Let $K$ be a non-empty convex open subset of $\mathbb{R}^d$, for some $d\in \mathbb{N}$.
Consider the family $\mathcal{F}$ of infinitely differentiable convex functions on $K$.
Let $C(K)$ be the set of continuous functions defined on $K$.
Equip $C(K)$ with the topology of uniform convergence on compact subsets of $K$ (for a standard definition, see e.g., Walter Rudin – Functional Analysis – Section 1.44).
Is it true that the set of convex functions defined on $K$ is the closure of $\mathcal{F}$ in $C(K)$ equipped with the topology of uniform convergence on compact subsets of $K$?
What if we consider $\mathcal{F}$ embedded in $\mathbb{R}^K$ with the topology of pointwise convergence? I.e., is it true that the set of convex functions defined on $K$ is the closure of $\mathcal{F}$ in $\mathbb{R}^K$ equipped with the topology of pointwise convergence?
Since the pointwise limit of a convex function is convex, what we have to prove is just the density of $\mathcal{F}$ in the convex functions.
I tried to work using convolutions. Specifically, if $f$ is a convex function defined on the whole $\mathbb{R}^d$ and $\varphi $ is a non-negative $C^\infty$ compactly supported function with $\int_{\mathbb{R}^d} \varphi(x) \mathrm {d}x = 1$, then, if $\varphi_{\varepsilon}$ is the $\varepsilon$-approximation of the identity associated to $\varphi$, we have that $f*\varphi_\varepsilon$ is a convex $C^\infty$ function, which converges uniformly on compact sets to $f$ as $\varepsilon \to 0^+$. However, our function $f$ is defined only on $K$, and may explode as $x$ approaches to the boundary of $K$, so we can't directly extend $f$ to a globally defined convex function (whenever $K$ is not the whole $\mathbb{R}^d$ already) and use the previous approach to prove that the closure of $\mathcal{F}$ contains the convex functions.
Can we salvage the convolution approach? Otherwise, what can we do?
Best Answer
Let $K_1, K_2, \dots$ a sequence of compact convex sets that are closures of open sets, such that $K_1 \subset K_2 \subset \dots$ and $K = \bigcup_{i \in \mathbb{N}} K_i$. Let $f$ be any real-valued convex function defined on $K$. Then $f \mid_{K_i}$ is a real-valued convex Lipschitz function for every $i \in \mathbb{N}$. Extend $f\mid_{K_i}$ to a real-valued convex function $f_i$ defined on the whole $\mathbb{R}^d$ (see Extension of Convex Function, Theorem 4.1). Convolving $f_i$ against a suitable approximation of the identity, we can build a real-valued convex $C^\infty$ function $g_i$ defined on the whole $\mathbb{R}^d$ such that $\|f_i - g_i\|_{L^\infty(K_i)} \le 1/i$. It is then clear that for each $i \in \mathbb{N}$ we have that $\|f-g_j\|_{L^\infty(K_i)} \to 0, j \to \infty$. Being $f$ arbitrarily chosen, it follows that $C^\infty(K)$ convex functions are dense in the real-valued convex functions defined on $K$ equipped with the $C(K)$ topology (hence, even with the pointwise topology)