Real Analysis, Folland Problem 6.1.2 $L^p$ spaces

Question

Background Information: In this chapter we work on a fixed measure space $(X,M,\mu)$. If $f$ is measurable on $X$ and $0 < p < \infty$, we define $$\|f\|_{L^p} = \left[\int |f|^p d\mu\right]^{1/p}$$ and we define $$L^p(X,M,\mu) = \{f:X\rightarrow \mathbb{C}: f \ \text{is measurable and} \ \|f\|_p <\infty\}$$

Holder's Inequality - If $p > 1$ and $\frac{1}{p} + \frac{1}{q} = 1$ then $$\|fg\|_{L^1} \leq \|f\|_{L^p}\|g\|_{L^q}$$

Theorem 6.8

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

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

c.) $\|f_n - f\|_{L^\infty}$ 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 $L^{\infty}$

Attempted proof a.): Let $(X,M,\mu)$ be a fixed measure space. Suppose $f$ and $g$ are measurable functions on $X$. Let $p\in\{1,\infty\}$ from Holder's inequality it follows that $$\|fg\|_{L^1}\leq\|f\|_{L^1}\|g\|_{L^\infty}$$ I am not sure if this is rigourous enough and I am not sure how to prove the remaining parts.

Attempted proof b.) We have $$L^{\infty} = L^{\infty}(X,M,\mu) = \{f:X\rightarrow \mathbb{C}: f \ \text{is measurable and } \|f\|_{L^\infty} < \infty\}$$ It is obvious that $\|f\|_{L^\infty} = 0$ iff $f = 0$ a.e. and $\|f\|_{L^\infty} = |c|\|f\|_{L^\infty}$ so now we just need to show that the triangle inequlaity holds in order to prove the claim that $\|\cdot\|_{L^\infty}$ is a norm in $L^{\infty}$. Let $p = \infty$, by Minkowski's inequality it follows that $$\|f + g\|_{L^\infty} \leq \|f\|_{L^\infty} + \|g\|_{L^\infty}$$ thus we are done.

Attempted proof c.) If $\|f_n - g\|_{L^{\infty}}$ then $f_n\rightarrow f$ in $L^{\infty}$. Now for $\epsilon > 0$ there exists an $N$ such that $$E_n = \bigcup_{m\geq N}\{|f_m - f| > \frac{1}{b}\}$$ Take $E = \bigcup_{1}^{\infty}E_n$ then clearly $\mu(E^c) = 0$. Now we need to show that $f_n\rightarrow f$ uniformly on $E$ which I am not exactly sure how to do.

I know these are "simple" to prove, but I am pretty lost on how to begin with each of these. I just read the section yesterday so maybe I just need to digest it more. Any help would be great, thanks.

Answer 1

Assuming you're using the essential supremum norm, here's an answer to part b) and part c). Here is a link to wikipedia's page on the essential supremum and below the proof of part c) is a link to a proof that $(L^\infty, \|\cdot\|_\infty)$ is a Banach space.

Proof of b): Relying on the fact that a linear combination of essentially bounded and integrable functions is again essentially bounded and integrable (hence $L^\infty$ is a vector space) we shall only verify that $ess\, \sup |f|$ is indeed a norm on $L^\infty$ and the metric induced by this norm is complete. We first prove the essential supremum of the absolute value of an essentially bounded and integrable functions satisfies the three properties of a norm.

Lemma (for you to prove, ask for more details if necessary): It is important to note that for any $f\in L^\infty$, we have $|f(x)|\leq \|f\|_\infty$ a.e. on $X$.

First suppose $\|f\|_\infty=0$. Then if $\epsilon >0$ there exists $C\geq 0$ such that $|f(x)|\leq C$ almost everywhere on $X$ and $0\leq C< \epsilon$ since $\epsilon$ cannot be a lower bound of the set $\hat{U}_f=\{C\geq 0: |f(x)|\leq C \text{ a.e. on } X\}$. But then we have $|f(x)|\leq C< \epsilon$ a.e. on $X$ so that $|f(x)|<\epsilon$ a.e. on $X$ which in turn implies $f(x)=0$ a.e. $X$ (since $\epsilon >0$ was arbitrary), so the only set where $f$ is not zero is a set of a measure zero, so by the definition of $L^\infty$ (remember elements of $L^\infty$ aren't functions, but classes of functions, where two functions in a given class differ only on sets of measure zero, but are identical else where) we see that $f(x)\equiv 0$, i.e. the class of functions of which are zero almost everywhere. Thus $\|f\|_\infty=0$ implies $f=0$. The other direction is less work; if $f(x)=0$ almost everywhere on $X$ then it follows that $0=\inf \{C\geq 0: |f(x)| \leq C \text{ a.e. on }X\}$, so $\|f\|_\infty=0$. Thus, we've proved $\|f\|_\infty=0$ if and only if $f=0$.

Now let $\alpha \in \mathbb{C}$. We wish to show $\|\alpha f\|_\infty = |\alpha | \cdot \| f\|_\infty$ or in other words, $|\alpha | ess\, \sup |f|=ess\, \sup |\alpha f|$. We may assume $\alpha \neq 0$ else it is trivial. Now note, that $$|\alpha f(x)|\leq |\alpha ||f(x)|\leq |\alpha |\cdot \|f\|_\infty \text{ a.e. on } X$$ hence by definition of $ess\, \sup |\alpha f|$ we have $$\|\alpha f\|_\infty \leq |\alpha |\cdot \|f\|_\infty. $$ On the other hand, we know $$|\alpha f(x)|\leq \|\alpha f\|_\infty \text{ a.e. on } X$$ which implies $$| f(x)|\leq \frac{\|\alpha f\|_\infty}{|\alpha|} \text{ a.e. on } X$$ so that by the definition of $ess \, \sup |f|$, we have $$\|f\|_\infty \leq \frac{\|\alpha f\|_\infty}{|\alpha|}$$ and hence $$|\alpha|\|f\|_\infty \leq \|\alpha f\|_\infty$$ this together with the first inequality we proved shows that $|\alpha | \cdot \| f\|_\infty = \|\alpha f\|_\infty$.

