Let $E\neq \{0\}$ be a Banach space.
For each $p\in[1,\infty), $ we define 
$$E\oplus\_p E = \{(x,y): x\in E, y\in E, \|(x,y)\| = \sqrt[p]{\|x\|^p + \|y\|^p}\}.$$
Let $F$ be another Banach space.
By $E\cong F,$ I mean that $E$ and $F$ are isometrically isomorphic.



> 
> Question: Suppose that $p,q\in [1,\infty).$ If 
>  $$E\oplus\_p E \cong E\oplus\_q E\,$$
>  then is it true that $p=q$?
> 
> 
> 


If $E$ is of finite-dimensional, then the question is affirmative. 
However, I do not know what will happen if $E$ is of infinite-dimensional.
I would be glad to see a proof if it is true or a counterexample if it is false. 




---


We say that a norm $\phi:\mathbb{R}^2\to\mathbb{R}$ is *normalized* if 
$$\phi(0,1) = \phi(1,0) = 1.$$


Also, $\phi$ is *monotone* if for $|a\_1|\leq |b\_1|$ and $|a\_2|\leq |b\_2|,$ then 
$$\phi(a\_1,a\_2) \leq \phi(b\_1,b\_2).$$


We define 
$$E\oplus\_\phi E = \{(x,y): x\in E, y\in E, \|(x,y)\| = \phi(\|x\|, \|y\|) \}.$$


A more general question: 



> 
> Suppose that $\phi,\psi:\mathbb{R}^2\to \mathbb{R}$ are norms that satisfy normalization and monotonicity.
>  Assume that $\phi$ and $\psi$ are not $\ell^\infty$ norm.
>  If 
>  $$E\oplus\_\phi E \cong E\oplus\_\psi E,$$
>  then is it true that $\phi = \psi?$?
> 
> 
> 



If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?
It is a sketch of the proof in the case of $p\ne q$ (a complete
proof on these lines is rather lengthy). Something similar can be
done in more general case, possibly with some exceptions.


Our plan is the following: We assume that $p,q\in[1,\infty)$,
$p\ne q$ and $W=E\_1\oplus\_p E\_2=F\_1\oplus\_q F\_2$ (isometrically),
where $E\_1,E\_2,F\_1,F\_2$ are all isometric to a Banach space $E$,
and get a contradiction.


By $S(X)$ we denote the unit sphere of a Banach space $X$, if $Z$
and $Y$ are subspaces of $X$, we set $\delta(Y,Z)=\inf\{||y-z||:~
y\in S(Y), z\in Z\}$ and call it an *inclination* of $Y$ to
$Z$.


If we can find $i\ne j\in\{1,2\}$ such that $F\_i$ has zero
inclination to $E\_1$ and $F\_j$ has zero inclination to $E\_2$, we
get a contradiction by observing that it means that
two-dimensional $\ell\_p$-sphere approximates $\ell\_q$-sphere with
an arbitrary precision, which is not true ($p$ and $q$ are fixed).


[This paragraph is added on 8/8/18 according to the request below] Assume without loss of generality that $\delta(F\_1,E\_1)=\delta(F\_2,E\_2)=0$. This
implies that there are sequences $\{x\_i\}\in S(F\_1)$, $\{y\_i\}\in
S(E\_1)$, $\{z\_i\}\in S(F\_2)$, and $\{w\_i\}\in S(E\_2)$, such that
$$\lim\_{i\to\infty}||x\_i-y\_i||=0=\lim\_{i\to\infty}||z\_i-w\_i||=0$$
Observe that our assumptions imply that the subspace $A\_i$ spanned
by $\{x\_i,z\_i\}$ is isometric to $\ell\_q^2$ and the subspace $B\_i$
spanned by $\{y\_i,w\_i\}$ is isometric to $\ell\_p^2$. Therefore,
using the argument of Proposition 1.a.9 in Lindenstrauss-Tzafriri,
Classical Banach spaces, v. I we see that there should be
isomorphisms $T\_i$ between $\ell\_p^2$ and $\ell\_q^2$ mapping unit
vector basis of $\ell\_p^2$ onto the unit vector basis of
$\ell\_q^2$ and being arbitrarily close to the identity. This is
clearly false if $p\ne q$.


It remains to consider the case where both $F\_1$ and $F\_2$ have
nonzero inclination to $E\_1$ (or to $E\_2$, the cases are
similar).


By $P\_1$ and $P\_2$ we denote projections on $W$ corresponding to
the decomposition $E\_1\oplus\_p E\_2$. Nonzero inclination to $E\_1$
implies that the restriction of $P\_2$ to both $F\_1$ and $F\_2$ are
isomorphic embeddings. If there are nonzero points $y\_1$ and
$y\_2$ in $F\_1$ and $F\_2$ which have the same image in $E\_2,$ we
get a contradiction by considering the space spanned by $y\_1$ and
$y\_2$: on one hand its unit sphere is $\ell\_p$-sphere, and on the
other hand - $\ell\_q$-sphere.


In a similar way (but using approximations) we get a contradiction
in the case where the inclination of $P\_2(F\_1)$ to $P\_2(F\_2)$ is
zero. Finally, if the inclination of $P\_2(F\_1)$ to $P\_2(F\_2)$ is
nonzero, we get a contradiction because in this case $P\_2$ would
be an isomorphic embedding of $W$ into $E\_2$, which is obviously
false.


