I'm self-learning Functional Analysis in Rudin's book and found some following statement hard to understand. Hope someone can help me clarify this.
1) $X, Y$ are Banach spaces, $T \in B(X,Y)$, let $U$ and $V$ be the open unit balls in $X,Y$. If $V \subset \lambda T(U)$ for some $\lambda > 0$, is it true that $T$ is an open mapping? (this statement is in proof of theorem 4.14, page 102)
2)$X, Y$ are Banach spaces, $T \in B(X,Y)$, let $U$ and $V$ be the open unit balls in $X,Y$. If $V \subset \overline{T(U)}$, then is this statement true: To every $y \in Y$ and every $\epsilon \gt 0$ corresponds an $x \in X$ with $||x|| \le ||y||$ and $||y - Tx|| \le \epsilon$ (this statement is in proof of theorem 4.13, page 100)
About the first one, after the author proves that $||T^{*}y^{*}|| \ge \lambda||y^{*}||$, then he states that $T$ is an open mapping (Use closed range theorem, from $||T^{*}y^{*}|| \ge \lambda||y^{*}||$, we have $\lambda V \subset {T(U)}$). About the second one, I got no clue because we even can't assure that $T(X)$ is dense in $Y$, how can we find $x$ such that $Tx$ is closed arbitrarily to $y$ for every $y \in Y$. Did I miss something?
Thanks. I really appreciate.
If $T(x) = y$, then $y + \lambda^{-1} \epsilon V \subset T(x+\epsilon U)$. So the image of any open neighbourhood of $x$ contains an open neighbourhood of $T(x)$.
Given $y \in Y\setminus\{0\}$, then $v = \frac{1}{\|y\|+\epsilon} y \in V$. Since $V \subset \overline{T(U)}$, there exists $z \in U$ such that $\|T(z) - v\| \le \epsilon \|y\|^{-1} $. Let $x = \|y\| z$. Then $\|x\| = \|y\| \|z\| \le \|y\|$. Furthermore $$ \|T(x) - y\| \le \| T(\|y\|z) - (\|y\| v)\| + \| (\|y\|v) - y\| \le \epsilon + \left\|\frac{\|y\|}{\|y\|+\epsilon}y - y\right\| \le 2\epsilon. $$