Banach spaces with simple best approximate solutions

Let $\langle V,||.||\rangle$ be a Banach space such that:

$\;\;$ for all continuous linear maps $\: L : V\to V \:$ and members $v$ of $V$, there exists a unqiue member $u$ of $V$
$\;\;$ that minimizes $\langle ||L(u)+(-v)||,||u||\rangle$ in the lexicographic order

$\;\;$ and

$\;\;\;$ for all continuous linear maps $\: L : V\to V \:$, $\:$ the function $\: L^{\dagger} : V\to V \:$ given by
$\;\;$ $\big[L^{\dagger}(v)$ minimizes $\langle ||L(L^{\dagger}(v))+(-v)||,||L^{\dagger}(v)||\rangle$ in the lexicographic order$\big]$
$\;\;\;$ is continuous and linear

Does it follow that $\langle V,||.||\rangle$ is

$\;\;$ 1. $\:$ isometrically
$\;\;$ 2. $\:$ homeomorphically

isomorphic to a Hilbert space?

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Isometry is false, I think. Consider $L^p$ of a finite set for $2<p<\infty$. Projection onto a subspace (the image of a map) is well-defined and linear, as is finding the closets point to $0$ in the inverse image of a set. But it's not a Hilbert space. –  Will Sawin Jan 17 '12 at 20:30
Do you have a reference or proof for either part of your second sentence? $\;$ –  Ricky Demer Jan 17 '12 at 20:42
I think also it's not true for an infinite dimensional Hilbert space. Let $H$ have orthonormal basis $(e_n)$ and let $V(e_n) = e_n/n$ (so $V$ is compact, but has unbounded inverse). Let $v=(1,1/2,1/3,1/4,\cdots)$. Then you can make $\| V(u) - v\|$ arbitrarily small (let $u=(1,1,1,\cdots,1,0,\cdots)$) but not $0$. So no $u$ minimise $\langle \|V(u)-v\|, \|u\| \rangle$ in the lexicographic order. –  Matthew Daws Jan 17 '12 at 20:47