# $L^{\infty}$ is a Banach Space

Show that $L^{\infty}$ is a Banach Space with respect to the norm $||.||_{\infty}$ where $||f||_{\infty}=\inf\{ a\ge 0: \mu\left(\{x: |f(x)| \gt a\}\right)=0\}$

I am going to prove the following lemmas and then use them in the proof.

Lemma-1: $||.||_{\infty}$ is a norm on $L^{\infty}$

Proof: Suppose that $||f||_{\infty}=0$. Let $A=\{x \in X : f(x) \ne 0\}$ and $A_n=\{x \in X: |f(x)| \gt \frac{1}{n}\}$ . Then $A=\cup_{n}A_n$. Then there exists $'a'$ such that $\mu\left(\{x: |f(x)| \gt a\}\right)=0$ and $a \lt \frac{1}{n}$ (by the definition of $||f||_{\infty}$). Thus $$\mu\left(\{x: |f(x)| \gt \frac{1}{n}\}\right) \le \mu\left(\{x: |f(x)| \gt a\}\right)=0$$

$\implies \mu(A_n)=0$ and hence $\mu(A)=0$. Thus $f =0 ,\mu$-a.e.

Now $$|f(x)+g(x)| \le |f(x)| +|g(x)| \le ||f||_{\infty}|+||g||_{\infty} ,\mu-\text{a.e}$$

Thus $||f+g||_{\infty} \le ||f||_{\infty}+||g||_{\infty}$

Now $$||\lambda f||_{\infty}=\inf\{a \ge 0: \mu\left(\{x: |\lambda f(x)| \gt a\}\right)=0\}$$ $$=\inf\{|\lambda|a \ge 0:\mu\left(\{x: |\lambda||f(x)| \gt |\lambda|a\}\right)=0\}$$ $$=|\lambda|\inf\{a \ge 0:\mu\left(\{x: |f(x)| \gt a\}\right)=0\}=|\lambda|||f||_{\infty}$$

Lemma-2: $||f_n-f||_{\infty} \to 0$ iff there exists $E \in \mathcal{M}$ such that $\mu(E^c)=0$ and $f_n \to f$ uniformly on $E$.

Proof: $(\implies)$ For every $k \in \mathbb{N}$, there exists $n_0(k) \in \mathbb{N}$ such that for all $n \ge n_0(k)$, $||f_n-f||_{\infty} \lt \dfrac{1}{k}$. Then for each $n \ge n_0(k)$ there exists $E_n^{k}$ such that $\mu(E_n^k)=0$ and $|f_n(x)-f(x)| \lt \dfrac{1}{k}$ for all $x \in (E_n^k)^{c}$. Let $$E^k=\cup_{n \ge n_0(k)} E_n^k$$. Then $\mu(E^k)=0$ and for all $x \not \in E^k, |f_n(x)-f(x)| \lt \frac{1}{k}, \forall n \ge n_0(k)$. Now Let $E=\cup_{k=1}^{\infty} E^k$. Then $\mu(E)=0$. Let $\epsilon \gt 0$. Then there exists a $k_0 \in \mathbb{N}$ such that $\frac{1}{k} \lt \epsilon$ for all $k \ge k_0$. Let $x \in E^c$. In Particular $x \not \in E^{k_0}$ . Then for all $n \ge n_0(k_0), |f_n(x)-f(x)| \lt \frac{1}{k_0} \lt \epsilon$

($\impliedby$) Let $\epsilon \gt 0$. Then there exists $n_0 \in \mathbb{N}$ such that for all $n \ge n_0, |f_n(x)-f(x)| \lt \epsilon, x\in E$. Then for all $n \ge n_0$,

$$\left(\{x: |f_n(x)-f(x)| \gt \epsilon\}\right) \subset E^c$$ $$\implies \mu\left(\{x: |f_n(x)-f(x)| \gt \epsilon\}\right)=0$$ which in turn gives us that for all $n \ge n_0$, we have $$||f_n-f||_{\infty} \lt \epsilon$$.

(Proof that $L^{\infty}$ is a Banach Space) :

Let $\{f_n\}_{n \in \mathbb{N}} \in L^{\infty}$ be a cauchy sequence. Then for $\epsilon \gt 0$, there exists a $n_0(\epsilon) \in \mathbb{N}$ such that for all $n,m \ge n_0(\epsilon)$, we have $||f_n-f_m||_{\infty} \lt \dfrac{\epsilon}{2}$. Then there exists $E^{\epsilon} \in \mathcal{M}$ such that $\mu(E^{\epsilon})=0$ and for $x \in (E^{\epsilon})^{c} ,|f_n(x)-f_m(x)| \lt \dfrac{\epsilon}{2}$.

