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edit approved Oct 14, 2017 at 15:27
Thomas Klimpel
edit approved Oct 14, 2017 at 15:27
Thomas Klimpel
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There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure thereon. Then a closed, bounded subetsubset $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).

There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure thereon. Then a closed, bounded subet $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).

There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure. Then a closed, bounded subset $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).

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don
don
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There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure thereon. Then a closed, bounded subet $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).

There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure thereon. Then a closed subet $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).

There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure thereon. Then a closed, bounded subet $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).

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don
don
  • 51
  • 3

There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure thereon. Then a closed subet $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).