Re: cl(bigcap_{1<=q<=infty} L^q) = L^p

On Fri, 27 May 2005 01:29:02 GMT, Kira Yamato <no@xxxxxxxx> wrote:

>Ok, thanks to everyone who replied. So, the proof is ok for 1<=q<infty,
>but not when q=infty, even though the conclusion is true including
>q=infty if instead we consider Cc(R) as stated by Martin.

No, the conclusion is _false_ for q = infinity. It remains false
regardless of what we consider.

Look. Say f(x) = 1 for all x. Then f is in L^infinity(R). Suppose
that g is in L^1(R) and ||f-g||_infinity < 1/2. Can you see a
problem here?

>-kira
>
>Martin wrote:
>> If we denote by Cc(R) all continuous functions with compact support,
>> then the set Cc(X) is dense subset of L^p. (Rudin: Real and Complex
>> Analysis, Theorem 3.14)
>> Since Cc(X) belongs to each L^p, we arrive to your conclusion - this is
>> another possibility for the proof.
>> Martin
>>


************************

David C. Ullrich
.

Relevant Pages

  • Re: cl(bigcap_{1<=q<=infty} L^q) = L^p
    ... If we denote by Ccall continuous functions with compact support, ... then the set Ccis dense subset of L^p. ... (Rudin: Real and Complex ...
    (sci.math)
  • Re: cl(bigcap_{1<=q<=infty} L^q) = L^p
    ... >If we denote by Ccall continuous functions with compact support, ... >then the set Ccis dense subset of L^p. ... (Rudin: Real and Complex ...
    (sci.math)
  • Re: cl(bigcap_{1<=q<=infty} L^q) = L^p
    ... Martin wrote: ... (Rudin: Real and Complex Analysis, ... Since Ccbelongs to each L^p, we arrive to your conclusion - this is another possibility for the proof. ...
    (sci.math)