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Suppose $A,B$ and $C$ are Banach spaces such that $A \subset B \subset C$. There are a few different wasy that I am aware of to think about the set $A$ being dense in $B$. One is to say that $A$ is dense in $B$ if and only if $\bar{A} = B$ where $\bar{A}$ denotes the set $A$ together will all of its accumulation points which, alternately, can be considered as the intersection of all closed sets containing $A$. An equivalent way to say that $A$ is dense in $B$ is to say that for every point of $B$ one can find a sequence in $A$ that converges to $B$. I particularly like this characterization because it avoids having to distinguish between 'closed in B' versus 'closed in C'.

Now lets suppose that the spaces involved are Banach function spaces with the supremum norm. My question regards how the distinction between pointwise convergence and uniform convergence comes into play when considering whether $A$ is dense in $B$. For example, suppose one has a sequence of functions $(f_n)$ in $A$ that converges uniformly to a function $f$ in $B$. Considerations of whether $A$ is dense in $B$ in this case don't depend upon the point at which the function is evaluated so the same definitinition of density as applied to general Banach spaces as indicated above can be applied in this instance. What can be said though if the convergence of $(f_n)$ to $f$ is not uniform?

My thoughts are that the property of $A$ being dense in $B$ would vary from point to point and thus would not be well-defined. Is this the right way to think about this?

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Why is $C$ mentioned? – Jonas Meyer Mar 15 '12 at 20:29
Because in the first paragraph I was making a distinction between closed in $B$ versus closed in $C$ (in which I made a typo so the closed in $C$ part wasn't there) – ItsNotObvious Mar 15 '12 at 20:34
If you are considering $A$ and $B$ with the restricted norm from $C$, then because $B$ is a Banach space, it is closed in $C$. That implies that a subset of $B$ is closed in $B$ if and only if it is closed in $C$. Similarly, $A$ is closed, so it is dense in $B$ if and only if $A=B$. Did you mean to say that they are complete, or do you really want to consider normed (not necessarily complete) spaces? – Jonas Meyer Mar 15 '12 at 21:07
@JonasMeyer I am only considering complete (Banach) spaces; I did not consider that a vector subspace of a Banach space is complete itself if and only if it is closed, but that makes sense. What I'm really trying to understand though is whether the concept of density has any meaning in relation to pointwise convergence. I think perhaps my first paragraph detracts from this intent.. – ItsNotObvious Mar 15 '12 at 21:22

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