A question about the sets used in the proof of $L^2$ is meager in $L^1$

Question

To prove that $L^2$ is meager in $L^1,$ I got a hint that I can prove that $L^2 \subseteq \cup \{f \in L^1 : \|f\|_{1} \leq n\}$ or that $L^2 \subseteq \cup \{f \in L^2 : \|f\|_{1} \leq n\}$ or that $L^2 \subseteq \cup \{f \in L^2 : \|f\|_{1} < n\}$ or that $L^2 \subseteq \cup \{f \in L^2 : \int |f|^2 dx < n\}.$

My questions are:

1- why are those sets showing the same thing?

2- why in the last set we used the square root of $\|.\|_{2}$?

3- Do not I need in my proof the closure of the set I will use so it is preferable for me to use sets with $\leq$ so that in that case I will not proof that it is closed (as the proof in the statement of the question in the following link $L^2$ is meager in $L^1$ ) as the set will be clearly closed?

Could anyone help me in answering this question?

score 2 · Accepted Answer · answered Mar 19 '20 at 12:10

2

Let's do this step by step. In order to show that $L^2$ is meager in $L^1$, we need to show that $L^2$ is a countable union of nowhere dense subsets of $L^1$ i.e. $L^2 = \cup_{n = 1}^{\infty} U_n$. So, in particular we need to find sets $U_n$ s.t. their countable union covers $L^2$. That is something that can be accomplished by any of the above sets:

Since $L^2$ is a normed vector space, a standard choice for such a cover is the open balls around some point, but we want nowhere dense sets in $L^2$, so we choose the sets $B_n(f)$ for some fixed $f \in L^2$ (in your example this $f$ was chosen to be $0$). This covers $L^2$ since for any $g \in L^2$ we have $\Vert f - g \Vert_1 < n$ for some $n \in \mathbb{N}$. Why? Because $f-g \in L^1$ and thus has finite 1-norm.

Here we can also see that is is irrelevant whether we consider $< n$ or $\leq n$. Cleary, if $\Vert f - g \Vert_1 < n$, then $\Vert f - g \Vert_1 \leq n$ and if $\Vert f - g \Vert_1 \leq n$, then $\Vert f - g \Vert_1 < n+1$.

It is also irrelevant if we choose the fixed $f \in L^2$ or $f \in L^1$, since in both cases $f-g \in L^1$ and thus $\Vert f-g \Vert_1 < \infty$ and thus there exists some $n \in \mathbb{N}$ s.t. $\Vert f-g \Vert_1 \leq n$.

It is also irrelevant whether we require $\Vert f-g \Vert_1 < n$ or $\Vert f-g \Vert_1^2 \leq n$.

It is also irrellevant whether we choose $\{g \in L^2 : \Vert f - g \Vert < n\}$ or $\{g \in L^1 : \Vert f - g \Vert < n\}$ since $\{g \in L^2 : \Vert f - g \Vert < n\} \subseteq \{g \in L^1 : \Vert f - g \Vert < n\}$ and $\{g \in L^2 : \Vert f - g \Vert < n\}$ is already sufficient to cover $L^2$.

The point is, all of the above unions cover $L^2$.

Regarding your last question, yes it is preferable to consider closed sets to begin with. However, $\overline{\{ f \in L^1 \vert \Vert f \Vert_1 < n\}} = \{ f \in L^1 \vert \Vert f \Vert_1 \leq n\}$ and recall that we want to show that all the $U_n$ are nowhere dense. So if one is nowhere dense then so is the other.

answered Mar 19 '20 at 12:10

G. Chiusole

5,594

I do not understand which example are you referring to that has $f=0$ in line 7 – Mar 19 '20 at 13:21
I am saying that the standard choice for such a covering is the open balls around some fixed $f$. These are precisely sets given in the first three options in your questions when $f \equiv 0$. – G. Chiusole Mar 19 '20 at 13:24
So if I chosen the closed sets I need not write the proof in the link I mentioned? – Mar 19 '20 at 13:25
You would still need to show that those sets have empty interior – G. Chiusole Mar 19 '20 at 13:25
yes I mean the proof in the statement of the post of this question https://math.stackexchange.com/questions/1159432/l2-is-meager-in-l1?noredirect=1&lq=1 – Mar 19 '20 at 13:28
What 3 options ? do we mean the first 3 sets in the statement of the question? – Mar 19 '20 at 13:30
Still did not get why in this 3 options $f $ identically $0$? are you saying that the center of these balls is taken zero? – Mar 19 '20 at 13:31
Yes, the first 3 sets – G. Chiusole Mar 19 '20 at 13:32
what about the last set? – Mar 19 '20 at 13:33
I believe this question shows why the interior is empty. – G. Chiusole Mar 19 '20 at 13:33
The last set is not an open or closed ball – G. Chiusole Mar 19 '20 at 13:34
I am speaking about the closure proof not the proof that the interior is empty – Mar 19 '20 at 13:34
Yes, I am saying that the center is $0$ – G. Chiusole Mar 19 '20 at 13:34
So it is preferable not to use the last set? – Mar 19 '20 at 13:36
If it is clear that the sets you choose are closed, then of course you do not need to show that they are closed. However, you somehow need to justify that. – G. Chiusole Mar 19 '20 at 13:37
I don't know which one is most preferable, since one still needs to prove that the interior is empty. Showing this may be more convenient had one chosen another – G. Chiusole Mar 19 '20 at 13:37
nowhere dense means its closure is hollow according to Royden …. but I think the closure here means to be inside $L^1$ or $L^2$ not just $\leq$ …. am I correct? – Mar 19 '20 at 13:40
They turn out to be the same. Try proving it. – G. Chiusole Mar 19 '20 at 13:44
Can we use in our sets $B_{n}$, $$|.|_{2}$$ instead of $|.|_1$ ? I can see that in the link you provided for nowhere dense proof – Mar 19 '20 at 13:57
Do those sets then still cover $L^2$? – G. Chiusole Mar 19 '20 at 13:58
yes …. am I correct? – Mar 19 '20 at 14:08
Do you have a proof? – G. Chiusole Mar 19 '20 at 14:09
No … maybe because the whole set will be subset of $L^1$ … frankly I do not know – Mar 19 '20 at 14:12
Is it a typo there? – Mar 19 '20 at 14:12
Try writing down the proof. If you cannot do it, then you need to take a step back and go over some basics before tackling the problem in your original post – G. Chiusole Mar 19 '20 at 14:13
So it is correct? according to what you explained it is not a typo …. am I correct? – Mar 19 '20 at 14:14

A question about the sets used in the proof of $L^2$ is meager in $L^1$

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