Let $$E= \cup_n E^{\frac{1}{n}}.$$ Then $\mu(E)=0$ and for all $x \in E^c, \{f_n(x)\}$ is a cauchy sequence. Let $f(x)=\lim_n f_n(x)$ for $x \in E^c$. Let $\epsilon \gt 0$. Then there exists $k_0 \in \mathbb{N}$ such that for all $k \ge k_0,\frac{1}{k} \lt \frac{\epsilon}{2}$. Then for all $n, m \ge n_0(k_0)$, we have $$|f_n(x)-f_m(x)| \lt \frac{1}{k} \lt \frac{\epsilon}{2}$$ Letting $m \to \infty$ we have $$|f_n(x)-f(x)| \lt \epsilon, x\in E^c.$$ By Lemma-2, we have $||f_n-f||_{\infty} \to 0$. From here by use of triangle inequality, we see that $f\in L^{\infty}$.

Is this proof alright?

Thanks for the help!!

## 2 Answers

I think your proof if correct. But here is a theorem in Folland's real analysis book I proved, (chapter 6.1, exercise 2)

Theorem 6.8

a.) If $f$ and $g$ are measurable functions on $X$, then $\|fg\|_{1}\leq\|f\|_{1}\|g\|_{\infty}$. If $f\in L^1$ and $g\in L^{\infty}$, $\|fg\|_{1} = \|f\|_{1}\|g\|_{\infty}$ if and only if $|g(x)| = \|g\|_{\infty}$ a.e. on the set where $f(x)\neq 0$.

b.) $\|\cdot\|_{\infty}$ is a norm on $L^{\infty}$.

c.) $\|f_n - f\|_{\infty}\to 0$ if and only if there exists $E\in M$ such that $\mu(E^c) = 0$ and $f_n\rightarrow f$ uniformly on $E$.

d.) $L^{\infty}$ is a Banach space.

e.) The simple functions are dense in $L^{\infty}$.

Proof a.) Let $f,g$ be measurable functions on $X$ such that $f\in L^1$ and $g\in L^{\infty}$. Then, since $g\in L^{\infty}$ then by definition $|g(x)|\leq \|g\|_{\infty}$ a.e. on $X$. So, $$|f(x)||g(x)|\leq |f(x)|\|g\|_{\infty}$$ Recall that $\int |f(x)g(x)|d\mu(x) = \|fg\|_{1}$. Now integrating over all $X$ we have \begin{align*} \int_X |f(x)||g(x)|d\mu(x) &= \int |f(x)g(x)|d\mu(x)\\ &\leq \|g\|_{\infty}\int_{X} |f(x)|d\mu(x)\\ &= \|g\|_{\infty}\|f\|_{1} \end{align*} Therefore, $$\|fg\|_{1}\leq\|f\|_{1}\|g\|_{\infty}$$ Second part:\ $\Leftarrow$ Let $A = \{x\in X: f(x)\neq 0 \}$. Suppose $|g(x)| = \|g\|_{\infty}$ for every $x\in A$. Then, $$\|fg\|_{1} = \int_{X}|f(x)||g(x)|d\mu = \|g\|_{L^\infty}\int_{A}|f(x)|d\mu = \|g\|_{L^\infty}\|f\|_{L^1}$$ since the integral is zero everywhere outside $A$, since $f(x)$ is zero on $X\setminus A$.\ $\Rightarrow$ Suppose $\|fg\|_{1} = \|f\|_{1}\|g\|_{\infty}$. So we have $$\int |fg| d\mu = \int |f|\,\|g\|_{\infty} d\mu$$ Since $|g| \leq \|g\|_{\infty}$, we have $|fg| = |f|\,|g|\leq |f|\,\|g\|_{\infty}$, and so we must have $|f|\,|g| = |f|\,\|g\|_{\infty}$ a.e.. So $|g(x)| = \|g\|_{\infty}$ a.e. on the set where $f(x)\neq 0$.

Proof b.) Clearly $\|f\|_{\infty} = 0$ if and only if $|f| = 0$. Now, suppose $\lambda\in K$ then $$\|\lambda f\|_{\infty} = ess \sup_{x\in X}|\lambda f(x)| = ess \sup_{x\in X} |\lambda||f(x)| = |\lambda| ess \sup_{x\in X} |f(x)| = |\lambda|\|f(x)\|$$ Finally we know that $|f(x)|\leq \|f\|_{\infty}$ and $|g(x)|\leq \|g\|_{\infty}$ a.e. on $X$, so that for a.e. on $X$ we have $$|f(x) + g(x)|\leq |f(x)| + |g(x)| \leq \|f\|_{\infty} + \|g\|_{\infty}$$ Then by definition of the essential supremum of $|f + g|$ we have $$\|f + g\|_{\infty}\leq \|f\|_{\infty} + \|g\|_{\infty}$$

