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?

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

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  • $\begingroup$ 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) $\endgroup$ – yoshi Jul 28 '19 at 15:31
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    $\begingroup$ @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. $\endgroup$ – Theo Bendit Jul 28 '19 at 15:33

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