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From measure theory we know that if $G$ is a finite measure space then $p \leq p^\prime$ implies $L^{p^\prime}(G) \subset L^p(G)$ where $L^p$ is the space of all $p$-integrable functions. So let $G$ be a finite measure space that is also a compact topological group and let $p>2$.

Now endow $L^p(G)$ with the $L^2$ norm.

Could someone give me an example of a Cauchy sequence in $L^p$ such that its limit s not in $L^p$?

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    I'm asking this because I've been thinking about for which functions $f$ I can obtain Fourier series that converge to $f$ in norm on all of the domain of $f$. This is true for any Hilbert space $H$ with a countable orthonormal subset spanning $H$. So I'm looking for Hilbert spaces and I tried to turn $L^p$ into a Hilbert space (which of course, isn't possible).2012-08-02
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    The title do not seem consistent with the body of the question. Also, is not a Banach space complete by definition?2012-08-02
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    @enzotib Yes, but he's taking the Banach space $L^p$ which is complete under the $p$-norm, and putting a different norm on it. A priori, there is no reason such a thing should make it complete.2012-08-02
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    @ronno Thank you! : ) (enzotib: what ronno said)2012-08-02
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    You know at least one "simple" subspace of $L^2 \cap L^p$ which is dense in both $L^2$ and $L^p$. Now take a function in $L^2 \smallsetminus L^p$ and approximate it in the $L^2$-norm by functions in that dense subspace.2012-08-02
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    @t.b. Oh, $C(G)$ for example? ($\cong$ compactly supported functions) I have to look it up. I haven't figured out how to remember which of them are dense in which spaces : ( There seems to be an infinitude of properties like smooth, compactly supported, vanishing at infty *etc-etc*... *head asplode* Thank you!2012-08-02
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    Well, yes, $C_c(G)$ works if you have a locally compact topology on $G$ (and your measure is nice enough), or smooth functions with compact support if you have a smooth structure. But you could just take the simple functions supported on a set of finite measure. However, a function vanishing at infinity need not be in any $L^p$ with $p \lt \infty$.2012-08-02
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    @t.b. Ok, got it: (Sorry, forgot to write that my $G$ is compact) Take $f \in L^2 \setminus L^p$. Take a sequence in $C(G) \subset L^2 \cap L^p$ converging to it in $\|\cdot\|_{L^2}$. Et voilà! Thanks for the help. I knew that actually. : )2012-08-02

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As discussed in the comments:

Remember that compactly supported functions are dense in $L^p$. Then $C(G)$ is dense in both $L^2$ and $L^p$. So pick any $f$ in $L^2$ and a sequence $g_n$ of continuous functions (yep, since $G$ is compact they're automatically compactly supported) converging to $f$ in $\|\cdot\|_{L^2}$. Then $g_n$ is in $L^p$ but its limit is not.

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    Which still leaves me with the question how to find function spaces on which I can define an inner product so that they are Hilbert spaces...2012-08-02
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    Three things: 1) In order for this to work you need to assume that $\mu$ has sets of arbitrarily small measure (so $\mu$ isn't purely atomic). 2) All separable Hilbert spaces are isometrically isomorphic (use the decomposition with respect to orthonormal bases) 3) I don't like it that you still seem to conflate the decomposition with respect to an onB with a Fourier series: Fourier series are a decomposition with respect to a *very special* onB, namely the one given by the characters on a compact group. There are many others, e.g. the [Haar system](http://en.wikipedia.org/wiki/Haar_wavelet).2012-08-02
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    I think of function spaces as Looking-glass cakes - you have to hand one round before you can cut it. The mathematical version is: you define a norm and then find the space for it.2012-08-02
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    @t.b. Thank you for the comment. Point 3) is noted.2012-08-02
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    @LeonidKovalev I didn't phrase my monologue very well: before I know the answer to the question I want to do both in order to find it: look at spaces and try different norms on them and on the other hand define a norm and see what spaces I can put it on.2012-08-02
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    @t.b. You'll be 40 soon : ) Old and wise.2012-08-10
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    yeah, at the moment I'm 39,9-40 :)2012-08-10