Proof c.) Suppose $\|f_n - f\|\to 0$ for all $n$. Then given $\epsilon > 0$ there exists an $N\in\mathbb{N}$ such that for $n\geq N$ $$\|f_n - f\|_{\infty} < \epsilon$$ Thus for any $n\geq N$, we have $$|f_n(x) - f(x)|\leq \|f_n - f\|_{\infty} < \epsilon \tag{1}$$ a.e. on $X$. Let $M_n = \|f_n - f\|_{\infty}$ for all $n\geq N$ and set $$A_n = \{x\in X: |f_n(x) - f(x)| > M_n \}$$ Then we have $\mu(A_n) = 0$. Now, let $A = \bigcup_{n\geq N}A_n$ then $\mu(A) = 0$. Let $E = A^c$ Then for $x\in E$ we have from $(1)$ $$|f_n(x) - f(x)| \leq \|f_n - f\|_{\infty} < \epsilon \ \ \forall n\geq N$$ Thus, $$|f_n(x) - f(x)| < \epsilon \ \ \forall n \geq N$$ and so $f_n\to f$ uniformly on $E$ and clearly $\mu(E^c) = 0$. \ Conversely, suppose $E\in M$ and $\mu(E^c) = 0$ and $f_n\to f$ uniformly on $E$, then for each $\epsilon > 0$ there exists an $N\in\mathbb{N}$ such that $$|f_n(x) - f(x)| < \epsilon \ \ \forall n\geq N \ \ \text{and} \ \ x\in E$$ Hence we also have $$|f_n(x) - f(x)| < \epsilon \ \ \text{a.e. on} \ \ X$$ Thus by definition of the essential supremum of $|f_n - f|$ we have $$\|f_n - f\|_{\infty} < \epsilon$$

Proof d.) We want to show that $L^{\infty}$ is complete as a metric space in the metric associated with $\|\cdot\|_{\infty}$. Let $\{f_n\}_{n\in\mathbb{N}}$ be a Cauchy sequence in $L^{\infty}$. Thus given $\epsilon > 0$, there exists an $N\in\mathbb{N}$ such that $$\|f_m - f_n\| < \epsilon \ \ \forall m,n\geq N \tag{1}$$ For each $m,n\in\mathbb{N}$, set $$F_{m,n} = \{x\in X: |f_m(x) - f_n(x)| > \|f_m - f_n\|_{\infty} \}$$ Then clearly $\mu(F_{m,n}) = 0$ for all $m,n\in\mathbb{N}$. Set $F = \bigcup_{m,n\in\mathbb{N}}F_{m,n}$ and $E = F^c$. Note that $\mu(E^c) = \mu(F) = 0$. Moreover, \begin{align*} E &= \bigcap_{m,n\in\mathbb{N}}\{x\in X: |f_m(x) - f_n(x)|\leq \|f_m - f_n\|_{\infty} \}\\ &= \{x\in X: |f_m(x) - f_n(x)\leq \|f_m - f_n\|_{\infty} \ \forall \ m,n\in\mathbb{N} \} \end{align*} Let $\epsilon > 0$ and choose $N\in\mathbb{N}$ such that $(1)$ holds. Then for $x\in E$ and for all $m,n\geq N$ we have $$|f_m(x) - f_n(x)| \leq \|f_n - f_m\| < \epsilon \tag{2}$$ This shows that for every $x\in E$, $\{f_n(x)\}$ is a Cauchy sequence in $K = (\mathbb{R}$ or $\mathbb{C}$). Since $K$ is a complete metric space, there exists a limit $f(x) = \lim_{n\to \infty}f_n(x)$. Note $f(x)$ is defined in $E$ (i.e. outside of $F$). Thus for $x\in F$, $f(x) = 0$. Note that $f = \lim_{n\to \infty}f_n(x)\chi_{E}$ is measurable. Now recall earlier that for $\epsilon > 0$ there is an $N\in\mathbb{N}$ such that for $x\in E$ we have for all $m,n\geq N$ $$|f_m(x) - f_n(x)| \leq \epsilon$$ Let $n\to \infty$ then we have for $m\geq N$ $$|f_m(x) - f(x)| \leq \epsilon$$ Thus for $m\geq N$ $$\|f_m(x) - f(x)\|_{\infty} = ess\sup |f_m - f| \leq \epsilon$$ This shows that $f_m \to f$ in $L^{\infty}$ norm.\ Finally, we note that $f\in L^{\infty}$ from the triangle inequality: $$\|f\|_{\infty} \leq \|f_m\|_{\infty} + \|f_m - f\|_{\infty} \leq \|f_m\|_{\infty} + \epsilon < \infty$$ for $m\geq N$. Thus $L^{\infty}$ is a Banach space.

Proof e.) Let $f\in L^{\infty}$ where $f$ is bounded. Then by Theorem 2.10 there exists a sequence $\{\phi_n\}$ of simple functions that converge uniformly to $f$. Thus given $\epsilon > 0$ $$\|f - \phi_n\|_{\infty} = ess\sup|f(x) - \phi_n(x)| \leq \epsilon$$ Thus since $\epsilon$ is arbitrary we are done.

• For part d), conclusion from part c) can be used by noting that $f_n \to f$ uniformly as their sup, $||.||_\infty$ on E is lower than $\epsilon$. Dec 2, 2019 at 18:07

I would probably add one more lemma, namely:

If $f \in L^{\infty}(X)$, then $|f(x)|\leq||f||_{\infty}$, a.e. on $X$.

This can be done by definition of infimum, like you did in Lemma 1.

Very nice proof that $L^{\infty}(X)$ is Banach space you can find in Real Analysis, Theory of measure and integration, by James Yeh.