Linear surjective operator's property proof

Question

Let $X,Y$ be normed spaces and $T\in \mathcal{L}(X,Y)$. I want to prove that if $T$ is surjective and for all $y\in Y$ $$\|y\|=\inf\{\|x\|\ \colon x\in X, \ Tx=y\},$$ then $$\{Tx \ \colon \|x\|<1\}=\{y\ \colon \|y\|<1\}.$$ Any ideas on how to approach this proof?

Answer 1

You know that $\|Tx\| \leq \|x\|$ (set $y = Tx$ in the given identity). This implies the inclusion $\{Tx \ \colon \|x\|<1\} \subseteq \{y\ \colon \|y\|<1\}$. For the other direction, take $y \in Y$ with $\|y\| < 1$. You find $\varepsilon > 0$ such that $\|y\| < 1 - \varepsilon$. Now the given identity shows there exists $x \in X$ with $\|y\| \leq \|x\| < \|y\| + \varepsilon$ and $T x = y$. It follows that $\|x\| < 1$ and so you obtain the other inclusion.