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Suppose $X$ and $Y$ are Banach Spaces and $T$ is a surjective bounded linear operator in $B(X,Y).$ Let $$L = \{T(x):x\in X \text{ and }\|x\|\leq 1\}$$ with closure $\bar{L}.$ Then show that there exists $r>0$ such that $\{y\in Y:\|y\|\leq r\}\subset \bar{L}.$

(Note: This one of three statements that one is required to prove.)

The proof of this fact starts as follows. Since $T$ is surjective we have that for every $y\in Y$ an $x\in X$ such that $y = T(x).$ Thus $\frac{y}{\|x\|}\in L$ and so $$Y = \bigcup_{n=1}^{\infty}n\bar{L}.$$ I don't understand how $$\frac{y}{\|x\|}\in L \implies Y = \bigcup_{n=1}^{\infty}n\bar{L}.$$ Perhaps someone can elaborate?

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1 Answers1

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We know that for every $y \in Y$, $$ \frac{y}{\| x \|} \in L \subseteq \bar L. $$

In particular, this means that $y \in \|x \| \bar L$. Hence, if we choose $n \ge \| x \|$, we have that $y \in n \bar L$. (*)

It then follows that $$ y \in \bigcup_{n=1}^\infty n\bar L. $$

(*) This is probably the only part here that isn't very obvious. I'll leave out the justification for it for now, because you can probably work it out. However, please leave a comment if you find it needs elaboration.

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