Banach mean value estimate on a convex set

If $x=y$ the estimate reads $\mathrm{\parallel }f(x)-f(x)\mathrm{\parallel }\le 0$ and holds; if $w:=f(y)-f(x)=0$ it reads $0\le M\mathrm{\parallel }y-x\mathrm{\parallel }$ and holds. Hence we may assume $x\mathrm{\ne }y$ and $w\mathrm{\ne }0$, and then [L3] provides a norming functional for this nonzero $w$.

By [L3] fix $\phi \in Y^{\ast }$ with $\mathrm{\parallel }\phi \mathrm{\parallel }=1$ and $\phi (w)=\mathrm{\parallel }w\mathrm{\parallel }$; then $\phi (f(y))-\phi (f(x))=\mathrm{\parallel }f(y)-f(x)\mathrm{\parallel }\ge 0$.

Define $\gamma :\mathbb{R}\to X$ by $\gamma (t):=x+t(y-x)$ and put $J:={\gamma }^{-1}[U]=\{t\in \mathbb{R}:\gamma (t)\in U\}\mathrm{.}$ By [L2], $[0,1]\subseteq J$. The set $J$ is open: if $t_0\in J$, openness of $U$ gives $\rho >0$ with $B(\gamma (t_0),\rho )\subseteq U$, and $\mathrm{\mid }t-t_0\mathrm{\mid }<\rho \mathrm{/}\mathrm{\parallel }y-x\mathrm{\parallel }$ implies $\mathrm{\parallel }\gamma (t)-\gamma (t_0)\mathrm{\parallel }=\mathrm{\mid }t-t_0\mathrm{\mid }\mathrm{\parallel }y-x\mathrm{\parallel }<\rho$ (recall that $x\mathrm{\ne }y$). Define the well-typed function $g:J\to \mathbb{R}$ by $g(t):=\phi (f(\gamma (t)))$.

The map $\gamma$ is differentiable at every $t_0\in \mathbb{R}$ with constant derivative $D\gamma (t_0)=(s\mapsto s(y-x))$, because $\gamma (t_0+s)-\gamma (t_0)=s(y-x)$ exactly, so the remainder vanishes.

The restriction $\gamma {\mathrm{\mid }}_J:J\to U$ is differentiable at every $t_0\in J$ with the derivative in [step 1.4]. Since $J$ and $U$ are open and $\gamma [J]\subseteq U$, two applications of [L4], first to $f\circ \gamma {\mathrm{\mid }}_J$ and then to the bounded linear functional $\phi$, show that $g$ is differentiable on $J$ with $g^{\mathrm{\prime }}(t_0)=\phi (Df(\gamma (t_0))(y-x))\mathrm{.}$ In particular $g$ is continuous on $J$, hence its restriction to $[0,1]$ is continuous there and differentiable on $(0,1)$.

For every $t_0\in [0,1]$ the bound in the statement gives $\mathrm{\parallel }Df(\gamma (t_0))\mathrm{\parallel }\le M$ because $\gamma (t_0)\in [x,y]$, so by [step 2.1], [L6] and $\mathrm{\parallel }\phi \mathrm{\parallel }=1$, $\mathrm{\mid }{g}^{\mathrm{\prime }}(t_0)\mathrm{\mid }=\mid \phi (Df(\gamma (t_0))(y-x))\mid \le \mathrm{\parallel }Df(\gamma (t_0))\mathrm{\parallel }\cdot \mathrm{\parallel }y-x\mathrm{\parallel }\le M\mathrm{\parallel }y-x\mathrm{\parallel }\mathrm{.}$

By [L5] applied to $g$ on the interval $[0,1]$, whose hypotheses were verified in [step 2.1], there is $c\in (0,1)$ with $g(1)-g(0)=g^{\mathrm{\prime }}(c)$.

Combining [step 1.2], [step 1.3] and [step 2.1], $\mathrm{\parallel }f(y)-f(x)\mathrm{\parallel }=\phi (f(y))-\phi (f(x))=g(1)-g(0)=g^{\mathrm{\prime }}(c)=\mathrm{\mid }{g}^{\mathrm{\prime }}(c)\mathrm{\mid }\le M\mathrm{\parallel }y-x\mathrm{\parallel },$ the second equality because $g(1)-g(0)=\mathrm{\parallel }f(y)-f(x)\mathrm{\parallel }\ge 0$ forces $g^{\mathrm{\prime }}(c)\ge 0$.

The estimate is [step 4.1] under the assumptions made there, and [step 1.1] disposes of the two degenerate cases; hence it holds in general.