0
$\begingroup$

Rudin defined the essential range of a function in $L^{\infty}$ as the set $R_{f}$ of complex numbers $w$ such that $\mu(x:|f(x)-w|<\epsilon)>0,\forall \epsilon$

He asked the reader to prove that $R_{f}$ is compact. But since $f\in L^{\infty}$, we would have existence of functions like $f(x)=x$, whose $L^{\infty}$ norm is also infinity. Then any point in $\mathbb{R}$ could be in $R_{f}$, so $R_{f}$ cannot be bounded. I can show $R_{f}$ is closed, but I am not clear where I was wrong at here.

  • 0
    oh, I was dumb. Thanks.2012-12-28

1 Answers 1

1

The counterexample you've provided does not apply as if $\|x\|_{L^{\infty}(\mu)} = \infty$ then $x\notin L^\infty(\mu)$ just by the definition of this linear space. Since you know how to show that $R_f$ is closed, you only have to show that it's bounded - this is necessary and sufficient for the compactness in $\Bbb C$.