Rudin's RCA $5.15$ Theorem.

Question

Premise. These are the definitions which we need for the proof of the theorem:

$1.39$ Theorem: suppose $f$ $\in$ $L^1(\mu)$ and $\int_E f d\mu$ $=$ $0$ for every $E$ $\in$ $\mathfrak M$ ( $\sigma$ - algebra ). Then $f$ $=$ a.e. on $X$.

Lusin's theorem with its corollary :

$4.25$ Theorem : If $f$ $\in$ $C(T)$ and $\epsilon$ $\gt$ $0$, there is a trigonometric polynomial $P$ such that

$|f(t) - p(t)|$ $\lt$ $\epsilon$ for every real $t$.

For any $f$ $\in$ $L^1(T)$, we define the Fourier coefficients of $f$ by the formula

$\hat f(n)$ $=$ $\frac {1} {2\pi}$ $\int_{-\pi}^{\pi}$ $f(t) e^{-int} dt $ ($n$ $\in$ $Z$).

$\hat f(n)$ $\to$ $0$ as $|n|$ $\to$ $\infty$, for every $f$ $\in$ $L^1$.

Let $c_0$ be the space of all complex functions $\varphi$ on $Z$ such that $\varphi(n)$ $\to$ $0$ as $n$ $\to$ $+-$$\infty$, with the supremum norm

||$\varphi||_\infty$ $=$ sup {$|\varphi(n)|$ : $n$ $\in$ $Z$}.

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There is the theorem :

Theorem. The mapping $f$ $\to$ $\hat f $ is a one-to-one bounded linear transformation of $L^1(T)$ into (but not onto) $c_0$.

There is the proof:

Define $\Lambda$ by $\Lambda f $ $=$ $\hat f$. It is clear that $\Lambda$ is linear. We have already known that $\Lambda$ maps $L^1(T)$ into $c_0$, and formula $\hat f(n)$ $=$ $\frac {1} {2\pi}$ $\int_{-\pi}^{\pi}$ $f(t) e^{-int} dt $ ($n$ $\in$ $Z$) shows that $|\hat f(n) |$ $\leq$ $||f||_1$, so that $||\Lambda||$ $\leq$ $1$. ( Actually, $||\Lambda||$ $=$ $1$, to see this, take $f$ $=$ $1$.) Let us now prove that $\Lambda$ is one-to-one. suppose $f$ $\in$ $L^1(T)$ and $\hat f(n)$ $=$ $0$ for every $n$ $\in$ $Z$. Then

$\int_{-\pi}^{\pi}$ $f(t)g(t)$ $dt$ $=$ $0$

if g is any trigonometric polynomial.

By Theorem $4.25$ and the dominated convergence theorem, $\int_{-\pi}^{\pi}$ $f(t)g(t)$ $dt$ $=$ $0$ holds for every $g$ $\in$ $C(T)$. Apply the dominated convergence theorem once more, in conjunction with the Corollary to Lusin's theorem, to conclude that $(1)$ holds if g is the characteristic function of any measurable set in $T$. Now Theorem $1.39$ shows that $f$ $=$ $0$ a.e. .

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I have a few questions :

1. how does the $\int_{-\pi}^{\pi}$ $f(t)g(t)$ $dt$ $=$ $0$ holds for every $g$ $\in$ $C(T)$ by using $4.25$ and the dominated convergence theorem.

2. How do we conclude that $\int_{-\pi}^{\pi}$ $f(t)g(t)$ $dt$ $=$ $0$ holds if $g$ is the characteristic function of any measurable set in $T$ ?

3. Why do we have the right to apply $1.39$ theorem on this ? ( Integral in this one depends on $\mu$ measure ) .

Any help would be appreciated.

Answer 1

1. Because of the norms in the domain and codomain, the inequality we are looking for is $$\tag1\|\Lambda f\|_\infty\leq\|f\|_1.$$ Since in $c_0$ the supremum norm is just the maximum of the absolute values of the entries, the inequality we need is $$\tag2 |\hat f(n)|\leq\|f\|_1 $$ for all $n$. It looks like Rudin is using the $1$-norm defined as $\|f\|_1=\frac1{2\pi}\int_{-\pi}^\pi|f|$. We have $$ |\hat f(n)|=\frac1{2\pi}\Bigg|\int_{-\pi}^\pi f(t)\,e^{-int}\,dt\Bigg| \leq\frac1{2\pi}\int_{-\pi}^\pi|f(t)|\,|e^{-int}|\,dt =\frac1{2\pi}\int_{-\pi}^\pi|f(t)|\,dt =\|f\|_1. $$ So $(2)$ holds, which means that $(1)$ holds. If you look at the definition 5.3, the symbol $\|\cdot\|$ has three different meanings: $\|x\|$ is the norm in the domain, $\|\Lambda x\|$ is the norm in the codomain, and $\|\Lambda\|$ is the one being defined. Since $(1)$ can be seen as $\|\Lambda(f/\|f\|_1)\|_\infty\leq1$, it shows that $\|\Lambda\|\leq1$. And it cannot be strictly less than $1$, because $\|\Lambda 1\|=\frac1{2\pi}\int_{-\pi}^\pi1=1$.

2. Theorem 4.25 shows that given $g\in C(T)$, there exists a sequence $\{g_n\}$ of trigonometric polynomials such that $g_n\to g$ uniformly (which we may write as $\|f-g_n\|_\infty\to0$). Since $g$ is bounded (being continuous on a compact set), the sequence $\{g_n\}$ is uniformly bounded. So if $|g_n|\leq c$, we have $|fg_n|\leq |f|\,c$, and this function you can use for dominated convergence. That is, $$ \int_{-\pi}^\pi f(t)g(t)\,dt=\lim_n\int_{-\pi}^\pi f(t)\,g_n(t)\,dt=0. $$ Next you use the corollary to Lusin, to write $1_E=\lim h_n$ for a given measurable set $E$, with $h_n$ continuous. Since $|f\,1_E|\leq|f|$, again you can apply Dominated convergence to get $$ \int_E f(t)\,dt=\int_{-\pi}^\pi f(t)\,1_E(t)\,dt=\lim_n\int_{-\pi}^\pi f(t)\,h_n(t)\,dt=0. $$ Now 1.39 gives you that $f=0$ a.e.

3. The measure $\mu$ in 1.39 is some measure. In particular Lebesgue measure as above.