Does uniform convergence of suitable functions yield pathwise convergence of their convex envelopes?

Question

For each $k\ge 1$, let $f_k:\mathbb R\to\mathbb R_+$ be $1-$Lipschitz, increasing such that $f_k(x)\ge x^+$ for $x\in\mathbb R$, $f_k(-\infty)=0$ and $$\lim_{x\to+\infty} \big(f_k(x)-x\big)=0.$$ Assume that $f_k$ converges uniformly to $f:\mathbb R\to\mathbb R_+$. Does it hold $$\lim_{k\to\infty} co(f_k)(x)=co(f)(x),\quad \forall x\in\mathbb R?$$ Here $co(f_k)$, $co(f)$ denote the so-called convex envelopes of $f_k$ and $f$, see e.g. https://en.wikipedia.org/wiki/Lower_convex_envelope

PS : Set $g_k:=co(f_k)$ and $g:=co(f)(x)$ (of coures there existence can be verified easily). Then one has $x^+\le g_k(x)\le f_k(x)$ and $x^+\le g(x)\le f(x)$.

Answer 1

$\newcommand\ep\varepsilon$The answer is yes, and we only need the uniform convergence of $f_k$ to $f$. Moreover, then $co(f_k)\to co(f)$ uniformly as well.

Indeed, for each real $x$, $$co(f)(x)=\sup\{g(x)\colon g\in G_f\},$$ where $G_f$ is the set of all convex functions $g$ such that $g\le f$.

Take now any real $\ep>0$. Then, by the uniform convergence of $f_k$ to $f$, for some natural $K_\ep$ and all natural $k\ge K_\ep$ we have $$f_k-\ep\le f\le f_k+\ep.$$ So, for any $g_k\in G_{f_k}$ we have $f\ge g_k-\ep$ and $g_k-\ep$ is convex, so that $g_k-\ep\in G_f$ and hence $co(f)\ge g_k-\ep$, for any $g_k\in G_{f_k}$. So, $$co(f)\ge co(f_k)-\ep. $$

Now take any $g\in G_f$. Then (for $k\ge K_\ep$) $g\le f_k+\ep$ and hence $g-\ep\in G_{f_k}$, so that $co(f_k)\ge g-\ep$, for any $g\in G_f$. So, $$co(f_k)\ge co(f)-\ep.$$

Thus, $|co(f_k)-co(f)|\le\ep$ for all $k\ge K_\ep$. $\quad\Box$