# How does $X = Y \oplus U = Y \oplus V$ for a Banach space $X$ and some subspaces imply that $U$ isometric $V$?

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.

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.