Proofs of Hahn-Banach

Kreyszig Functional Analysis

Here are Kreyszig’s proofs of the Hahn-Banach theorems, rewritten for clarity and motive

Hahn-Banach over Real Vector Spaces Let $X$ be a real vector space and $p$ be a sublinear functional on $X$. Furthermore, let $f$ be a linear functional which is defined on a subspace $Z$ of $X$ and satisfies

$$\forall x\in Z:f(x)\le p(x)$$

Then $f$ has a linear extension $\tilde{f}$ from $Z$ to $X$ satisfying

$$\forall x\in X:\tilde{f}(x)\le p(x)$$

that is, $\tilde{f}$ is a linear functional on $X$, is majorized by $p$ on $X$, and $f=\tilde{f}$ in $Z$.

Proof Note that a linear functional, by definition, must have a vector space as its domain. Thus, if we imagine a “linear extension of linear functional $f$,” this linear extension must have some subspace of $X$ as its domain.

Let $\mathcal{E}$ be the set of linear extensions of $f$ that are majorized by $p$ on their domains. $\mathcal{E}\ne \varnothing$ as $f\in \mathcal{E}$. $\mathcal{E}$ can be made a partially ordered set by defining $\le$ as $h_1\le h_2$ iff $h_2$ is a linear extension of $h_1$. We aim to use Zorn’s Lemma on $\mathcal{E}$.

Take an arbitrary chain $C=\{c_i\}\subseteq \mathcal{E}$ (this chain need not be countable). Define the functional $h$ defined on the subspace ${\bigcup }_i\mathcal{D}(c_i)$ (note the union of nested subspaces is itself a vector subspace) such that for any $x\in \mathcal{D}(h)$, $h(x)=c_i(x)$ for any $c_i$ such that $x\in \mathcal{D}(c_i)$. We must first show this construction is well-defined; obviously, as $\mathcal{D}(h)={\bigcup }_i\mathcal{D}(c_i)$, every $x\in \mathcal{D}(h)$ has some prescribed value $h(x)$. It suffices to show these prescribed values are consistent if an $x$ exists in the domain of multiple $c_i$s. FTSOC, assume not. I.e. there exists $c_1,c_2\in C$ such that $c_1(x)\ne c_2(x)$. As $C$ is a chain, $\le$ defines a total order, and thus either $c_1\le c_2$ or $c_2\le c_1$. WLOG assume $c_1\le c_2$; as $c_2$ is an extension of $c_1$, $c_2(y)=c_1(y)$ for all $y\in \mathcal{D}(c_1)$. However, $x\in \mathcal{D}(c_1)$ and $c_1(x)\ne c_2(x)$, creating an immediate contradiction.

Clearly $h$ is majorized by $p$ on its domain, as each element $c_i$ is majorized by $p$. Furthermore we can clearly see $h$ is an extension of $c_i$ for any $c_i\in C$, and thus to prove $h$ is an upper bound of $C$ it suffices to show $h$ is linear. For any $x_1,x_2\in \mathcal{D}(h)$, we can take $c_1$, $c_2$ such that $x_1\in \mathcal{D}(c_1)$ and $x_2\in \mathcal{D}(c_2)$. As $\le$ defines a total order on $C$, either $c_1\le c_2$ or $c_2\le c_1$; assume the former. Then, $x_1,x_2\in \mathcal{D}(c_2)$ and more specifically $Span\{x_1,x_2\}\subset \mathcal{D}(c_2)$. For any scalars $\alpha ,\beta \in \mathbb{R}$, we can conclude

$$h(\alpha x_1+\beta x_2)=c_2(\alpha x_1+\beta x_2)=\alpha c_2(x_1)+\beta c_2(x_2)=\alpha h(x_1)+\beta h(x_2)$$

Thus, $h$ is linear, and constitutes an upper bound of the chain $C$.

As we have shown any arbitrary chain has an upper bound, we can use Zorn’s lemma to take some maximal element $\tilde{f}$ in $\mathcal{E}$. It suffices to show this maximal element $\tilde{f}$ satisfies $\mathcal{D}(\tilde{f})=X$, as then it becomes our desired extension of $f$.

