Computation of tangent functional

Question

In Measures Which Agree on Balls by Hoffmann-Jørgenson, the tangent functional is defined as follows.

If $x \in S$, we define the tangent functional $\tau(x,\cdot)$ at $x$ as \begin{equation} \tau(x,y)=\lim_{t \to 0^+} \frac{\|x+ty\|-1}{t}, \end{equation}

where $S$ is the unit sphere in a Banach space $E$, and $y \in E$. It can be seen that $\tau(x,y)=\sup_{x^* \in T(x)} \langle x^*,y \rangle$ as well.

However, in the case that $E=C_0(\Omega)$, the set of real continuous functions which vanish at infinite, where $\Omega$ is a metrizable locally compact space, it is claimed (page 324) that "an easy computation shows" \begin{equation} \tau(x,y)=\sup_{\omega \in N(x)}x(\omega)y(\omega) \quad \forall x \in S, \forall y \in E \end{equation} where $N(x)=\{\omega \in \Omega : x(\omega)=\pm 1\}$.

It seems as if this computation might not be so easy after all. Does anyone see why the claim holds?

Answer 1

$\newcommand{\om}{\omega}\newcommand{\Om}{\Omega} \newcommand{\de}{\delta}\renewcommand{\th}{\theta}$For each real $t>0$ and some $\om_t\in\Om$, \begin{equation*} \|x+ty\| =|(x+ty)(\om_t)|=|x(\om_t)+ty(\om_t)|, \end{equation*} because $x$ and $y$ are in $C_0(\Om)$ and hence $x+ty\in C_0(\Om)$. For this reason and because $\Om$ is a metrizable locally compact space, we moreover have \begin{equation*} x(\om_s)\to x(\om_0)\quad\text{and}\quad y(\om_s)\to y(\om_0) \end{equation*} for some $\om_0\in\Om$ and some subnet $(\om_s)$ of the net $(\om_t)_{t\in(0,\infty)}$ (directed downwards, from $\infty$ to $0$).

If we had $|x(\om_0)|\le1-\de$ for some real $\de>0$, then we would also have \begin{equation*} \tau(x,y)=\lim_s\frac{|x(\om_s)+sy(\om_s)|-1}s \le\lim_s\frac{1-\de/2+s|y(\om_s)|-1}s=-\infty. \end{equation*} However, since $\|x\|=1$ and $x\in C_0(\Om)$, we have $|x(\om_x)|=1$ for some $\om_x\in\Om$ and hence \begin{equation*} \tau(x,y)\ge\lim_t\frac{|x(\om_x)|-t|y(\om_x)|-1}t=-|y(\om_x)|>-\infty. \end{equation*}

So, $|x(\om_0)|=1$. So, by symmetry, without loss of generality (wlog) $x(\om_0)=1$. Therefore and because $\|x\|=1$ and $\|y\|<\infty$,
\begin{equation*} \begin{aligned} \tau(x,y)&=\lim_s\frac{|x(\om_s)+sy(\om_s)|-1}s \\ &=\lim_s\frac{x(\om_s)+sy(\om_s)-1}s \\ &\le\lim_s\frac{1+sy(\om_s)-1}s \\ &=y(\om_0)=x(\om_0)y(\om_0)\le\sup_{N(x)}xy=:\th(x,y). \end{aligned} \end{equation*}

On the other hand, again because $x$ and $y$ are in $C_0(\Om)$, for some $\om_1\in\Om$ such that $|x(\om_1)|=1$ we have $\th(x,y)=x(\om_1)y(\om_1)$. By symmetry, wlog $x(\om_1)=1$. So, \begin{equation*} \th(x,y)=y(\om_1)=\lim_{t\downarrow0}\frac{x(\om_1)+ty(\om_1)-1}t \le\lim_{t\downarrow0}\frac{\|x+ty\|-1}t=\tau(x,y). \end{equation*}

The latter two displays show that $\tau(x,y)=\th(x,y)$. $\quad\Box$

Answer 2

The idea is that, for $|x(\omega)|\sim 1$ and $t$ small, $$|x(\omega)+ty(\omega)|\sim|x(\omega)^2+tx(\omega)y(\omega)|\sim 1+tx(\omega)y(\omega)$$

Indeed, if $|x(\omega)|<1-\varepsilon$, then for all sufficiently small $t$ one has that the quantity $|x(\omega)+ty(\omega)|$ is smaller than $1-\varepsilon-t\|y\|$, which is a lower bound for $\|x+ty\|-\varepsilon$; consequently, $\omega$ is eventually irrelevant to the norm $\|x+ty\|$. On the other hand, at a point $\omega$ such that $x(\omega)\in\{\pm 1\}$, one concludes the exact manipulation $$|x(\omega)+ty(\omega)|=|x(\omega)^2+tx(\omega)y(\omega)|=1+tx(\omega)y(\omega)$$ for all small $t$. A few estimates will then suffice to reach the desired result.