Non empty interior in the image implies open map

Question

I am looking at the proof showing that $L^2(0,1)$ is meager in $L^1(0,1)$. Define $B_n = \{f\in L^2 : \|f\|_2 \leq n\}$. With the continuous identity map $T:L^2 \rightarrow L^1$, if one of $T(B_n)$ has non empty interior, then $T$ is an open map, thus $T$ is onto which is a contradiction. Therefore, all of the image $T(B_n)$ has empty interior (and closed) thus $T(L^2) = T(\bigcup_n B_n)$ is meager in $L^1$.

My question is, how do we get from one $T(B_n)$ has non empty interior in $L^1$ to $T$ is an open map? The solution makes this looks very trivial, but I just couldn't see it.

It is a general fact of linear mappings. Consider a linear mapping $T\colon X\to Y$, where $X$ and $Y$ are normed spaces. Let $B\subset X$ denote the open unit ball. You can prove that $T$ is open if and only if $T(B)$ is a neighborhood of the origin in $Y$. — Giuseppe Negro, Jul 24 '16 at 20:01
Why not use the fact that every $f\in L^2$ is the $L^1$ limit of a sequence of functions in $L^1\setminus L^2?$ — zhw., Jul 24 '16 at 20:11
Thank you guys for the hint, I posted one method of seeing this. @GiuseppeNegro I agree with your statement, but I do not see how to derive that the image of a ball centered at $f$ in $B_n$ is a open neighborhood of $T(f)$ in $L^1$ from $B_n$ has non empty interior. Unless preimage of open balls are also open balls for continuous linear maps. — Xiao, Jul 24 '16 at 21:06
Xiao and @GiuseppeNegro is the 'IV' in Mindlack's answer here (Does open map imply dimension of domain is greater than or equal to dimension of range? (please don't use sard's theorem)) related? it says open (and smooth) implies image has non-empty interior. — BCLC, May 24 '21 at 04:34
Xiao and @zhw. is the 'IV' in Mindlack's answer here (Does open map imply dimension of domain is greater than or equal to dimension of range? (please don't use sard's theorem)) related? it says open (and smooth) implies image has non-empty interior. — BCLC, May 24 '21 at 04:34
@BCLC: frankly, I don't know. My old comment relates to a standard and easy fact of linear maps, whereas that answer concerns invariance of domain, which is a much more complicated theorem of differential topology. I wouldn't see too many connections between these two things. — Giuseppe Negro, May 24 '21 at 11:14

Xiao · Answer 1 · 2016-08-27T02:27:28.253

Okay, I think I have the answer.

Method One: Take $f\in B_n$ and $g\in L^1\setminus L^2$, then $f+\frac{g}{n}$ would be a sequence in $L^1\setminus L^2$ which converges to $f$ in $L^1$. This would show that each $T(B_n)$ has empty interior, because the tail of this sequence is contained in any $L^1$ ball of $T(f)$.

Okay, here is the second method, Let $T:L^2\rightarrow L^1$ be the inclusion map. And denote $B_1(f,r)$ and $B_2(f,r)$ be open ball in $L^1$ and $L^2$.

Now if $T(\overline{B_2(0,n)})$ has non-empty interior in $L^1$, then there exists a $L^1$-open ball that is contained in $T(\overline{B_2(0,n)})$, let us write this $L^1$-open ball as $f_0 + rB_1(0,1)$, note that $f_0$ is in $\overline{B_2(0,n)}$, thus $$rB_1(0,1) \subset T(\overline{B_2(0,n)})- f_0\subset T(\overline{B_2(0,2n)})$$ and this means for some constant $M$, we have $$MB_1(0,1) \subset T(B_2(0,1)).$$ This says the image of the unit ball in $L^2$ is a open neightborhood of $0$ in $L^1$.

DanielWainfleet · Answer 2 · 2016-08-27T03:48:45.610

2

Anther way to show that if some $T(B_n)$ has non-empty interior in $L^1,$ then $T$ is an open map :

For $r>0$ let $B_r=\{f\in L^2: \|f\|_2\leq x\}.$ Then $T(B_r)=(r/s)T(B_s)=\{(r/s)T(f): f\in B_s\}$ for all positive $r,s$. So if int$_{L^1}\;T(B_r)\ne \phi$ for some $r>0$ then int$_{L^1}\;T(B_r)\ne \emptyset$ for all $r>0.$ Suppose this happens.

Let $U$ be open in $L^2$. For each $f\in U$ take $r(f)>0$ such that $V_f=\{g\in L^2:\|g-f\|_2<r(f)\}\subset U.$ Take $g_f\in B_{r(f)/2}$ such that $T(g_f)\in W_f\subset T(B_{r(f)/2})$ where $W_f$ is open in $ L^1.$

Then $T(f)\in W_f+T(f-g_f) \subset T(V_f)\subset T(U),$ while $W_f+T(f-g_f)=\{h+T(f-g_f):h\in W_f\}$ is open in $L^1.$

So $T(U)=\cup_{f\in U}(\;W_f+T(f-g_f)\;)$ is open in $L^1.$

Note: This applies if we replace $L^2, L^1$ with any two normed linear spaces $X, Y$ and any linear $T:X\to Y.$

edited Aug 27 '16 at 03:48

answered Aug 27 '16 at 03:43

DanielWainfleet

57,985

DanielWainfleet, is the 'IV' in Mindlack's answer here (Does open map imply dimension of domain is greater than or equal to dimension of range? (please don't use sard's theorem)) related? it says open (and smooth) implies image has non-empty interior. – BCLC May 24 '21 at 04:33
1

@BCLC . That is a Q about finite-dimensional manifolds. I don't see an immediate relation. – DanielWainfleet May 24 '21 at 17:17
thanks DanielWainfleet! – BCLC May 26 '21 at 06:40

Non empty interior in the image implies open map

2 Answers2