Proving that the norm of a linear and surjective operator is 1 (Step in Figiel's Theorem)

Question

In the book Geometric Nonlinear Functional Analysis by Y.Benyamini and J.Lindenstrauss, in Theorem 14.2 (page 343), due to Figiel, the authors prove that a certain operator has norm $1$.

This operator $T\colon F\rightarrow E$, where $E$ is a finite dimensional Banach space and $F$ is an arbitrary Banach space, has the following properties:

1. It is linear and surjective.
2. There exists $f:E\rightarrow F$ isometry (nonlinear) such that $f(0)=0$ and $T\circ f=\text{Id}_E$.
3. There exists a norm dense subset $D$ of the unit sphere $S_E\subset E$ such that if a point $y\in F$ verifies $Ty\in D$, then $\|y\|\geq 1$.

From property (2) it is clear that the norm of $T$ is bigger than $1$. To prove the other inequality the authors argue that it follows from (1) and (3). However I'm not able to prove this.

I am only able to prove that if $y\in F$ is such that $Ty\in D$, then $\|Ty\|\leq \|y\|$ (this is trivial since $D$ is a subset of $S_E$, so $1=\|Ty\|$ and by $(3)$ we have $\|y\|\leq 1$).

I tried by contradiction, supposing that there exists $h\in S_F$ such that $\|Th\|=K>1$ and then approximating $T(\frac{h}{K})\in S_E$ by a point $Ty\in D$ (here we use $T$ is surjective), but I'm not able to produce a contradiction. Any suggestion would be greatly appreciated.

Answer 1

It is clear that if we could replace condition $3$ with the same condition but with $D = S_E$ we would be done, so let's try to do that.

First, observe that condition $1$ together with the open mapping theorem implies that $T$ is an open map. Equivalently, there is some fixed constant $M$ such that for every $y \in E$, there is an $x \in E$ with $Tx = y$ and $\|x\| \leq M \|y\|$.

Suppose $Ty \in S_E$ so that we aim to show that $\|y\| \geq 1$. Pick $z \in D$ (where here $D$ is our original dense set and not all of $S_E$) such that $\|Ty - z\|< \varepsilon$.

Then there is an $x$ such that $Tx = Ty - z$ and $$\|x\| \leq M \|Ty - z\| < M \varepsilon$$ Note that $T(y-x) = z \in D$ so that $\|y-x\| \geq 1$ by condition $3$. We can now combine our estimates. Indeed, we have $$1 - M\varepsilon \leq \|y-x\| - \|x\| \leq \|y\|.$$ Since $\varepsilon$ is arbitrary, we can send it to $0$ to conclude that $\|y\| \geq 1$.