I have to prove that, given $L=L(g, [a, b])$ the Lipschitz constant of $g:[a, b]\subset\mathbb{R}\longrightarrow\mathbb{R}$, we have that if $$ \frac{|g(b)-g(a)|}{|b-a|}=L,\quad (\star) $$ then $g(a+t(b-a))=g(a)+t(g(b)-g(a))$ for $0\leq t\leq1$.
My attempt. By setting $x=a+t(b-a)$ and $y=g(a)+t(g(b)-g(a))$, we have to prove that $g(x)=y$. I'm trying to show this using the inequalities $$ |g(x)-g(a)|<L(x-a),\qquad |g(b)-g(x)|<L(b-x) $$ (do these inequality come from the hypothesis $(\star)$?). In particular, undoing the absolute value of the first inequality we get $$ g(a)+\frac{|g(b)-g(a)|}{|b-a|}(a-x)<g(x)<g(a)+\frac{|g(b)-g(a)|}{|b-a|}(x-a)= $$ $$ =g(a)+\frac{|g(b)-g(a)|}{|b-a|}t(b-a) $$ But how can I conclude that $g(x)=y$?
Thank You
All points $(x,g(x))$ in the graph of $g$ lie in the region cut out by the inequalities $$ a\le x\le b$$ $$-L \le \frac{ g(x)-g(a)}{x-a} \le L, \text{ if } a<x.$$ Also, all points in the graph lie in the region cut out by the inequalities $$ a\le x\le b$$ $$-L \le \frac{ g(b)-g(x)}{b-x} \le L, \text{ if } x<b.$$ The intersection of the two regions is the line segment connecting $(a,g(a))$ and $(b,g(b))$.