$f:\mathbb R\to\mathbb R$ 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 one of the tangent line $L_n$ at $x_n$.

(If at a point $f$ is not differentiable, then multiple lines can be the tangent; $L_n$ 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 exists, but can it contain a set of points that was precisely the points on the graph of $f$?

$f:\mathbb R\to\mathbb R$ 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 $f$?