# Question about the proof open mapping theorem

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?

• what's $\bar L$ ? I suspect you use something like $x\in \bar L$ implies $-x\in \bar L$? – Calvin Khor Mar 6 at 16:38
• $\overline{L}$ is simply the closure of $L$. Yes I tried that, but that only gives us something like $-y - p \in \overline{L}$ right? – Sigurd Mar 6 at 16:39
• 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) – Calvin Khor Mar 6 at 16:40
• 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? – Sigurd Mar 6 at 16:44
• Hmm, the difference of two norm 1 things can have norm 2 e.g. in $\mathbb R$, $\| 1 - (-1)\|_{\mathbb R} = 2$ – Calvin Khor Mar 6 at 16:48

Collection of comments. But first, let me prominently display the definition of $$L$$ to reduce possible confusion

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

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.