We say that a function $f\colon (a,b)\to \mathbb{R}$, where $-\infty\leq a<b\leq +\infty$, is convex (resp. strictly convex) if for all $x,y\in (a,b)$ with $x<y$ and for every $t\in (0,1)$ it holds that $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$ $\Big($ resp. $f(tx+(1-t)y)< tf(x)+(1-t)f(y)\Big)$.
The simplest example of a convex function that is not stricly convex is a linear function $f(x)=\alpha x+\beta$ on any interval. However, linear functions do not exhaust all variants, since the function $f(x)=|x|$ on interval $(-1,1)$ is convex but not strictly convex. At the same time, this function is piecewise linear, so, in particular, linear at "almost every" neighborhood.
My question is as follows: can a convex function $f$ that is not stricly convex be nowhere linear, that is, have the property that for every $x_0\in (a,b)$ and for every $\delta>0$ the function $f\big|_{(x_0-\delta,x_0+\delta)\cap (a,b)}$ is not linear, i.e. there is no $\alpha,\beta\in \mathbb{R}$ such that $f(x)=\alpha x+\beta$ for every $x\in (x_0-\delta,x_0+\delta)\cap (a,b)$?
Best Answer
That is not possible: If $f$ is convex on an interval, and equality holds in the convexity condition for some $x < y$ and some $0 < t < 1$, then $f$ is linear on $[x, y]$.
For a proof set $z = tx+(1-t)y$ and let $l(x)$ be the linear function with $l(x)=f(x)$ and $l(y) = f(y)$. Note that $l(z) = f(z)$. Then show that
This shows that $f(u) = l(u)$ on $[x, y]$.