Recently, I've been studying convexity in Banach spaces. In particular, I've seen the characteristic of convexity of a Banach space $X$, given by $\varepsilon_0(X)=\sup_{\delta_X(\varepsilon)=0}\varepsilon$, where $\delta_X(\varepsilon)$ is the modulus of convexity of $X$. Geometrically, it is said that the characteristic of convexity bounds the length of segments which lies entirely on the unit sphere. However, I'm unable to conclude this from the definition given before. Could anyone explain me how?
Thank you very much!!
To interpret the modulus of convexity $\delta_X (\epsilon)$ we can think of $\epsilon$ as the length of line segment that we're going to attach to points on the unit ball. The infimum is then taken so that we identify where the midpoint is closest to the unit ball. The modulus is then the distance from the midpoint to the ball.
Now $\epsilon_0(X)$ takes the supremum of the $\epsilon$ such that $\delta_X(\epsilon)=0$. For $\delta_X(\epsilon)$ to be zero we need the midpoint between two points on the ball to be on the ball as well. This implies that the entire segment from $x$ to $y$ is on the ball. If that never happen we have $\epsilon_0(X)=0$. However if there is a segment on the boundary, say between points $x$ and $y$, then for $0 < \epsilon \leq \|x -y\|$ we have $\delta_X(\epsilon) = 0$. This implies $\epsilon_0(X) \geq \epsilon$. By considering all possible lengths we see that $\epsilon_0(X)$ will the the length of the longest segment on the boundary. This is the bound we were looking for.