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Let $(X,\|\cdot\|)$ be a Banach space, and $B_{1}$ its closed unit ball. Recall that the modulus of convexity of $X$ is the function $\delta_{X}:[0,2]\longrightarrow [0,1]$ given by

$$ \delta_{X}(\varepsilon):=\inf\{1 -\frac{\|x+y\|}{2}:x,y\in B_{1},\|x-y\|\geq \varepsilon \}. $$

There is a lot of literature related to this function, but I have found nothing for $X:=C([0,1])$, the Banach space of the continuous functions defined on $[0,1]$, endowed its usual supremum norm.

Somebody know some result for $\delta_{C([0,1])}(\varepsilon)$? Or at least a no trivial lower bound for $\delta_{C([0,1])}(\varepsilon)$?

Many thanks in advance for your comments.

user123043
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1 Answers1

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For $X=C([0,1])$ we should have $\delta_{X}(\varepsilon) = 0$.

Indeed, by definition $\| x+y \|= \sup\left\{ |x(t)+y(t)|: t\in [0,1] \right\}$. Take $x,y\in B_1$ s.t. $x(0.25)=y(0.25)=1$ and $|x(0.75)-y(0.75)|\ge \varepsilon$. In this case we have $1-\frac{\| x+y \|}{2}=0$.

Manlio
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  • Indeed, the modulus of convexity is a measure for how much the unit ball is convex. The supremum (uniform) norm is analogous to $\ell_\infty$ norm. For example, in the plane, the unit ball in $\ell_\infty$ is the square $[-1,1]^2$ with a totally flat boundary. Thus, the zero convexity. – A.Γ. Apr 18 '20 at 10:27
  • Many thanks for your comments! – user123043 Apr 19 '20 at 09:48