# Inclusion of open balls gives an inequality

The proof I am reading contains the following statement:

... we obtain: $$\overline{B_Y} \subset r\cdot \overline{T(B_X)}$$ (Here $$B_X$$ and $$B_Y$$ are unit balls centered at the origin in Banach spaces $$X$$ and $$Y$$, resp. Also $$r > 0$$ and $$T$$ is linear and continuous)

Therefore, since $$\overline{B_Y}$$ is the closed unit ball in $$Y$$, for each $$y\in Y$$ and $$\epsilon > 0$$, there is an $$x \in X$$ for which: $$\|y - T(x)\| < \epsilon \text{ and } \|x\| \leq r \cdot \|y\|$$

Certainly, by inclusion of the balls, the first inequality is true. I don't understand where the second inequality is coming from though -- where does it come from?

• Is $T$ linear?$\hspace{0pt}$ – Theo Bendit Jul 28 '19 at 14:21
• yes! sorry I'll add that – yoshi Jul 28 '19 at 14:29
• In this terminology, should a unit ball contain the origin? – diplodoc Jul 28 '19 at 14:52
• It does not say so explicitly, but let us assume so. i updated the prompt – yoshi Jul 28 '19 at 14:54

Given the provided set inclusion, take $$y \in Y \setminus \{0\}$$. Then $$\frac{y}{\|y\|} \in \overline{B_Y} \subseteq r \cdot\overline{T(B_X)}$$. Therefore, there exists some $$z \in \overline{T(B_X)}$$ such that $$\frac{y}{\|y\|} = rz$$. Given $$z \in \overline{T(B_X)}$$, there must be some $$v \in B_X$$ such that $$\|T(v) - z\| < \frac{\varepsilon}{r\|y\|}$$. Putting this together, $$\|T(v) - z\| < \frac{\varepsilon}{r\|y\|} \implies \Big\|\|y\|rT(v) - \|y\|rz\Big\| < \varepsilon \implies \|T(r\|y\|v) - y\| < \varepsilon.$$ Take $$x = r\|y\|v \in X$$ such that $$\|x\| = \Big\|r\|y\|v\Big\| = r\|y\| \cdot \|v\| \le r\|y\|$$ as required.
If $$y = 0$$, then take $$x = 0$$.
• Why does $r\|y\| \cdot \|x\| \leq r\|y\|$ imply the inequality? It looks like this just means $x$ is in $B_X$ (by canceling) – yoshi Jul 28 '19 at 15:31
• @yoshi The $x$ I have and the $x$ in the question are not the same. I've edited my answer to make it more clear, hopefully. – Theo Bendit Jul 28 '19 at 15:33