# Isometries of direct sums of Hilbert spaces

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.

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 $$(x/2,x/2)$$.
• 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. Commented Nov 4, 2022 at 10:28
• @MartinArgerami Also, I think your argument (with appropriate modifications) works for any $l_p$-sums, while the extreme points argument only works for $p=1$.