The L1(E) Normed Linear Space

# The L1(E) Normed Linear Space

Definition: The $E$ be a Lebesgue measurable set. Then the $L^1(E)$ Space is the set $\displaystyle{L^1(E) = \{ f \: \mathrm{measurable} : \int_E |f| < \infty \}}$ with the norm $\| \cdot \|_1 : L^1(E) \to [0, \infty)$ defined for each $f \in L^1(E)$ by $\displaystyle{\| f \|_1 = \int_E |f|}$. |

Observe that $L^1(E)$ is the set of measurable functions that are Lebesgue integrable on $E$, and for every $f \in L^1(E)$, the $1$-norm of $f$ is simply the value of the integral of $|f|$ on $E$.

Proposition 1: $(L^1(E), \| \cdot \|_1)$ is a normed space. |

**Proof:**The set of measurable functions is a linear space and $L^1(E)$ is a subset of that space. To show that $L^1(E)$ is a linear space, all we need to show is that it is closed under addition, closed under scalar multiplication, and contains the zero function. But clearly, if $f, g \in L^1(E)$ then $f$ and $g$ are measurable and hence $f + g$ is measurable. Furthermore, $\int_E |f + g| \leq \int_E |f| + \int_E |g| < \infty$ and so $f + g$ is Lebesgue integrable on $E$, so $(f + g) \in L^1(E)$. If $\alpha \in \mathbb{R}$ and $f \in L^1(E)$ then $f$ is measurable and so $\alpha f$ is measurable. Furthermore, $\int_E |\alpha f| = \int_E |\alpha||f| =|\alpha| \int_E |f| < \infty$ and so $\alpha f$ is Lebesgue integrable on $E$, so $\alpha f \in L^1(E)$. Lastly, $0$ is a measurable function and is clearly Lebesgue integrable on $E$, so $0 \in L^1(E)$. Hence $L^1(E)$ is a linear space.

- We now show that $\| \cdot \|_1$ is a norm on $L^1(E)$.

**Showing that $\| f \|_1 = 0$ if and only if $f = 0$ a.e. on $E$:**Suppose that $\| f \|_1 = 0$. Then $\int_E |f| = 0$ which implies that $f = 0$ a.e. on $E$. Conversely, suppose that $f = 0$ a.e. on $E$. Then $\| f \|_1 = \int_E |f| = \int_E 0 = 0$. Therefore $\| f \|_1 = 0$ if and only if $f = 0$ a.e. on $E$.

**Showing that $\| \alpha f \|_1 = |\alpha| \| f \|_1$.:**Let $\alpha \in \mathbb{R}$ and let $f \in L^1(E)$. Then:

\begin{align} \quad \| \alpha f \|_1 = \int_E |\alpha f| = \int_E |\alpha||f| = |\alpha| \int_E |f| = |\alpha| \| f \|_1 \end{align}

**Showing that $\| f + g \|_1 \leq \| f \|_1 + \| g \|_1$:**Let $f, g \in L^1(E)$. Then by the triangle inequality and the additivity of the Lebesgue integral:

\begin{align} \quad \| f + g \|_1 = \int_E |f+g| \leq \int_E [|f| + |g|] = \int_E |f| + \int_E |g| = \| f \|_1 + \| g \|_1 \end{align}

- Therefore $\| \cdot \|_1$ is a norm. Hence $(L^1(E), \| \cdot \|_1)$ is a normed linear space. $\blacksquare$