Let $U$ and $V$ be the open unit balls in the Banach spaces $X$ and $Y$, respectively. If $T \in \mathcal{B}(X,Y)$ and $\delta > 0$, the the implications
$$
(a)\to(b)\to(c)\to(d)
$$
hold among the following statements(a) $\lVert T^* y^* \rVert \geq \delta \lVert y^* \rVert$
(b) $\delta V \subset \overline{T(U)}$
There're (c) and (d) as well, but I focus only on the first two.
Proof : Assume (a), and pick $y_0 \notin \overline{T(U)}$. Since $\overline{T(U)}$ is convex, closed and balanced, theorem 3.7 shows that there's a $y^*$ such that $|<y,y^*>|\leq 1$ for every $y \in \overline{T(U)}$, but $|<y_0,y^*>| > 1$. If $x \in U$, it follows that
$$
\left| \left\langle x,T^* y^* \right\rangle \right| = \left|\left\langle Tx, y^* \right\rangle \right| \leq 1
$$
Thus $\lVert T^* y^* \rVert \leq 1$, and now (a) gives
$$
\delta < \delta \left| \left\langle y_0,y^* \right\rangle \right| \leq \delta \lVert y_0 \rVert \lVert y^* \rVert \leq \lVert y_0 \rVert \lVert T^* y^* \rVert \leq \lVert y_0 \rVert.
$$
It follows $y \in \overline{T(U)}$ if $\lVert y \rVert \leq \delta$
First question, why the very last statement? namely why:
It follows $y \in \overline{T(U)}$ if $\lVert y \rVert \leq \delta$
?
For (b) part I'll just write the bit I don't get, not the full proof.
Next, assume (b). Take $\delta = 1$, without loss of generality. Then $\overline{V} \subset \overline{T(U)}$. To every $y \in Y$ and every $\epsilon > 0$ corresponds an $x \in X$ with $\lVert x \rVert \leq \lVert y \rVert$ and $\lVert y – Tx \rVert < \epsilon$.
Basically I don't understand the full sentence after the period, does it follow from continuity of $T$ somehow?
Best Answer
In a) implies b) it is shown that $y_0 \notin \overline {T(U)}$ implies $\delta <\|y_0\|$. The contrapositive of this statement is $\|y\| \leq \delta$ implies $y \in \overline {T(U)}$.
In b) implies a) if $y \neq 0$ then $\frac y {\|y\|} \in \overset {-} V$. Hence $\frac y {\|y\|} \in \overline {T(U)}$. Hence for every $\epsilon >0$ there exists $x_1 \in U$ such that $\|\frac y {\|y\|} -Tx_1\| <\epsilon /\|y\|$. Now take $x=\|y\| x_1$. Then $\|y-Tx\| <\epsilon $ and $\|x\| \leq \|y\|$.