Let $f : \Bbb{R^n} \to \Bbb{R} \cup \{ +\infty \} $ be a convex function, then the domain of $f$ is the convex set $ D= \{ x \in \Bbb{R^n} ~ |\quad f(x) < \infty \}.$
Questions:
$1)$ Can we find a convex function $g: \Bbb{R^n} \to \Bbb{R} \cup \{ +\infty \} $ such that $g=f$ on $D$ and the domain of $g$ has nonempty interior ?
$2)$ In above question can we additionally assume for all $ x \in
\text{int[ }
\text{Domain} (g) ]\setminus D $
$g$ is differentiable at $ x $ ?
As an example Take $$f(x,y)=\left\{\begin{matrix} |x| & (x,y) \in \Bbb{R} \times \{ 0 \} \\ +\infty & O.W \end{matrix}\right.$$
Then one can take $g(x,y) = \sqrt{x^2 + y^2 } $.
P.S: Question $1$ geometrically is clear since we can extend $f$ along $ D^{\bot} $.
Basically I am interested the answer for locally convex space $X$, however I think the answer of the question $2$ even is not clear in $\Bbb {R^n} $.
The answer to both questions is yes. The support $D$ of $f$ is a convex set. Let ${\rm aff}(D)$ be its affine hull. Then, ${\rm aff}(D)=x_0+U$ for a linear subspace $U$ of $\mathbb R^n$ and any $x_0\in {\rm aff}(D)$. Let $V$ be a complementary subspace, so $\mathbb R^n=U\oplus V$.
Starting with the first question, define $g$ on $D+V$ by $g(x+v)=f(x)$ for any $x\in D$ and $v\in V$, and set $g$ to be $\infty$ elsewhere. This is convex with support $D+V$, which has nonempty interior ${\rm relint}(D)+V$ (${\rm relint}(D)$ is the relative interior).
For the second question, choose a nonnegative smooth function $\theta\colon U\to\mathbb R$ with support in the interior of the unit ball, symmetric so that $\theta(u)=\theta(-u)$ and with unit integral $\int_U\theta(u)\,du=1$. Let $D_g$ consist of the points $x+v\in D+V$ such that the closed ball of radius $\lVert v\rVert$ in $U$ entered at $x$ is contained in $D$. This contains $D$ and has interior containing ${\rm relint}(D)$. Define $g\colon D_g\to\mathbb R$ by $$ g(x+v)= \int_U f(x+\lVert v\rVert z)\theta(z)\,dz, $$ with $\int_U\,\cdot\,dz$ denoting the standard Lebesgue integral on $U$. Then, $g$ is smooth on $D_g\setminus D$. For $x+v,y+w\in D_g$ and $p,q\ge0$ with $p+q=1$, \begin{align} pg(x+v)+qg(y+w)&=\int_U\left(pf(x+\lVert v\rVert z)+qf(y+\lVert w\rVert z)\right)\theta(z)\,dz\\ &\ge\int_Uf(px+qy+(p\lVert v\rVert+q\lVert w\rVert) z)\theta(z)\,dz \end{align} Using the inequality $p\lVert v\rVert+q\lVert w\rVert\ge\lVert pv+qw\rVert$ and the fact that convexity of $f$ implies that $\int_Uf(x+tz)\theta(z)\,dz$ is increasing in $t$ to get $$ pg(x+v)+qg(y+w)\ge g(px+qy+pv+qw). $$ So, $g$ is convex. Entend to all of $\mathbb R^n$ by setting $g(x)=\infty$ outside of $D_g$.
I'll clarify a few properties which have been asked in the comments:
Convexity of $D_g$: Suppose that $x,y\in D$ and $v,w\in V$ are such that $x+v,y+w\in D_g$. For $p,q > 0$, $p+q=1$, we need to show that $z+u\in D_g$ where $z=px+qy$ and $u=pv+qw$. We have $\lVert u\rVert\le p\lVert v\rVert+q\lVert w\rVert$. By definition of $D_g$, we need to show that if $a$ is in the closed ball of radius $\lVert u\rVert$ about 0 in $U$ then $z+a\in D$. Setting $$ b = \frac{\lVert v\rVert}{p\lVert v\rVert+q\lVert w\rVert}a,\ c = \frac{\lVert w\rVert}{p\lVert v\rVert+q\lVert w\rVert}a, $$ then $\lVert b\rVert\le\lVert v\rVert$ and $\lVert c\rVert \le \lVert w\rVert$. So, $x+b,y+c\in D$ and, by convexity of $D$, $$ z+a=px+qy+pb+qc=p(x+b)+q(y+c)\in D $$ QED
$g$ is differentiable on $D_g\setminus D$: Rearranging the definition of $g$, $$ g(x+v)=\int_Uf(z)\lVert v\rVert^{-d}\theta((z-x)/\lVert v\rVert)\,dz $$ where $d$ is the dimension of $U$. You now only need to use that $\lVert v\rVert^{-d}\theta((z-x)/\lVert v\rVert)$ is differentiable over $v\not=0$. QED
$\int_Uf(x+tz)\theta(z)\,dz$ is increasing in $t\ge0$: using the symmetry of $\theta$, $$ \int_Uf(x+tz)\theta(z)\,dz=\frac12\int_U\left(f(x+tz)+f(x-tz)\right)\theta(z)\,dz $$ As $\theta$ is non-negative, it only needs to be shown that $t\mapsto f(x+tz)+f(x-tz)$ is increasing in $t$. In fact, using convexity of $f$, this is convex and symmetric in $t$. QED