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We first prove the algebraic extension step needed for Hahn-Banach separation theorem. Suppose $p$ is a finite sublinear functional on a real vector space and $g$ is a linear functional on a vector subspace $W$, with $g\le p$ there. For $v\notin W$, choose a real number $c$ betweenThis interval is nonempty: for any $w,z\in W$,Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\tilde{g}(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\tilde{g}(w+tv)\le p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated linear functional extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so Zorn's lemma supplies a maximal extension. The one-direction argument shows its domain is the whole vector space. This proves the required algebraic Hahn-Banach theorem, rather than assuming the separation result.

Choose $a\in A$, and put $C=A-a$. This is a radially open convex set containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its Minkowski functionalis finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in (s+t)C$, proving subadditivity by taking infima. Moreover,Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+ϵ)x\in C$ for some $ϵ>0$, giving $p(x)<1$.

Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a linear functional on $U+\mathbb{R}a$ byIt is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=sp(u/s-a)\ge s=g(u+ta)$. For $t\ge 0$, domination follows from $g(u+ta)\le 0\le p(u+ta)$. The extension step therefore gives $L:V\to \mathbb{R}$ with $L\le p$, $L{|}_U=0$ and $L(a)=-1$. If $x\in A$, thenso $L(x)<0$. ConsequentlyBecause $L(a)\ne 0$, this kernel of a linear map has codimension one and is a hyperplane. This establishes the geometric Hahn-Banach theorem for radially open sets without assuming a norm or a topology on $V$.

For the convex function, use its strict epigraph in $V\times \mathbb{R}$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite convex function is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta -ts$ is continuous, and a strict negative value persists near $t=0$.

Apply the separation theorem with the zero subspace. A nonzero separating linear functional has form $L(x,\beta )=g(x)+c\beta$. Its values on the convex epigraph cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the epigraph. At $x=0$ and $\beta >0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\ge 0$. ThereforeThe resulting linear functional need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb{R}$.

Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ givesIf $x_n\in \ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The kernel of a linear map is therefore closed.

Finally, a linear functional with closed kernel on a normed vector space is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. For $l(x)\ne 0$, $x/l(x)-v\in M$, so $d\le \Vert x\Vert /|l(x)|$. The same resulting bound holds when $l(x)=0$:This proves continuity and explains exactly how closedness of the kernel is used.