$f:\mathbb R\to\mathbb R$ 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 $n$.




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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 $f$?



"Mollification" of a convex function with a finite set of points unchanged
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., $\inf\_{|x\_i|<R}\varepsilon\_i > 0$).


