Hahn-Banach theorem

Every bounded linear functional on a linear subspace of a real normed space extends to the whole space without increasing its norm.

A bounded complex-linear linear functional on a complex linear subspace of a normed space extends to the whole space with the same operator norm. To deduce this from the real Hahn-Banach theorem, extend the real part $u$ under the bound $\Vert \ell \Vert \Vert x\Vert$ on the underlying real vector space, then put $L(x)=u(x)-iu(ix)$. This is complex-linear and agrees with $\ell$ on its original domain. Given $x$, choose a complex number $a$ of modulus one with $aL(x)=|L(x)|$. Then $|L(x)|=\mathrm{Re}L(ax)=u(ax)\le \Vert \ell \Vert \Vert x\Vert$. Restriction gives the reverse operator norm inequality.

A linear functional $f$ dominated by a finite real convex function $\varphi$ on a vector subspace $W$ extends with the same domination to the whole real vector space. To add $v\notin W$, choose $g(v)$ between the supremum of $(f(w)-\varphi (w-tv))/t$ and the infimum of $(\varphi (w+sv)-f(w))/s$, over $w\in W$ and $s,t>0$. Convexity, applied to $(s(w_1-tv)+t(w_2+sv))/(s+t)$, makes every lower candidate at most every upper candidate. The candidates at $w=0,s=t=1$ bound the interval by finite numbers. Chain unions preserve domination, so Zorn's lemma gives a full extension. No homogeneity of $\varphi$ is assumed.

If $p$ is a sublinear functional, $q$ a superlinear functional and $q\le p$ on a real vector space, there exists a linear functional $U$ between them. More generally, a linear functional $S$ on a subspace extends with this domination whenever $S(y)\le p(x+y)-q(x)$ for all ambient $x$ and subspace $y$. In a one-dimensional extension the admissible values for a sandwich extension form a nonempty real interval. Positive homogeneity checks the extension on both signs of the new coordinate. Chain unions preserve linearity and the mixed domination, so Zorn's lemma gives a maximal extension, whose domain must be the whole space. Finally $x=0$ in the mixed inequality gives the upper bound; using the pair $(x,-x)$ gives the lower bound. Conversely, any linear $U$ between $q,p$ obeys the mixed inequality by $U(y)=U(x+y)-U(x)$.

Suppose $S$ satisfies the mixed domination on a subspace $Y$, and $z\notin Y$. Apply that domination to $y+y^{\prime }$ and $x+x^{\prime }$, split the $p$ argument as $(x+y+z)+(x^{\prime }+y^{\prime }-z)$, and use superadditivity of $q$. This proves every lower candidate in the displayed interval is at most every upper candidate. The zero-vector candidates bound the lower supremum below by $-p(-z)$ and the upper infimum above by $p(z)$, so both endpoints are finite. Choosing any intervening value $a$ permits $U(y+tz)=S(y)+ta$. For $t>0$, scale the upper comparison by $t$; for $t<0$, scale the lower comparison by $-t$. Both give the required mixed domination, proving the interval characterizes the permitted extension values.

In a real normed vector space with a specified countable dense subset, a bounded linear functional on any vector subspace admits a norm-preserving extension without the axiom of choice. Add the dense vectors one at a time using the supremum endpoint in the one-dimensional dominated extension of a real linear functional. This makes the recursion deterministic. Extend from the resulting dense subspace by continuity, using the least index in the dense list which approximates a given vector to within $2^{-n}$.

If $Y$ is a closed linear subspace of a Hausdorff locally convex space and $x_0\notin Y$, choose a continuous seminorm $q$ with $q(x_0-y)\ge 1$ for all $y\in Y$, using a basic neighborhood disjoint from $Y$. The linear functional $g(y+t{x}_0)=t$ is bounded by $q$. The Hahn-Banach theorem extends it to a continuous linear functional with the displayed values. This is the linear-subspace form of separation, sharper than merely distinguishing the two sets.

A real linear functional $g$ on a vector subspace, dominated by a sublinear function $p$, can be extended across one new vector $v$ by setting $\tilde{g}(m+tv)=g(m)+tc$. The displayed lower and upper bounds are compatible because $g(m+n)\le p(m-v)+p(n+v)$. A value between them preserves domination for both signs of $t$. Chain unions and Zorn's lemma then yield the full Hahn-Banach theorem.

A real separable Banach space has a sequence $({\phi }_n)$ in its dual unit ball with $\Vert x\Vert ={\sup }_n{\phi }_n(x)$ for every $x$. Choose a dense sequence $(u_n)$ in the unit sphere and apply the Hahn-Banach theorem to obtain ${\phi }_n(u_n)=1$. In the complex case use ${\sup }_n\mathrm{Re}{\phi }_n(x)$.

For a closed subspace $Y$ of a normed space $X$,This is the dual norm formula in the quotient $X/Y$.

For a normed space $X$, the map $J:X\to X^{\prime \prime }$ defined byis a linear isometry. The easy inequality is $\Vert Jx\Vert \le \Vert x\Vert$; Hahn--Banach extends the norm-one functional on $\mathrm{span}\{x\}$ that takes $x$ to $\Vert x\Vert$, proving the reverse inequality.

Point evaluation at zero is bounded on $C([0,1])$ with the essential-supremum norm, and Hahn--Banach extends it to a bounded functional on $L^{\infty }([0,1])$. No $g\in L^1$ represents this extension: continuous functions supported in shrinking neighbourhoods of zero have value one at zero while their integrals against $g$ tend to zero. Hence $(L^{\infty }{)}^{\prime }$ is strictly larger than $L^1$.

A point outside a nonempty closed convex subset of a normed space can be strictly separated from it by the real part of a bounded linear functional.

If a convex set $E$ in a real normed vector space contains $B(0,ϵ)$, every $x\notin E$ admits a continuous linear functional $T$ with $Tx\ge 1\ge Te$ for all $e\in E$, even if $E$ is not closed. Its Minkowski functional $p(v)=\inf \{t>0:v\in tE\}$ is sublinear functional, satisfies $p(v)\le \Vert v\Vert /ϵ$, $p(e)\le 1$, and $p(x)\ge 1$. Extend $tx\mapsto tp(x)$ by the dominated Hahn-Banach theorem. The resulting domination $T\le p$ applied to $v$ and $-v$ makes $T$ continuous and supplies the separation inequalities.

For a nonempty radially open convex set $C$ disjoint from a vector subspace $U$, there is a nonzero linear functional $L$ whose kernel of a linear map contains $U$ and avoids $C$. Choose $a\in C$ and the Minkowski functional $p$ of $C-a$. On $U+\mathbb{R}a$, set $g(u+ta)=-t$. Disjointness implies $p(u-a)\ge 1$, which gives $g\le p$ for negative $t$; nonnegativity of $p$ gives it for nonnegative $t$. The algebraic Hahn-Banach theorem extends $g$ to $L\le p$. For $x\in C$, $L(x)+1=L(x-a)<1$, so $L(x)<0$. Also $L(a)=-1$ and $L{|}_U=0$.