Why does Rudin say that $f$ is in $L^\infty$?

Question

The setting is that we have $L^\infty([0,1],m)$, $m$ Lebesgue measure, and we have shown the Gel'fand transform is an isometry onto $\mathcal{C}(\Delta)$, $\Delta$ being the space of maximal ideals / nonzero multiplicative functionals (that is, characters). Then he says the following.

Next, $\hat f\mapsto\int fdm$ is a bounded linear functional on $\mathcal{C}(\Delta)$. By the Riesz representation theorem, there is a regular probability measure $\mu$ on $\Delta$ that satisfies:

$$\int_\Delta\hat fd\mu=\int_0^1f dm, \tag{10}$$

for every $f\in L^\infty(m)$. This measure is related to the topology of $\Delta$ in the following way:

(i) $\mu(V)>0$ if $V$ is open and nonempty;

(ii) To every bounded Borel function $\phi$ on $\Delta$ there corresponds an $\hat f\in\mathcal{C}(\Delta)$ such that $\hat f=\phi$ $\mu$-a.e.;

(iii) If $V$ is open, so is $\overline V$;

(iv) If $E$ is a Borel set in $\Delta$, then:

$$\mu(E)=\mu(E^0)=\mu(\overline E); \tag{11}$$

If $V$ is as in (i), Urysohn's lemma implies that there is an $\hat f\in\mathcal{C}(\Delta)$, $\hat f\geq0$, such that $\hat f=0$ outside $V$ and $\hat f(p)=1$ at some $p\in V$. Hence $f$ is not the zero element of $L^\infty(m)$, and the integrals in (10) are positive. This gives (i).

In (ii), assume $|\phi|\leq1$. Since $\mathcal{C}(\Delta)$ is dense in $L^2(\mu)$ (recall that $\mu$ is a regular Borel probability measure), there are functions $\hat f_n\in\mathcal{C}(\Delta)$ such that $\int|\hat f_n-\phi|d\mu\to0$, and which can be so adjusted that $|\hat f_n|\leq1$. Then $\|f_n\|_\infty\leq1$, and (10) implies that $\{f_n\}$ is a Cauchy sequence in $L^2(m)$. Hence there is an $f\in L^\infty(m)$ such that:

$$\int_\Delta|\hat f_n-\hat f|d\mu=\int_0^1|f_n-f|^2dm\to0, \tag{12}$$

as $n\to\infty$. Thus $\phi=\hat f$ $\mu$-a.e.

Now I understand that the $\hat f_n$ exist, and can be adjusted in that way; that $\hat f_n$ converges in $L^2(\mu)$, and hence is Cauchy therein; that $f_n$ is therefore Cauchy in $L^2(m)$, and hence converges to some $f\in L^2$. But I just cannot find a way of proving that there is an $f\in L^\infty$ that is the limit of the $f_n$'s in $L^2$. I mean, the measure is finite, so $L^2\subseteq L^\infty$, not the other way round, so $f\in L^2$, OK, but how is it in $L^\infty$? I feel like I have to exploit the fact that $|\phi|\leq1$ somehow, but can't quite see how… I did figure out we have $f_{n_k}\to f$ pointwise $m$-a.e., and that $\int|f_n|^2\to\int|f|^2$, but how do I use these facts to conclude?

Answer 1

I figured it out while editing the image in. $f_n\in L^\infty$, and since the measure is finite this implies $f_n\in L^p$ for all $p\in[1,\infty]$. By Fatou's lemma (poor Fatou, always forgotten), we have:

$$\int|f|^p=\int\liminf|f_{n_k}|^p\leq\liminf\int|f_{n_k}|^p\leq\liminf\|f_n\|_\infty\leq1,$$

hence $f\in L^p$ for all $p$. But since the measure is finite (indeed, a probability), we have:

$$\|f\|_\infty=\lim_{p\to\infty}\|f\|_p\leq1,$$

by this question's proof, as we wanted. So $f\in L^\infty$ after all.

Update

@Hamza's comment prompted me to reflect, and indeed I can conclude $\|f\|_\infty\leq1$ in another way. Indeed, take the $f_{n_k}\to f$ pointwise $m$-a.e.. This produces a set $E$ such that $m(E)=0$ and $f_{n_k}\to f$ pointwise outside $E$. $\|f_{n_k}\|_\infty\leq1\implies|f_{n_k}|\leq1$ $m$-a.e., so this produces $E_k$ of measure zero such that $|f_{n_k}|\leq1$ outside $E_k$. Take $A=\bigcup_kE_k\cup E$. This is a coutable union of measure-zero set, hence has measure zero, and outside it we have $|f_{n_k}|\leq1$ and $f_{n_k}\to f$, so $|f|\leq1$ outside $E$, hence $\|f\|_\infty\leq1$.