Problem 4 Barry Simon a comprehensive course in analysis part 1.

Question

(a) For any bounded Baire function, $f$, on a compact Hausdorff space $X$, prove that $||f||_{\infty}:=\inf\left\{\sup_x |g(x)|: f-g=0\text{ for a.e. } x\right\}$ exists, defines a seminorm, and equals $\sup_{x} |f(x)|$ if $f\in C(X)$. Prove that

$||f||_{\infty}=\sup\left\{a:\mu(\left\{x:|f(x)|>a\right\})>0\right\}$

(b) For any bounded Baire function $f$ and any $\mu_{l}$ for normalized positive functional $l$, we have that

$|l(f)|\leq ||f||_{\infty}$

(c) If $f_n,f$ are Bounded Baire functions and $||f_n-f||_{\infty}\to 0$, prov that $l(f)\to l(f_n)$.

For b) I have: $|f|\leq |f|_{\infty}$ then $|l(f)|\leq l(|f|)\leq ||f||_{\infty}l(1)=||f||_{\infty}$ because $l$ is normalized.

For c) I have, Using (b), $|l(f-f_n)|\leq ||f-f_n||_{\infty}\to 0 $

How prove $||f||_{\infty}:=\inf\left\{\sup_x |g(x)|: f-g=0\text{ for a.e. } x\right\}$ exists? Some hint?

Answer 1

It is obvious that $f=g$ a.e. implies $|f(y)| \leq \sup_x|g(x)|$ a.e. so $|f(y)| \leq RHS$ a.e. Hence LHS $\leq $RHS. On the other hand if $g(x)=f(x)$ when $|f(x)| \leq \|f\|_{\infty}$ and $0$ otherwise then $f=g$ a.e. and $\sup_x|g(x)| \leq \|f\|_{\infty}$ so RHS $\leq$ LHS.