The L∞(E) Normed Linear Space

# The L∞(E) Normed Linear Space

Definition: The $E$ be a Lebesgue measurable set. Then the $L^{\infty}(E)$ Space is the set $L^{\infty}(E) = \{ f \: \mathrm{measurable} : \exists M > 0 \: \mathrm{where} \: |f(x)| \leq M \: a.e. \: \mathrm{on} \: E \}$ with the norm $\| \cdot \|_{\infty} : L^{\infty}(E) \to [0, \infty)$ defined for each $f \in L^{\infty}(E)$ by $\| f \|_{\infty} = \inf \{ M : |f(x)| \leq M \: a.e. \: \mathrm{on} \: E \}$. |

Observe that $L^{\infty}(E)$ is the set of measurable functions that are essentially bounded on $E$. That is, $f \in L^{\infty}(E)$ if $f$ is bounded except possibly on a set of measure zero. The $\infty$-norm of $f$ is simply the infimum of all $M > 0$ that bound $f$ except possibly on a set of measure zero.

Proposition 1: $(L^{\infty}(E), \| \cdot \|_{\infty})$ is a normed space. |

**Proof:**Like before, $L^{\infty}(E)$ is a subset of the set of measurable functions (which is a linear space), so to show that $L^{\infty}(E)$ is a linear space it suffices to show that $L^{\infty}(E)$ is closed under addition, closed under scalar multiplication, and contains the $0$ function.

- Let $f, g \in L^{\infty}(E)$. Then $f$ and $g$ are measurable and so therefore $f + g$ is measurable. Since $f$ is bounded almost everywhere on $E$ there exists an $M_f > 0$ and a set $E_f \subseteq E$ with $m(E_f) = 0$ such that $|f(x)| \leq M_f$ on $E \setminus E_f$. Similarly, since $g$ is bounded almost everywhere on $E$ there exists an $M_g > 0$ and a set $E_g \subseteq E$ with $m(E_g) = 0$ such that $|g(x)| \leq M_g$ on $E \setminus E_g$. Observe that a finite union of measure zero sets is also of measure zero. In particular, $m(E_f \cup E_g) = 0$, and we have that:

\begin{align} \quad |f(x) + g(x)| \leq |f(x)| + |g(x)| \leq M_f + M_g \end{align}

- on $(E \setminus E_f) \cap (E \setminus E_g) = E \setminus (E_f \cup E_g)$. Therefore $(f + g) \in L^{\infty}(E)$.

- Now let $\alpha \in \mathbb{R}$ and let $f \in L^{\infty}(E)$. Then $f$ is measurable and so therefore $\alpha f$ is measurable. Since $f$ is bounded almost everywhere on $E$ there exists an $M > 0$ and a set $E^* \subseteq E$ with $m(E^*) = 0$ such that $|f(x)| \leq M$ on $E \setminus E^*$. Therefore:

\begin{align} \quad |\alpha f(x)| = |\alpha||f(x)| \leq |\alpha|M \end{align}

- on $E \setminus E^*$. Therefore $\alpha f \in L^{\infty}(E)$.

- Lastly, the zero function $0$ is measurable and clearly bounded everywhere on $E$, and so $0 \in L^{\infty}(E)$. Hence $L^{\infty}(E)$ is a linear space.

- All that remains to show is that $\| \cdot \|_{\infty}$ is indeed a norm on $L^{\infty}(E)$.

**Showing that $\| f \|_{\infty} = 0$ if and only if $f = 0$ a.e. on $E$:**Suppose that $\| f \|_{\infty} = 0$. Then $|f(x)| \leq 0$ a.e. on $E$ which implies that $f = 0$ a.e. on $E$. Conversely, suppose that $f = 0$ a.e. on $E$. Then $|f(x)| \leq 0$ a.e. on $E$. So $0 \leq \| f \|_{\infty} = \inf \{ M : |f(x)| \leq M \: a.e. \: \mathrm{on} \: E \} \leq 0$, thus $\| f \|_{\infty} = 0$.

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

\begin{align} \quad \| \alpha f \|_{\infty} &= \inf \{ M : |\alpha f(x)| \leq M \: a.e. \: \mathrm{on} \: E \} \\ &= \inf \{ M : |\alpha||f(x)| \leq M \: a.e. \: \mathrm{on} \: E \} \\ &= |\alpha| \inf \{ M^* : |f(x)| \leq M^* \: a.e. \: \mathrm{on} \: E \} \\ &= |\alpha| \| f \|_{\infty} \end{align}

- Lastly, let $f, g \in L^{\infty}(E)$. Then:

\begin{align} \quad \| f + g \|_{\infty} &= \inf \{ M : |f(x) + g(x)| \leq M \: a.e. \: \mathrm{on} \: E \} \\ &\leq \inf \{ M : |f(x)| + |g(x)| \leq M \: a.e. \: \mathrm{on} \: E \} \\ &\leq \inf \{ M_1 : |f(x)| \leq M_1 \: a.e. \: \mathrm{on} \: E \} + \inf \{M_2 : |g(x)| \leq M_2 \: a.e. \: \mathrm{on} \: E \} & \leq \| f \|_{\infty} + \| g \|_{\infty} \end{align}

- Therefore $\| \cdot \|_{\infty}$ is a norm on $L^{\infty}(E)$ and so $(L^{\infty}(E), \| \cdot \|_{\infty}$ is a normed linear space. $\blacksquare$