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I'm struggling to understand a particular implication in the proof of Corollary 5 of this paper involving Gateaux differentiability of the norm. The claim is that Gateaux differentiability of the norm implies that for all $x^* \in V^*:=\{x^* \in S^* : \exists x \in S \text{ such that } \langle x^*, x \rangle=1\}$, there exists a $x \in S$ such that $T(x)=\{x^*\}$, where $T(x)$ is the set of normals to $S$ at $x$. Here, $S$ and $S^*$ are the unit spheres in the underlying Banach space $E$ and its dual $E^*$, respectively. Why is this? It does not seem clear.

Edit: Maybe this is deserving of its own question, but I'm also unsure why the $L^p$ norm is Gateaux differentiable when $1<p<\infty$. Are there any references for this?

Edit 2: I think this actually answers the question raised in the edit.

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    $\begingroup$ Please re-read your post carefully and edit it accordingly. $\endgroup$ Commented Nov 14, 2023 at 17:27
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    $\begingroup$ There is something fishy in the question, especially in the definition of the set $V^*$. $\endgroup$ Commented Nov 14, 2023 at 17:39
  • $\begingroup$ @IosifPinelis I think I fixed it. Thank you! $\endgroup$ Commented Nov 14, 2023 at 17:45
  • $\begingroup$ @DelioMugnolo Thank you for pointing this out! I think I fixed it. $\endgroup$ Commented Nov 14, 2023 at 17:45
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    $\begingroup$ What is $S^*$ ? $\endgroup$ Commented Nov 14, 2023 at 17:47

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We have the following definitions: \begin{equation*} V^*:=\{f\in S^*\colon\exists x\in S\ f(x)=1\}, \end{equation*} where $S$ and $S^*$ are the unit spheres in the underlying Banach space $E$ and its dual $E^*$, respectively, and \begin{equation*} T(x):=\{f\in E^*\colon\|f\|\le1,f(x)=1\} \end{equation*} for $x\in S$.

We want to show that \begin{equation*} f\in V^*\overset{\text{(?)}}\implies\exists x\in S\ T(x)=\{f\} \tag{1}\label{1} \end{equation*} provided that the norm on $E$ is Gateaux differentiable.

Since $f\in V^*$, we have $\|f\|=1$ and $f(x)=1$ for some $x\in S$; fix any such $x$. Then \begin{equation*} f\in T(x). \end{equation*} Take any $g\in T(x)$. It suffices to show that $g=f$.

To obtain a contradiction, suppose that $g\ne f$. Then \begin{equation*} f(y)=0\ne g(y) \end{equation*} for some $y\in E$. Fix any such $y$.

By the Gateaux differentiability of the norm on $E$, for some real $l_y$ we have
\begin{equation*} \|x+ty\|=\|x\|+l_y t+o(t)=1+l_yt+o(t) \end{equation*} (as $t\to0$).

Case 1: $l_y\ne0$. Then \begin{equation*} 1=f(x)=f(x)+tf(y)=f(x+ty)\le\|x+ty\|=1+l_y t+o(t)<1 \end{equation*} if $|t|$ is small enough and $l_y t<0$. So, Case 1 results in a contradiction. (Note that the condition $g(y)\ne0$ was not used in Case 1.)

Case 2: $l_y=0$. Then for any real $a\ne0$ and all real $t$ close enough to $0$ we have \begin{equation*} \|x+t(ax+y)\|=\|(1+ta)x+ty\|=(1+ta)\|x+\tfrac t{1+ta}\,y\| \\ =(1+ta)(1+0 t+o(t))=1+at+o(t), \end{equation*} so that \begin{equation*} l_{ax+y}=a\ne0. \end{equation*} Note also that $g(ax+y)=a+g(y)=0$ if $a=-g(y)\ne0$. So, replacing $y$ by $ax+y$ and interchanging the roles of $f$ and $g$, we reduce Case 2 to Case 1. $\quad\Box$

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