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Here $$ L = \{T(x) : x \in X \textrm{ and } \|x\| \leq 1\}. $$ While going through this proof, I don't understand the step in red, namely why $y-p \in \overline{L}$, could someone explain this?

I've figured out another possible way to infer that $y \in \overline{L}$. We have $p+y \in \overline{L}$ and $p \in \overline{L}$. Thus, there are sequences $\{x_n\},\{z_n\} \subset X$ with $Tx_n \to p+y$ and $Tz_n \to p$ with $\|x_n\| \leq 1, \|z_n\| \leq 1$. Then we have $T(x_n - z_n) = Tx_n - Tz_n \to p+y-p = y$, and also $\|x_n - z_n\| \leq 1$, therefore $y \in \overline{L}$, is this correct?

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  • $\begingroup$ what's $\bar L$ ? I suspect you use something like $x\in \bar L$ implies $-x\in \bar L$? $\endgroup$ – Calvin Khor Mar 6 at 16:38
  • $\begingroup$ $\overline{L}$ is simply the closure of $L$. Yes I tried that, but that only gives us something like $-y - p \in \overline{L}$ right? $\endgroup$ – Sigurd Mar 6 at 16:39
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    $\begingroup$ Yes, but note that if $y \in B$ then so is $-y$, then $p-y \in \bar L$, then $-p+y\in \bar L$ (Also I didn't see the definition of $L$ until later, sorry) $\endgroup$ – Calvin Khor Mar 6 at 16:40
  • $\begingroup$ Yeah I think that works, thanks! Here you use that if $y \in \overline{L}$, we have $Tx_n \to y$ for $\|x_n\| \leq 1$, and then $T(-x_n) \to -y$ for $\|-x_n\| \leq 1$ right? Do you think my alternative proof is also correct? $\endgroup$ – Sigurd Mar 6 at 16:44
  • $\begingroup$ Hmm, the difference of two norm 1 things can have norm 2 e.g. in $\mathbb R$, $\| 1 - (-1)\|_{\mathbb R} = 2$ $\endgroup$ – Calvin Khor Mar 6 at 16:48
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Collection of comments. But first, let me prominently display the definition of $L$ to reduce possible confusion

\begin{equation} L=\{T(x) : x \in X \text { and }\|x\| \leq 1\} \end{equation}

That is, $L$ is the image under $T$ of the closed unit ball in $X$.

How the author deduces that $y-p\in \overline L$.

This is because balls $B$ around 0 are symmetric ($B=-B$), so if $y \in B_{0,Y}(t)$, then $-y\in B_{0,Y}(t)$. By the same reasoning as the line prior,

$$ p-y \in \overline L.$$ Since $\overline L = \overline {-L} = - \overline L$ (this computation was carried out in the comments), $-p+y\in \overline L$, which is the claimed result.

Is your alternative proof correct?

Not as it stands; the reason being that $\| x_n \|\le 1, \|z_n \|\le 1$ does not imply $\|x_n - z_n\| \le 1$. Indeed the triangle inequality only gives $ \|x_n - z_n\| \le \|x_n\| + \|z_n\|\le 2$ which could be satisfied with equality without further restrictions on $x_n , z_n$. So something like the symmetry argument above seems necessary, though I do not know enough proofs of this theorem to make a definitive statement.

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