Now it remains to prove the triangle inequality, i.e. for $f,g\in L^\infty$ (again keeping in mind these are classes of functions), we wish to show, $$\|f+g\|_\infty \leq \|f\|_\infty + \|g\|_\infty.$$ Now, we know that $|f(x)|\leq \|f\|_\infty$ and $|g(x)|\leq \|g\|_\infty$ almost everywhere on $X$, so that $$|f(x)+g(x)|\leq |f(x)|+|g(x)|\leq \|f\|_\infty+ \|g\|_\infty \text{ a.e. on } X$$ and since the right hand side is just some real nonnegative number, it follows by the definition of $ess \, \sup |f+g|$ that $$\|f+g\|_\infty \leq \|f\|_\infty + \|g\|_\infty.$$

Part c): Convergence in $L^\infty$ is uniform convergence almost everywhere. More precisely $$\lim_{n\to \infty} \|f_n-f\|_\infty=0 \iff \exists \, E\in \Sigma \text{ such that } \mu(E^c)=0 \text{ and } (f_n)\to f \text{ uniformly on } E$$

Proof: If $\exists \, E\in \Sigma \text{ such that } \mu(E^c)=0 \text{ and } (f_n)\to f \text{ uniformly on } E$ then, we know that for any $\epsilon >0$ there exists $N\in \mathbb{N}$ such that $$|f_n(x)-f(x)|< \epsilon \text{ for all } n\geq N \text{ and all } x\in E.$$ Now from this we wish to show $\lim_{n\to \infty} \|f_n-f\|_\infty=0$. By the above inequality however, we have for any $n\geq N$, $$|f_n(x)-f(x)|< \epsilon \text{ a.e. on } X$$ since $\mu(E^c)=0$ and $E^c$ is at most where the uniform convergence doesn't hold. Thus by the definition of $ess\, \sup |f_n-f|$ we have $$\|f_n-f\|_\infty <\epsilon$$ for all $n\geq N$. Thus $f_n \to f$ as $n\to \infty$ (in the $L^\infty$ metric).

The converse will take a little more work. We assume $\lim_{n\to \infty} \|f_n-f\|_\infty=0$, i.e. for any $\epsilon >0$ there exists $N'\in \mathbb{N}$ such that $n\geq N'$ implies $\|f_n-f\|_\infty <\epsilon$. This implies that for any $n\geq N'$, $$|f_n(x)-f(x)|\leq \|f_n-f\|_\infty <\epsilon$$ a.e. on $X$. Let $M_n=\|f_n-f\|_\infty $ for each $n\geq N'$ and $$A_n=\{x\in X: |f_n(x)-f(x)|> M_n\}.$$ Since the above inequality holds almost everywhere on $X$, it is clear that $\mu(A_n)=0$. Let $A=\cup_{n\geq N} A_n$. Then $\mu(A)=0$ since a countable union of sets with measure zero is itself measurable and with measure zero. Finally let $E=A^c$ and it follows that this is the set where $f_n$ converges uniformly to $f$ and $\mu(E^c)=0$, completing the proof.

For a proof of d), see L^infinity is a Banach space

Let me know if you have any questions/see a typo I missed/etc... I'll try to update soon with hints for parts a) and e) but I must run now... I know e) has an analogous theorem for $L^p$ spaces so the proof should not be greatly different...

Edit 1: Part a) Here is a proof of the first claim in a), i.e. if $f$ and $g$ are measurable and $g$ is essentially bounded then $\|fg\|_1\leq \|f\|_1\|g\|_\infty$.

Proof: Recall the above lemma that says $|g(x)|\leq \|g\|_\infty$ a.e. on $X$ for any essentially bounded and measurable function $g: X\to \mathbb{C}$. Thus, $$|f(x)||g(x)|\leq |f(x)|\cdot \|g\|_\infty \text{ a.e. on } X.$$ Upon integrating over all of $X$, implicitly using the fact that if $h$ is an integrable function and $A\subset X$ is a set of measure zero then $\int_A h =0$, we obtain, $$\int_X |f(x)g(x)|d\mu \leq \|g\|_\infty \int_X |f(x)|d\mu$$ i.e. $\|fg\|_1\leq \|f\|_1\|g\|_\infty$.

Edit 2: Now for the second claim: If $\|fg\|_1=\|f\|_1\cdot \|g\|_\infty$ if and only if $|g(x)|=\|g||_\infty$ on the set where $f(x)\neq 0$.

Proof: Let $A=\{x\in X: f(x)\neq 0\}$. Suppose $|g(x)|=\|g||_\infty$ for every $x\in A$. Then, $$\|fg\|_1=\int_X |f(x)||g(x)|d\mu=\|g\|_\infty \cdot \int_A |f(x)|d\mu = \|g\|_\infty \|f\|_1$$ since the integral is zero everywhere besides $A$, since $f(x)$ is zero on $X\setminus A$.

I still can't figure out the converse...