L1 Space

Definition¹

The function space $L^{1}$ is defined as follows.

$$ L^{1} (E) := \left\{ f : E \to \mathbb{R} \Big \vert \int_{E} | f | dm \lt \infty \right\} $$

Properties

$L^{1}$ is a vector space.
$L^{1}$ is a normed space. The norm is defined as follows. $$ \left\| f \right\|_{1} := \int \left| f(x) \right| dx $$
$L^{1}$ is a complete space.

Explanation

The $L^{1}$ space is a special case of the $L^{p}$ space when $p=1$, and it was defined as the set of integrable functions when discussing Lebesgue integrability.

For the generalized proof concerning $L^{p}$ spaces, refer to here.

Proof

2.

Definition of a norm
Let $V$ be a vector space over $\mathbb{F}$. If a function $\left\| \cdot \right\| : V \to \mathbb{F}$ satisfies the following three conditions for $\mathbf{u}, \mathbf{v} \in V$ and $k \in \mathbb{F}$, then $\left\| \cdot \right\|$ is defined as a norm on $V$.
Positive definiteness: $\left\| \mathbf{u} \right\| \ge 0$ and $\mathbf{u} = \mathbb{0} \iff \left\| \mathbf{u} \right\| = 0$
Homogeneity: $\left\|k \mathbf{u} \right\| = | k | \left\| \mathbf{u} \right\| $
Triangle inequality: $\left\| \mathbf{u} + \mathbf{v}\right\| \le \left\|\mathbf{v} \right\| + \left\| \mathbf{u} \right\|$

Let us define the norm of $L ^{1}$ as $\displaystyle \left\| f \right\|_{1} := \int_{E} |f| dm$.

Part 1. Positive definiteness
Since $| f | \ge 0$, we have $\left\| f \right\|_{1} \ge 0$. If $f = 0$ almost everywhere, then $\left\| f \right\|_{1} = 0$. Conversely, if $\left\| f \right\|_{1} = 0$, then $f = 0$ must hold almost everywhere.
Part 2. Homogeneity
$$\left\| c f \right\| _{1} = \int_{E} | c f | dm = |c| \int_{E} | f | dm = |c| \left\| f \right\| _{1}$$
Part 3. Triangle inequality
$$ \left\| f + g \right\|_{1} = \int_{E} | f + g | dm \le \int_{E} | f | dm + \int_{E} | g | dm = \left\| f\right\|_{1} + \left\| g\right\|_{1} $$

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3.

Completeness
Let a norm $\left\| \cdot \right\|_{X}$ be defined on a vector space $X$. If for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ satisfying $$n, m \ge N \implies \left\| f_{n} - f_{m} \right\|_{X} \lt \varepsilon$$ then the sequence $f_{n} \in X$ is called a Cauchy sequence. If every Cauchy sequence converges to an element of $X$, then $X$ is said to be complete.

Let $f_{n} \in L^{1}$ be a Cauchy sequence. Then we can choose $N_{1}$ satisfying

$$ \left\| f_{n} - f_{N_{1}} \right\|_{1} \lt {{1} \over {2}} $$

and likewise we can choose $N_{2} > N_{1}$ satisfying

$$ \left\| f_{n} - f_{N_{2}} \right\|_{1} \lt {{1} \over {2^2}} $$

In this manner, we can choose $N_{n} > N_{n-1}$ satisfying $$ \left\| f_{n} - f_{N_{n}} \right\|_{1} \lt {{1} \over {2^n}} $$ By the triangle inequality, $$ \left\| f_{N_{n}} - f_{N_{n-1}} \right\|_{1} \lt \left\| f_{N_{n}} - f_{n} \right\|_{1} + \left\| f_{n} - f_{N_{n-1}} \right\|_{1} \lt {{1} \over {2^n}} + {{1} \over {2^{n-1}}} \lt {{3} \over {2^{n}}} $$

Levi’s theorem
If $\displaystyle \sum_{k=1}^{\infty} \int |f_{k}| dm \lt \infty$, then $\displaystyle \sum_{k=1}^{\infty} f_{k} (x)$ converges almost everywhere and the following holds.
$$ \int \sum_{k=1}^{\infty} f_{k} dm = \sum_{k=1}^{\infty} \int f_{k} dm $$

By Levi’s theorem, $\displaystyle \sum_{n=1}^{\infty} | f_{N_{n}} - f_{N_{n-1}} |_{1}$ converges. Therefore, the following converges almost everywhere.

$$ f_{N_{1}}(x) + \sum_{n=2}^{ k } \left[ f_{N_{n}} (x) - f_{N_{n-1}} (x) \right] = f_{N_{k}} $$

Here, if the right-hand side converges to $f(x)$, then $f_{N_{k}} (x)$ on the right-hand side also converges to $f(x)$.

Fatou’s lemma
For a sequence $\left\{ f_{n} \right\}$ of non-negative measurable functions,
$$ \displaystyle \int_{E} \left( \liminf_{n \to \infty} f_{n} \right) dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$

By Fatou’s lemma,

$$ \begin{align*} \left\| f - f_{n} \right\|_{1} =& \int |f - f_{n}| dm \\ \le & \liminf_{k \to \infty} \int | f_{N_{k}} - f_{n}| dm \\ =& \liminf_{k \to \infty} \left\| f_{N_{k}} - f_{n} \right\| \\ \lt& \varepsilon \end{align*} $$

Since $f_{n}$ is a Cauchy sequence by assumption, the above inequality holds for any $\varepsilon > 0$, and therefore $\left\| f_{n} - f \right\|_{1} \to 0$. In short, $f_{n}$ is Cauchy and its subsequence converges to $f$, so $f_{n}$ converges to $f$. Here $f - f_{n} \in L^{1}$, and since $L^{1}$ is a vector space,

$$ ( f - f_{n} ) + f_{n} = f \in L^{1} $$

Since every Cauchy sequence of $L^{1}$ converges to an element of $L^{1}$, $L^{1}$ is a complete space.

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