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If $H$ is a separable Hilbert space, then for any norm one vectors $x$ and $y$ we can find an surjective isometry $U$ such that $Ux=y$. Is the same true for $H\oplus H$, that is, the direct sum in $l_1$ sense? This space is isomorphic, but not isometric with $H$, so I am not sure the property still holds. Of course, if we consider the direct sum $\oplus_2$ in $l_2$ sense, then the space is isometric to $H$ and the property again holds.

I suspect it is not true, but I cannot come up with a pair of points for which it fails.

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Let $V$ be an isometry in your sense. Being a linear operator on $H\oplus H$, we can think of $V$ as a $2\times 2$ matrix of operators. That is, $$ V=\begin{bmatrix} A&B\\ C&D\end{bmatrix}, $$ with $A,B,C,D\in B(H)$. That is, $$ V(x,y)=(Ax+By,Cx+Dy),\qquad\qquad x,y\in H. $$ Note that since $\|(x,y)\|_2\leq\|(x,y)\|_1\leq\sqrt2\,\|(x,y)\|_2$, the bounded operators on $H\oplus H$ are the same regardless of the norm.

Since $V$ is an isometry, we have $$ \|x\|+\|y\|=V(x,y)=\|(Ax+By,Cx+Dy)\|=\|Ax+By\|+\|Cx+Dy\|. $$ Taking $y=0$, $$\tag1 \|x\|=\|Ax\|+\|Cx\|,\qquad\qquad x\in H. $$ Taking $x=0$, $$\tag2 \|y\|=\|By\|+\|Dy\|,\qquad\qquad y\in H. $$ Taking $y=\lambda x$ with $\|x\|=1$, and $|\lambda|=1$, $$\tag3 2=\|(A+\lambda B)x\|+\|(C+\lambda D)x\|,\qquad\qquad x\in H. $$ Combining $(1)$, $(2)$, and $(3)$, for $x\in H$ with $\|x\|=1$ we have $$\tag4 \|Ax\|+\|Bx\|+\|Cx\|+\|Dx\|\leq2=\|(A+\lambda B)x\|+\|(C+\lambda D)x\|. $$ This implies equality in both triangle inequalities, so for any $x\in H$ and $|\lambda|=1$ $$\tag5 \|Ax\|+\|Bx\|=\|(A+\lambda B)x\|,\qquad \|Cx\|+\|Dx\|=\|(C+\lambda D)x\|. $$ We now work with $A,B$ since the computations for $C,D$ are entirely analogous. Squaring, expanding, and cancelling square norms in $(5)$, $$\tag6 \operatorname{Re}\lambda \langle Bx,Ax\rangle=\|Bx\|\,\|Ax\|,\qquad\qquad x\in H,\ |\lambda|=1. $$ By using $\lambda=1,-1,i,-i$ we get that $$\tag7 \|Bx\|\,\|Ax\|=0,\qquad\qquad x\in H. $$ and analogously $$\tag8 \|Cx\|\,\|Dx\|=0,\qquad\qquad x\in H. $$ When $Ax=0$, we get from $(1)$ that $\|Cx\|=\|x\|$; then $(8)$ implies that $Dx=0$. Similarly, when $Bx=0$ we get that $\|Dx\|=\|x\|$ and then $Cx=0$. So either $$\tag9 \|Ax\|=\|Dx\|=0,\qquad\qquad \|Cx\|=\|Bx\|=\|x\|, $$ or $$\tag{10} \|Ax\|=\|Dx\|=\|x\|,\qquad\qquad \|Cx\|=\|Bx\|=0. $$ Suppose that $(9)$ occurs for a certain $x$ and $(10)$ for $y$, both nonzero. For $x+y$, we have $$ \|A(x+y)\|=\|Ay\|=\|y\|,\qquad \|B(x+y)\|=\|Bx\|=\|x\|, $$ so $x+y$ satisfies neither $(9)$ nor $(10)$. This proves that $V$ satisfies either $(9)$ or $(10)$ for all $x$. In other words, the possibilities for $V$ are $$ V=\begin{bmatrix} A&0\\0&D\end{bmatrix},\qquad\qquad\text{ or } \qquad\qquad V=\begin{bmatrix}0& B\\ C&0\end{bmatrix} $$ with $A,B,C,D\in B(H)$ isometries.

Now it is easy to find the counterexample. Fix $x\in H$ with $\|x\|=1$ and consider the elements $(x,0)$ and $(x/2,x/2)$. With $V$ of the first form we need to have
$$ (x/2,x/2)=V(x,0)=(Ax,0), $$ and this forces $x=0$. With $V$ of the second form the problem is the same: $$ (x/2,x/2)=V(x,0)=(0,Cx) $$ and we get $x=0$. So no isometry (surjective or not) can map $(x,0)$ to $(x/2,x/2)$.

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    $\begingroup$ I think the lack of an isometry in this case can be nicely understood from the fact that $(x,0)$ is an extreme point of the unit ball of $H\oplus_1 H$, while $(x/2,x/2)$ is not (at least that is true for $H=\mathbb R$, I did not check the general case carefully). Of course, an isometry would have to map extreme points of the unit ball to extreme points. $\endgroup$
    – MaoWao
    Commented Nov 4, 2022 at 10:28
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    $\begingroup$ That's a much nicer, conceptual way of looking at it, love it. But it only works for surjective isometries. A non-surjective isometry can send extreme points to non-extreme points. $\endgroup$ Commented Nov 4, 2022 at 15:51
  • $\begingroup$ @MartinArgerami Also, I think your argument (with appropriate modifications) works for any $l_p$-sums, while the extreme points argument only works for $p=1$. $\endgroup$
    – Theo
    Commented Nov 4, 2022 at 21:00

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