 \leq p \leq \infty$. If $(f_n)_{n=1}^{\infty}$ is a sequence of functions in $L^p(X, \mathfrak T, \mu)$ such that there exists a sequence of positive real numbers $(\epsilon_n)_{n=1}^{\infty}$ such that $\displaystyle{\sum_{n=1}^{\infty} \epsilon_n < \infty}$ and $\| f_n - f_{n+1} \|_p \leq \epsilon_n^2$ for all $n \geq 1$, then there exists a function $f \in L^p (X, \mathfrak T, \mu)$ such that: a) $\displaystyle{\lim_{n \to \infty} f_n(x) = f(x)}$ $\mu$-almost everywhere on $X$. b) $\displaystyle{\lim_{n \to \infty} \| f - f_n \|_p = 0}$.

The case when $p = \infty$ can be found on the Lemma to the Riesz-Fischer Theorem (p = ∞) page.

In the proof below we use the following results: Chebyshev's Inequality for Nonnegative Measurable Functions, The Borel-Cantelli Lemma, and Fatou's Lemma for Nonnegative Measurable Functions. These results can be found on the Measure Theory hub.

(1)
\begin{align} \quad \| f_n - f_{n+m} \|_p &= \| f_n - f_{n+1} + f_{n+1} - f_{n+2} + f_{n+2} - ... + f_{n+m-1} - f_{n+m} \|_p \\ & \leq \| f_n - f_{n+1} \|_p + \| f_{n+1} - f_{n+2} \|_p + ... + \| f_{n+m-1} - f_{n+m} \|_p \\ & \leq \sum_{k=n}^{n+m-1} \| f_k - f_{k+1} \|_p \\ & \leq \sum_{k=n}^{\infty} \| f_k - f_{k+1} \|_p \\ & \leq \sum_{k=n}^{\infty} \epsilon_k^2 \quad (*) \end{align}
• Observe that since $\displaystyle{\sum_{n=1}^{\infty} \epsilon_n < \infty}$ we must have that $\displaystyle{\lim_{n \to \infty} \epsilon_n = 0}$. So there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\epsilon_n < 1$. So for $n \geq N$ we have that $\epsilon_n^2 < \epsilon_n < 1$ and so $\displaystyle{\lim_{n \to \infty} \epsilon_n^2 = 0}$. From above, we see that as $n \to \infty$, the sum $\displaystyle{\sum_{k=n}^{\infty} \epsilon_k^2 \to 0}$. So from above, we see that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| f_n - f_{m+n} \|_p < \epsilon$. Therefore, $(f_n)_{n=1}^{\infty}$ is a pointwise Cauchy sequence of functions.
• Now suppose that $1 \leq p < \infty$. Consider the set:
(2)
\begin{align} \quad \{ x \in X : |f_n(x) - f_{n+1}(x)|^p \geq \epsilon_{n}^p \} \end{align}
• By Chebyshev's inequality for nonnegative measurable functions we have that since $\epsilon_{n}^p > 0$ that:
(3)
\begin{align} \quad \mu ( \{ x \in X : |f_n(X) - f_{n+1}(x)|^p \geq \epsilon_{n}^p \}) \leq \frac{1}{\epsilon_{n}^p} \int_X |f_n - f_{n+1}|^p \: d \mu = \frac{\| f_n - f_{n+1} \|_p^p}{\epsilon_{n}^p} \leq \frac{\epsilon_n^{2p}}{\epsilon_n^p} = \epsilon_n^p \end{align}
• For each $n \in \mathbb{N}$ we have that $\mu ( \{ x \in X : |f_n(x) - f_{n+1}(x)| \geq \epsilon_{n+1} \} \leq \epsilon_n^p < \infty$. By the Borel-Cantelli lemma, almost every $x \in X$ belongs to a finite collection of these sets. That is, there exists a measurable set $Y \in \mathfrak T$ such that every $x \in Y$ belongs to at most a finite collection of these sets and $\mu (X \setminus Y) = 0$.
• So for each $x \in Y$ there exists an $N_x \in \mathbb{N}$ such that if $n \geq N_x$ then $|f_n(x) - f_{n+1}(x)| < \epsilon_n$. So if $m, n \in \mathbb{N}$ are such that $m \geq n \geq N_x$ then:
(4)
\begin{align} \quad |f_m(x) - f_n(x)| &= |f_m(x) - f_{m-1}(x) + f_{m-1}(x) - ... + f_{n+1}(x) - f_n(x) | \\ & \leq |f_m(x) - f_{m-1}(x)| + ... + |f_{n+1}(x) - f_n(x)| \\ & \leq \sum_{k=n+1}^{m} |f_k(x) - f_{k-1}(x)| \\ & \leq \sum_{k=n}^{m-1} |f_{k+1}(x) - f_k(x)| \\ & \leq \sum_{k=n}^{m-1} |f_k(x) - f_{k+1}(x)| \\ & \leq \sum_{k=n}^{m-1} \epsilon_k \\ & \leq \sum_{k=n}^{\infty} \epsilon_k \end{align}
• As $n \to \infty$ we have that $\displaystyle{\sum_{k=n}^{\infty} \epsilon_k \to 0}$. So for each $x \in Y$, $(f_n(x))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{C}$ and hence converges. Define a function $f : X \to \mathbb{C}$ as:
(5)
\begin{align} \quad f(x) = \left\{\begin{matrix} \lim_{n \to \infty} f_n(x) & \mathrm{if} x \in Y \\ 0 & \mathrm{if} \: x \in X \setminus Y \end{matrix}\right. \end{align}
• Since $(f_n)_{n=1}^{\infty}$ is pointwise Cauchy in $L^p(X, \mathfrak T, \mu)$ it is bounded. So there exists an $M \in \mathbb{R}$, $M \geq 0$ such that $\| f_n \|_p \leq M$ for all $n \in \mathbb{N}$. Now by Fatou's Lemma we have that:
(6)
\begin{align} \quad \int_X |f|^p \: d \mu = \int_X \lim_{n \to \infty} |f_n|^p \: d \mu \leq \liminf_{n \to \infty} \int_X |f_n|^p \; d \mu = \liminf_{n \to \infty} \|f_n\|_p^p \leq M^p < \infty \end{align}
• Hence $f \in L^p (X, \mathfrak T, \mu)$. By definition, $f$ satisfies (a).
• Lastly we show that $f$ satisfies (b). We have that:
(7)
\begin{align} \quad \lim_{n \to \infty} \|f - f_n\|_p^p &= \lim_{n \to \infty} \int_X |f - f_n|^p \: d \mu \\ &= \lim_{n \to \infty} \int_X \lim_{m \to \infty} |f_m - f_n|^p \: d \mu \\ & \leq \lim_{n \to \infty} \liminf_{m \to \infty} \int_X |f_m - f_n|^p \: d \mu \\ & \leq \lim_{n \to \infty} \liminf_{m \to \infty} \| f_m - f_n \|_p^p \\ & \leq \lim_{n \to \infty} \liminf_{m \to \infty} \left (\sum_{k=n}^{m-1} \| f_k - f_{k+1} \|_p \right )^p \\ & \leq \lim_{n \to \infty} \left ( \sum_{k=n}^{\infty} \epsilon_k^2 \right )^p \\ & \leq 0 \end{align}
• Hence $\displaystyle{\lim_{n \to \infty} \| f - f_n \|_p^p = 0}$. So $\displaystyle{\lim_{n \to \infty} \| f - f_n \|_p = 0}$. $\blacksquare$