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I'm trying to prove the following:

Let $X$ be a Banach space and let $Y,U,V \subseteq X$ be subspaces such that $X = Y \oplus U = Y \oplus V$. Then $U$ and $V$ are isometrically isomorphic.

So far I managed to construct a simple linear bijection $U \to V$ using the bijections $p_U: U \to X/Y: u\mapsto u + Y$ and $p_V: V \to X/Y: v \to v + Y$, however this does not give rise to an isometry. (Consider e.g. $X = (\mathbb{R}^2,\|\cdot \|_2), Y = (1,0)^T \mathbb{R}, U = (0,1)^T \mathbb{R}, V = (1,1)^T \mathbb{R}$).

I would be thankful for any hints on how to construct an isometry.

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It's not true. Consider $X = \mathbb R^3$ with the norm

$$ \|(x,y,z)\| = \max\left(|x|, \sqrt{y^2 + z^2}\right) $$

Let $U = \{(x,0,z) \in X: x, z \in \mathbb R\}$, $V = \{(0,y,z) \in X: y,z \in \mathbb R\}$, $Y$ the span of $(1,1,0)$. Then $X = Y \oplus U = Y \oplus V$ but $U$ and $V$ are not isometric: e.g. the unit ball of $U$ is the convex hull of $4$ extreme points but that of $V$ is not.

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