# If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?

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 $\phi = \psi?$?

• Objection to your final statement since in the case where $E$ is the real line. then the cases $1$ and $\infty$ are isometric. Rather too silly to be worth mentioning but for the fact that it suggests looking at the extreme points of the unit balls. If they are what I suspect (i.e. situated on the axes and the diagonals---except for $p=2$), then this would prove your conjecture. Jul 31, 2018 at 13:25
• As you might have noticed, but just in case, I do not assume that $p$ and $q$ are holder conjugate. Jul 31, 2018 at 13:49
• In the infinite dimensional case, the unit balls need not have any extreme points.
– user114263
Jul 31, 2018 at 17:24
• @DuchampGérardH.E. If $E = \mathbb{R},$ then I guess we can look at their set of extreme points and deduce that $p=q?$ Because for two norms to be the same, they must have the same set of extreme points.I think this is a characterization, that is, two norms are the same iff they have the same set of extreme points. Aug 2, 2018 at 7:05
• Since $E\cong F$ implies $E\oplus_\psi E \cong F\oplus_\psi F$, you might as well phrase the question like this: does $E\oplus_\phi E \cong E\oplus_\psi E$ imply $\phi = \psi$? (Nitpick: and you may want to assume $E\neq 0$, since otherwise the answer is trivially no.) Aug 3, 2018 at 12:46

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.

• +1 but I'd like to know more about your argument in [if we can find ... ($p,q$ are fixed).] or a reference. Aug 8, 2018 at 7:26
• @ Duchamp Gérard H. E. I have inserted the corresponding argument. Aug 8, 2018 at 15:31
• @Idonknow It is a strange and impolite approach: to open a bounty and to pay no attention to an answer. Aug 8, 2018 at 20:15
• Thanks. I suppose that you note $\ell_p^2$, the space $\ell_p\oplus \ell_p$. Aug 8, 2018 at 20:25
• @ Duchamp Gérard H. E. I use the notation $\ell_p^2$ for two-dimensional $\ell_p$, it also can be written as $\mathbb{R}\oplus_p\mathbb{R}$. Aug 8, 2018 at 21:10

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.

• @ Dirk Werner. Welcome to MO! I did not know about this paper of Ehrhard, otherwise I would prefer to give your answer rather than mine. I was inspired to prove it myself because the proposer of the question was eager to know the answer and proposed a 100-point bounty. Aug 19, 2018 at 0:10