# “Mollification” of a convex function with a finite set of points unchanged

$$f:\mathbb R\to\mathbb R$$ is a convex continuous function. We have a finite or a countable set of triples: $$\{(x_n,f(x_n),D_n)\}_{n\in N}$$, where $$D_n$$ is the slope of a tangent line $$L_n$$ at $$x_n$$ (if at a point $$f$$ is not differentiable, then multiple lines can be tangents; $$L_n$$ is just one of those lines).

Assuming that, for any $$n,m,k$$, the intersection of $$L_n$$ and $$L_m$$ cannot be the point $$(x_k, f(x_k))$$, then we want to prove that there exists a smooth function $$g$$ such that $$g(x_n)=f(x_n)$$ and $$g'(x_n)=D_n$$ for any $$n$$.

The original problem that I am trying to solve involves multi-dimensional manifolds, but I think it is easy to generalize the 2-dimensional case.

By the mollification theorem, a smooth function approximating $$f$$ must exist, but can it contain a set of points that corresponds precisely to the points on the graph of $$f$$?

Making quantitative the assumption "the intersection of $$L_n$$ and $$L_m$$ cannot be the point $$(x_k, f(x_k))$$", it is possible to construct such a function $$g$$ (as pointed out by Jaume, the nonquantitative assumption is not sufficient). Let us consider the problem in $$\mathbb R^n$$.

Given a family of indices $$I$$, let $$(x_i)_{i\in I}\subset \mathbb R^n$$ be a family of points and let $$(L_i)_{i\in I}$$ be a family of affine functions. We assume that there is a convex function $$f:\mathbb R^n\to\mathbb R$$ such that $$f(x_i) = L_i(x_i)$$ and $$f\ge L_i$$.

Quantitative assumption: There is an $$\varepsilon > 0$$ such that for any $$i\not=j$$ we have $$L_i(x)\ge L_j(x)$$ for any $$x\in B(x_i, \varepsilon)$$ (when $$x = x_i$$ this follows from the convexity of $$f$$) (this assumption is equivalent to the original one if $$I$$ is finite).

Let $$h=\sup_{i\in I} L_i$$. The function $$h$$ is convex and $$h\le f < \infty$$. Let $$\rho\in C^{\infty}_c(B(0,\varepsilon))$$ be a convolution kernel ($$\rho\ge 0$$ and $$\int\rho=1$$). Define $$g=h\star\rho$$. It is not hard to check that $$g$$ is smooth, $$g(x_i)=L_i(x_i)$$ and $$g\ge L_i$$ for each $$i\in I$$.

Relaxing the assumption: It is possible, with an appropriate partition of unity argument, to prove the result even if the value of $$\varepsilon$$ is allowed to depend on $$i$$ (so we would have $$\varepsilon_i$$), provided that it remains locally bounded away from $$0$$ (i.e., $$\inf_{|x_i| 0$$).

• Very helpful! I think I understand most of your exposition but what is the variable $R$ in the last line? – High GPA Sep 26 at 9:56
• It should be understood as "For any $R>0$, $\inf_{|x_i|<R}\varepsilon_i > 0$". – dario2994 Sep 29 at 9:46

No, as long as the number of triples you are allowing is infinite. Pick any strictly convex function that is smooth anywhere but at zero (say $$f(x) = |x|+|x|^2$$). And take as your set of $$x_n$$ to be the set $$\{1/m, m \in \mathbb Z \setminus \{0\}\}$$.

Since there's points near zero where the slope is set to be (essentially) $$1$$ and points where it is set to be (essentially) $$-1$$ you cannot find a smooth extension.

• You might want to find some extra hypothesis that align with some variation of Tietze's extension theorem (such as the set $\{x_m, m\in 1, 2,\dots\}$ being closed, only asking for a finite number of derivatives..). – Jaume Sep 25 at 7:48