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Let $(X, \|.\|_1)$ is isometrically isomorphic to $(X, \|.\|_2)$ and $\|.\|_2\leq \|.\|_1$. Assume that $x_0$ is a smooth point of $(X, \|.\|_1)$ and $\|x_0\|_2=1$. According to the definition of a smooth point let the collection of support functionals at $x_0$ be

$$J(x_0)=\{\mu x_1^*:|\mu|=1\},\qquad x_1^*\in (X^*, \|.\|_1^*)\quad \text{and}\quad \|x_1^*\|=1.$$

My question: Is $x_0$ also smooth in $(X, \|.\|_2)?$ Is the comparison between two norms necessary to conclude this?

I haven't been able to proceed. Please help. Any help is appreciated. Thank you in advance.

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    $\begingroup$ Cross posted on Math.SE. While this is not forbidden, the accepted good practice is to wait at least a week in order to let the members of the other site to post a reasonably good answer, and to put a cross posting notice on the body of the question. Otherwise, it seems you want to issue a competition between the StackExchanges to get as soon as possible the sought for answer, which is not the purpose of this Q&A sites. $\endgroup$
    – Daniele Tampieri
    Commented Dec 5 at 7:54
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    $\begingroup$ Many thanks for the suggestion. I will remember this and definitely do accordingly. Pardon me as of now. I badly need a lead for the question. $\endgroup$
    – Tuh
    Commented Dec 5 at 8:05
  • $\begingroup$ I think you need to specify the map $X\to X$ that implements the isometric isomorphism. This map will send a smooth point for $\lVert\cdot\rVert_1$ to a smooth point for $\lVert\cdot\rVert$. However, I don't see why a point $x_0$ would be a smooth point for both norms simultaneously. $\endgroup$
    – Yemon Choi
    Commented Dec 5 at 17:26

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$\newcommand\R{\Bbb R}$Counterexample: $X=\R^2$, $\|x\|_1=|u|+|v|$ and $\|x\|_2=\max(|u|,|v|)$ for $x=(u,v)\in\R^2$.

Then $\|\cdot\|_2\le\|\cdot\|_1$ and the map $\R^2\ni(u,v)\mapsto\frac12\,(u-v,u+v)$ is an isometric isomorphism of $(X,\|\cdot\|_2)$ to $(X,\|\cdot\|_1)$. However, $x_0:=(1,1)$ (with $\|x_0\|_2=1$) is smooth in $(X,\|\cdot\|_1)$ but not in $(X,\|\cdot\|_2)$.

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  • $\begingroup$ But $\|(1, 1)\|_1\neq 1$. $\endgroup$
    – Tuh
    Commented Dec 5 at 16:04
  • $\begingroup$ What is your definition of a smooth point? $\endgroup$
    – Iosif Pinelis
    Commented Dec 5 at 17:01
  • $\begingroup$ A norm one point $x_0$ is called a smooth point if the collection of its support functionals contains all the unimodular scalar multiplication of only one support functional, i.e., $J(x_0)=\{\mu x_1^*:|\mu|=1\},\qquad x_1^*\in (X^*, \|.\|_1^*)\quad \text{and}\quad \|x_1^*\|=1.$ $\endgroup$
    – Tuh
    Commented Dec 5 at 18:00
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    $\begingroup$ @Tuh : The isomorphism is from $(X,\|\cdot\|_2)$ to $(X,\|\cdot\|_1)$, not vice versa. $\endgroup$
    – Iosif Pinelis
    Commented Dec 6 at 14:40
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    $\begingroup$ @Tuh : An isomorphism does not have to coincide with its inverse. In this case, the inverse of the isomorphism $(u,v)\mapsto\frac12\,(u-v,u+v)$ is $(s,t)\mapsto(s+t,t-s)$. $\endgroup$
    – Iosif Pinelis
    Commented Dec 6 at 14:54

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