Re: cl(bigcap_{1<=q<=infty} L^q) = L^p
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Fri, 27 May 2005 06:41:56 GMT
In article <9crb91t5er92k5koc4rs1a8i5j32pfpsbm@xxxxxxx>,
David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:
>On Thu, 26 May 2005 14:01:43 GMT, rob@xxxxxxxxxxxxxx (Rob Johnson)
>wrote:
>
>>In article <6hjle.8288$uR4.7524@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
>>Kira Yamato <no@xxxxxxxx> wrote:
>>>Ok, I'm trying to verify my often flawed logic, in this case, about
>>>density of L^p(R) spaces.
>>>
>>>Let f in L^p(R). Define the n-boxed function of f,
>>> f_n(x) = 0 if |x|>n,
>>> = n if f(x)>n,
>>> = -n if f(x)<n,
>>> = f(x) otherwise.
>>>Now, {f_n}, being bounded in domain and range, are clearly in L^q(R) for
>>>all 1<=q<=infty.
>>>
>>>Since clearly
>>> |f-f_n| <= |f|,
>>>Lebesgue's Dominated Convergence Theorem gives
>>> lim_n \int |f-f_n| = \int lim_n |f-f_n| = 0.
>>>So,
>>> f_n -> f strongly in L^p.
>>>
>>>But {f_n} are also in every other L^q spaces. So, we conclude that
>>> bigcap_{1<=q<=infty} L^q
>>>is dense in L^p.
>>>
>>>Is this correct or should I retake Real Variable I? Thanks ahead for
>>>helping me.
>>
>>The conclusion in the text of your message is correct. However, the
>>equality in the title of your post is not. Certainly, L^p is in the
>>closure you specify, but it is not all of that closure. Therefore,
>>the equality is not warranted.
>
>I assumed that "closure" must mean "closure in the L^p norm".
>What topology are you assuming?
Well, sure, if you work in the L^p topology, you get L^p since L^p is
complete under the L^p norm. However, if you haven't really thought
about it, you might make the statement that I made.
Rob Johnson <rob@xxxxxxxxxxxxxx>
take out the trash before replying
.
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- From: Kira Yamato
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- From: Rob Johnson
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- From: David C . Ullrich
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