Non empty interior in the image implies open map 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.
 A: 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 $L^1$.
A: 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 $T:X\to Y.$ 
