Restriction to dense subset of functions whose graph is dense Let $f: [0, 1] \to \mathbb R$ be a measurable function. A function $g: [0, 1] \to \mathbb R$ is said to be a condensation limit of $f$ if $g$ is continuous and agrees with $f$ on a dense subset of $[0, 1]$.
Let $k \geq 1$ be an integer, and $f: [0, 1] \to \mathbb R$ be a measurable function whose graph is dense in $[0, 1] \times \mathbb R$.
Question: Is the set of $k$ times continuously differentiable condensation limits of $f$ dense in $C^k$?
Note: Here $C^k$ denotes the space of $k$ times continuously differentiable functions under the norm $\| g \|_{C^k} := \sum_{i = 0}^k \sup _{x \in [0, 1]} \lvert g^{(i)} (x) \rvert$, where $g^{(i)}$ is the $i$’th derivative of $g$.
 A: Yes. This is true.
Proposition: Let $A\subseteq[0,1]\times\mathbb{R}$ be dense. Then for each $k\geq 0$ and $\epsilon>0$ and $g\in C^k([0,1])$, there is some $C^k$ function $f:[0,1]\rightarrow\mathbb{R}$ where $\{x:(x,f(x))\in A\}$ is dense in $[0,1]$ and where $\|f-g\|<\epsilon$.
Proof: We shall construct a sequence of functions by recursion. Let $f_0=g$. If $f_n$ has already been constructed, then let $U_n=(0,1)\setminus\overline{\{x:(x,f_n(x))\in A\}}$. If $U_n=\emptyset$, then we have already constructed our required function $f$ and we do not need to continue this construction. Otherwise, $U_n$ is a non-empty open set, so $U_n$ is the union of open intervals. Choose the interval $(a_n,b_n)$ in $U_n$ with the maximum length, and then let $f_{n+1}:[0,1]\rightarrow\mathbb{R}$ be a smooth function where $\|f_{n+1}-f_n\|_{C^k}<\epsilon 2^{-(n+1)}$ and where $f_{n+1}-f_n$ is supported on $(a_n,b_n)$ but where
$(c,f_{n+1}(c))\in A$ for some $c\in(a_n+(b_n-a_n)/3,a_n+2(b_n-a_n)/3)$. Then $f_n$ is Cauchy, so $f_n\rightarrow f$ for some $f$ in the norm, (and therefore $f_n\rightarrow f$ pointwise as well). We observe that $(U_n)_n$ is a decreasing sequence of open sets with $b_n-a_n\rightarrow 0$. We also observe that $|g-f\|<\epsilon$. Q.E.D.
One can also generalize this question in many ways (such as if we replace $[0,1]$ and $\mathbb{R}$ with differentiable manifolds, complex manifolds, topological spaces and if we require $f$ to be smooth, holomorphic, harmonic, continuous, etc.), and I suspect that in many of those generalized questions the answer is also yes.
