Preimage of bounded sets by linear transformations on normed vector spaces

Question

Let $E$ and $F$ be real normed vector spaces (not necessarily Banach) and $ \ T:E \to F \ $ be a linear transformation (not necessarily continuous nor open). Consider these two sentences:

(1) The preimage by $T$ of some open nonempty set is bounded.

(2) The preimages by $T$ of all bounded sets are bounded.

I proved that (2)$\Rightarrow$(1). I want to know if is true that (1)$\Rightarrow$(2) and some hint to prove it or a counterexample.

In addition I want an example of some linear transformation that satisfies neither.

Thanks in advance.

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EDIT $1$: I just answered my own question. Without further assumptions it is false that (1)$\Rightarrow$(2). Now I would like do know if this implication holds if I assume $im(T)$ dense in $F$. What can I say if $ \ im(T) = F$? Thanks again.

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EDIT $2$: To start a bounty I added other sentences

(3) $T$ is injective.

(4) The preimage by $T$ of some open neighbourhood of the origin $ \ 0_F \ $ is bounded.

I already have the implications (2)$\Rightarrow$(4)$\Rightarrow$(1), (2)$\Rightarrow$(4)$\Rightarrow$(3), (1)$\nRightarrow$(4) and (1)$\nRightarrow$(3). In order to receive the bounty, the answer will have to handle EDIT $1$ and to prove all the remaining implications between sentences 1, 2, 3 and 4 (prove the valid ones and give counterexamples to the false). Good luck!

Answer 1

If image of $T$ is dense, then (1) implies (2). Let $M\subset F$ be open and non-empty such that $T^{-1}(M)$ is bounded in $E$. Since $im T$ is dense in $F$, there is an interior point $y$ of $M$ such that $Tx=y$ for some $x\in E$. By linearity it holds $T^{-1}(M-\{y\}) = T^{-1}(M)-\{x\}$. However, $U:=M-\{y\}$ is an open neighborhood of the origin of $F$. Moreover, its preimage is bounded.

Let now $B\subset F$ be a bounded set. Then $B\subset tU$ for some $t>0$. This implies $T^{-1}(B)\subset T^{-1}(tU)=t T^{-1}(U)$, and $T^{-1}(B)$ is bounded.

The latter argument can be used to show (4) $\Rightarrow$ (2). If $im(T)$ is dense then also (1) implies (4).

(3) does not imply (2) and (4): Take $E=F=l^\infty$. Set $Tx:=(x_1,x_2/2,x_3/3,\dots)$. Then $T^{-1}$ is unbounded and $T^{-1}(U)$ is unbounded for all open sets $U$ containing the origin, which follows from $T^{-1}e_k=k e_k$ with $e_k$ being the unit vectors/sequences in $l^\infty$.

Answer 2

(1)$\nRightarrow$(2)

Let $ \ E = F = \mathbb{R}^2 \ $ with usual euclidean norm and $ \ T(x,y) = (x,0)$, $\forall (x,y) \in E$. Then clearly $T$ is linear. We have that $ \ B = B \big( (1,1) , 1 \big) \ $ is a nonempty open set in $F$ with $ \ T^{-1} [B] = \varnothing \ $ bounded. So, (1) is true. In addition, we have that $ \ L = B \big( (0,0) , 1 \big) \ $ is a bounded set in $F$ with $ \ T^{-1} [L] = \ ] \! -1,1[ \ \times \mathbb{R} \ \ $ unbounded. It follows that (2) is false. Therefore, the conditional (1)$\Rightarrow$(2) is false.

I would like to know if this implication holds if I impose that $im(T)$ is dense in $F$ or, a fortiori, if $im(T)=F$. I will edit the original question to address that.