Extending linear functionals dominated by a convex (or sublinear) function

Let $X$ be a vector space. A subset $C$ of $X$ is called a cone provided that $x+y \in C$ and $\lambda x \in C$ whenever $x, y$ belong to $C$ and $\lambda > 0$. Define a partial order in $X$ by defining $x \prec y$ to mean $y-x \in C$. A linear functional $f$ on $X$ is said to be positive (with respect to the cone $C$) provided that $f \geq 0$ on $C$. Let $Y$ be any subspace of $X$ with the property that for each $x \in X$ there is a $y \in Y$ with $x \prec y$. Suppose $f$ is a linear functional on $Y$ that is positive with respect to the cone $C \cap Y$. Show that $f$ may be extended to a linear functional on $X$ that is positive with respect to $C$.

Let $f$ be the convex function, taking values in $\mathbb {R} \cup \left\{ \infty \right\}$, defined by

Let $T_{0}$ denote the linear function defined on the $x$-axis by $T_{0}(x, 0): = 0$ for all $x \in \mathbb {R}$. Show that $T_{0}$ has no extension to a linear functional on $\mathbb {R}^{2}$ for which $T(x, y) \leq f(x, y)$ everywhere, even though $T_{0} \leq f$ along the $x$-axis.