I have encountered this result in a lecture note. Could you have a check if my proof is fine? Thank you so much for your help!
Let $L$ be a subspace of a vector space $X$ and $A$ an absorbing subset of $X$. Let $\ell: L \to \mathbb R$ be linear and $m \in \mathbb R$ such that $\ell \le m$ on $A \cap L$. Then there is a linear extension $\hat \ell: X \to \mathbb R$ of $\ell$ such that $\hat \ell \le m$ on $A$. If, moreover, $X$ is a t.v.s. and $A$ a neighborhood of $0$, then $\hat \ell$ is continuous.
My attempt: Because $A$ is absorbing, so $0 \in A$ and thus $0 \in A \cap L$. So $0 = \ell(0) \le m$. Let $p:X \to \mathbb R$ be the Minkowski functional of $A$, i.e., $p(x) := \inf \{t>0 \mid x \in tA\}$. Then $p$ is sublinear and $p \le 1$ on $A$. We define $p':X \to \mathbb R$ by $p' (x) := mp(x)$. Then $p'$ is sublinear and $p' \le m$ on $A$.
Let $x \in L$. Then for each $n \in \mathbb N^*$, there is $a_n \in A$ such that $x = (p(x)+1/n)a_n$. Then $a_n \in L$ and $\ell(x) = (p(x)+1/n)\ell(a_n) \le m(p(x)+1/n)$. Passing to the limit, we get $\ell (x) \le mp(x) = p'(x)$ for all $x \in L$. The result then follows by applying Hahn-Banach theorem. $\blacksquare$
Remark: If we have an additional assumption that $A$ is convex, then $\{x \in X \mid p(x)<1\} \subset A$.
As @copper.hat pointed out in a comment, for $p$ to be sub-additive, we need $A$ to be convex. This can be seen as follows. Let $x,y \in X$ and $x \in t_1 A$ and $y \in t_2 A$. Then $$ x+y \in t_1 A + t_2 A = (t_1+t_2) \left [ \frac{t_1}{t_1+t_2} A + \frac{t_2}{t_1+t_2} A \right ]. $$
If $A$ is convex, then $\frac{t_1}{t_1+t_2} A + \frac{t_2}{t_1+t_2} A \subset A$ and thus $x+y \in (t_1+t_2) A$.