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We know that every convex function $g:\mathbf{R}\rightarrow\mathbf{R}$ is locally Lipschitz, and therefore $g$ is differentiable almost everywhere (Alexandrov's theorem says that $g$ is in fact twice differentiable almost everywhere and so $g$ is ${\rm C}^1$ almost everywhere). For $n=1,2$ we define the lower (upper, resp.) smooth convex envelope of $g$ as follows:

$L_n(g):={\rm sup}\{h:\mathbf{R}\rightarrow\mathbf{R}: h\leq g,\; h\;\hbox{is convex},\;h\in{\rm C}^n(\mathbf{R})\}$

($U_n(g):={\rm inf}\{h:\mathbf{R}\rightarrow\mathbf{R}: h\geq g,\; h\;\hbox{is convex},\;h\in{\rm C}^n(\mathbf{R})\}$, resp.).

My question is: can we formulate necessary and/or sufficient conditions on $g$ which yield the following conclusions:

(1) $L_1(g)$ exists ($U_1(g)$ exists, resp.), meaning that there exists at least one lower (upper, resp.) function ${\rm C}^1$-smooth function $h$,

(2) $L_1(g)=g$ ($U_1(g)=g$, resp.),

(3) $L_2(g)$ exists ($U_2(g)$ exists, resp.), meaning that there exists at least one lower (upper, resp.) function ${\rm C}^2$-smooth function $h$,

(4) $L_2(g)=g$ ($U_2(g)=g$, resp.).

Ofcourse, we can extend the question to the case $h\in{\rm C}^n(\mathbf{R})$ for every $n\in\mathbf{N}$, or even to the case when $h$ is analytic functon on $\mathbf{R}$, but for my purposes it is enough to consider the case $n=1$ and $n=2$. If classical smoothness can not be considered, I would settle for weakly differentiable functions $h\in {\rm W}^{2,p}_{loc}(\mathbf{R})$ for some $p\geq 1$, but I have the feeling that this could be much harder question. One possibility of extra assumption on $g$ is, for instance, that $g$ is differentiable (or twice differenntiable) except on at most countable points $x\in\mathbf{R}$. Remark. Corollary 7.2.4 in Garling: A Course in Mathematical Analysis, Volume 1, provides that one-sided derivatives of convex function $g$, denoted by $g'(x-)$ and $g'(x+)$, exist everwhere, except on at most countably many points $x\in\mathbf{R}$. Here

A convex function is differentiable at all but countably many points

we have an improvement of the corollary from Garling's book, which says that every convex function is differentiable except at most countably many points, but the question still stands.

Andrija
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  • This paper might be relevant for the $\mathcal{C}^1$ case. It is also noteworthy that requiring the convex envelope to be in $\mathcal{C}^2$ is a fairly strong; there are simple examples of smooth functions that do not have a $\mathcal{C}^2$ envelope. – madnessweasley Feb 26 '18 at 14:36
  • Thanks for bringing my attention to the paper by Griewank and Rabier. If ${\rm C}^2$-envelope is to strong, I guess we are left with ${\rm W}^{2,p}$-envelope, but I have no idea how to build such Sobolev functions. – Andrija Feb 26 '18 at 16:04

1 Answers1

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A partial answer:

Always $L_1 (g) = L_2 (g) = g$, Think about the set of all affine functions majorized by $g$

Also the way you defined functions $L_i (g) ,$ and $U_i (g)$ doesn't make any sense! I am assuming you meant for all $x \in R$ $$L_n(g)(x) :={\rm sup}\{h(x): \quad h\leq g,\; h\;\hbox{is convex},\;h\in{\rm C}^n(\mathbf{R})\}$$

$$U_n(g)(x) :={\rm inf}\{h(x): \quad h\ge g,\; h\;\hbox{is convex},\;h\in{\rm C}^n(\mathbf{R})\}$$

Note that since $g$ is convex on all over $\Bbb R$ then $g = g^{**}$ where

$$ g(x) = g^{**} (x) = \sup \{\langle x , y \rangle\ - g^{*}(y) : \quad y \in R\} $$

Which clearly says $g$ is the pointwise supremum of some affine functions.

Red shoes
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  • Well, if there exists (at least one) affine functiion $h$ such that $h\leq g$, it makes sense to try to prove that $L_1(g)=L_2(g)=g$. Otherwise, I don't see the argument. I think that one possibility for additional condition on $g$ is $\lim_{x\rightarrow\pm\infty}g(x)=+\infty$. Then, by strict convexity of $g$, it follows $\lim_{x\rightarrow\pm\infty}|{{g(x)}\over{x}}|>0$, and there exists at least one lower affine bound, so it probably gives $L_1(g)=L_2(g)=g$, but I am not sure. Maybe you can be more explicit in your answer? Thanks. – Andrija Feb 28 '18 at 18:26
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    For sure there exist at least one affine function . Do you know what subdifferential means ? At every point on $g$ has at least one subgradient @Andrija – Red shoes Mar 01 '18 at 04:35
  • I see, we always have an affine lower bound. There was a typo in the definition of $U_i(g)$ (sup instead of inf), I corrected it. I guess this means that every convex function is supremum of piecewise affine and continuous lower bounds. From there, we can probably prove that $L_1(g)=L_2(g)=g$. Thanks... – Andrija Mar 01 '18 at 08:14