How accurate is $ f(px+(1-p)y)\leq pf(x)+(1-p)f(y)$ for a convex function $f$

Question

When $f:\mathbb{R}\rightarrow \mathbb{R}$ is a convex function, it is well known that $ f(px+(1-p)y)\leq pf(x)+(1-p)f(y)$ for any $p\in (0,1)$ and $x,y \in \mathbb{R}$. This inequality becomes an equality in case of a linear function. But what about functions which are close to linear on some interval $[a,b]$, what can be said about $|pf(x)+(1-p)f(y)-f(px+(1-p)y)$| for $x,y \in [a,b]$?

Answer 1

A crude but simple way to bound the quantity you describe can be found if $f$ is twice differentiable; in this case we have $$ |pf(x)+(1-p)f(y)-f(px+(1-p)y)|\leq p(1-p)|x-y|^2\sup_{z\in[a,b]}|f^{\prime\prime}(z)| $$ for all $x,y\in[a,b]$ and $p\in[0,1]$. This can be proven by thrice applying the mean value theorem: WLOG assume that $x<y$, then there exist $\alpha,\beta\in[a,b]$ with $x\leq\alpha\leq px+(1-p)y\leq\beta\leq y$ such that $f(x)-f(px+(1-p)y)=f^\prime(\alpha)(1-p)(x-y)$ and $f(y)-f(px+(1-p)y)=f'(\beta)p(y-x)$. Furthermore, there exists $\gamma$ with $\alpha\leq\gamma\leq\beta$ such that $f^\prime(\beta)-f^\prime(\alpha)=(\beta-\alpha)f^{\prime\prime}(\gamma)$. Assembling all this gives $$ |pf(x)+(1-p)f(y)-f(px+(1-p)y)|=|p(f(x)-f(px+(1-p)y))+(1-p)(f(y)-f(px+(1-p)y))|=|pf'(\alpha)(1-p)(x-y)+(1-p)f^\prime(\beta)p(y-x)|=p(1-p)|y-x||f^\prime(\alpha)-f^\prime(\beta)|=p(1-p)|y-x||\alpha-\beta||f^{\prime\prime}(\gamma)| $$ from which the bound follows.