TITLE: If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?
QUESTION [8 upvotes]: 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 $\phi = \psi?$?

REPLY [5 votes]: It was proved by E. Behrends in Studia Math. 55, 71-85 (1976) that apart from $E=\mathbb{R}^2$ with the sup norm, which is isometric to $E$ with the $\ell_1$-norm, a Banach space $E$ admits a decomposition $E_1\oplus_p E_2$ for at most one value of $p$. This theorem is what Misha has outlined above.