Let $(X,\| \cdot \|_X)$, $(Y,\| \cdot \|_Y)$ be normed linear spaces and $T: X \rightarrow Y$ be a surjective linear operator. Show that the following are equivalent:
(1) $T$ is open, i.e. for any open set $U$ in $X$, $T(U)$ is an open set in $Y$.
(2) There is $M>0$ such that for every $y \in Y$, there is a $x \in X$ with
$T(x) = y \text{ and } \| y \|_Y \ge M\| x \|_X$
My idea: If $T$ is also injective, then (1) implies $T^{-1}$ is continuous. Hence $T^{-1}$ is bounded and (2) is very similar with the form of boundness.
(I know this question is related to open mapping theorem. But I haven't learn it yet and it is unnecessary here.)
It's just a very initial idea but I don't know how to figure it out clearly. Thanks for your reply!
(1) implies (2): Take $U$ to be the open unit ball. Then $T(U)$ is an open set containing $T0=0$. So there exists $r >0$ such that $B(0,r) \subset T(U)$. [I will omit subscripts $X$ and $Y$ for the norms and for open balls for ease of typing]. Let $y \neq 0$ in $Y$. Choose $n$ such that $\frac 1 {n\|y\|}y \in B(0,r) \subset T(U)$. (i.e. $n >\frac 1 r$). We can write $\frac 1 {n\|y\|}y=Tx'$ with $x' \in U$. Let $x=n\|y\|x'$. Then $y=Tx$ and $\|x\|\leq n\|y\|$. [The case $y=0$ is trivial].
(2) implies (1): Let $U$ be open and $y_0\in T(U)$. We can write $y_0=Tx_0$ with $x_0 \in U$ Since $U$ is open there exists $t>0$ such that $B(x_0,t) \subset U$. Suppose $y \in B(y_0,Mt)$. Apply (2) to $y-y_0$. We can write $y-y_0=T(x)$ with $\|y-y_0\| \geq M\|x\|$. Now $y=T(x_0+x)$ and $x+x_0 \in B(x_0,t)$ because $Mt >\|y-y_0\|\geq M\|x\|$. Hence, $x+x_0 \in U$. We have proved that $y\in T(U)$. Thus $B(y_0,Mt) \subset T(U)$. This completes then proof of (1).
I have corrected some notations.