FTSOC assume $\mathcal{D}(\tilde{f})\ne X$, and that there exists some $y_1\notin \mathcal{D}(\tilde{f})$. We will construct a linear extension $g$ of $\tilde{f}$ such that $\mathcal{D}(g)=Span\{\mathcal{D}(\tilde{f}),y_1\}$ and such that $g$ is majorized by $p$. Then, $g\in \mathcal{E}$ and $\tilde{f}\le g$ while $g\ne \tilde{f}$, contradicting the maximality of $\tilde{f}$. Thus, $\mathcal{D}(\tilde{f})$ must be the entire space $X$.

How do we go about constructing such a $g$? Note that any element $z\in \mathcal{D}(g)$ has a unique representation $z=x+\alpha y_1$ where $\alpha \in \mathbb{R}$ and $x\in \mathcal{D}(\tilde{f})$. By linearity, if we take $c=g(y_1)$ then $g(z)=g(x)+\alpha c=\tilde{f}(x)+\alpha c$, and thus $c\in \mathbb{R}$ fully characterizes the linear extension $g$. It, therefore, suffices to show the existence of an appropriate value of $c=g(y_1)$ that ensures $g$ is majorized by $p$.

We refine the problem condition further. It suffices to show there exists some $c\in \mathbb{R}$ such that, for any $x\in \mathcal{D}(\tilde{f})$ and $\alpha \in \mathbb{R}$,

$$g(x+\alpha y_1)=\tilde{f}(x)+\alpha c\le p(x+\alpha y_1)$$

We hope to use the positive-homogeneity of $p$, and thus split into cases based on the sign of $\alpha$. Noting that the $\alpha =0$ case is trivially true, this is equivalent to showing there is some $c\in \mathbb{R}$ such that for any $\alpha >0$ and $x\in \mathcal{D}(\tilde{f})$,

$$\alpha \tilde{f}(x/\alpha )+\alpha c\le \alpha p((x/\alpha )+y_1)⟺\forall x\in \mathcal{D}(\tilde{f}):\tilde{f}(x)+c\le p(x+y_1)$$

and such that for any $\alpha <0$,

$$\alpha \tilde{f}(x/\alpha )+\alpha c\le -\alpha p(-(x/\alpha )-y_1)⟺\forall x\in \mathcal{D}(\tilde{f}):\tilde{f}(x)+c\ge -p(-x-y_1)$$

Let $A={\sup }_{x\in \mathcal{D}(\tilde{f})}(-p(-x-y_1)-\tilde{f}(x))$ and $B={\inf }_{x\in \mathcal{D}(\tilde{f})}(p(x+y_1)-\tilde{f}(x))$. It suffices to show there is some $c$ such that $A\le c\le B$; i.e. it suffices to show $A\le B$. This must clearly be true as for any $x\in \mathcal{D}(\tilde{f})$,

$$p(x+y_1)+p(-x-y_1)\ge p(0)=0\Rightarrow p(x+y_1)-\tilde{f}(x)\ge -p(-x-y_1)-\tilde{f}(x)$$

Generalized Hahn-Banach over Real/Complex Vector Spaces Let $X$ be a vector space over the real or complex scalar field and $p$ be a real-valued functional on $X$ that is subadditive ($p(x+y)\le p(x)+p(y)$) and for any scalar $\alpha$ satisfies $p(\alpha x)=|\alpha |p(x)$. Let $f$ be a linear functional defined on a linear subspace $Z$ of $X$ that satisfies

$$\forall x\in Z:|f(x)|\le p(x)$$

Then $f$ has a linear extension $\tilde{f}$ from $Z$ to $X$ such that

$$\forall x\in X:|\tilde{f}(x)|\le p(x)$$

Proof The case where $X$ is a real vector space is an immediate (weaker) corollary of the real Hahn-Banach theorem. If $X$ is a real vector space, then $p$ is a sublinear functional on $X$, and $f$ satisfies $f(x)\le p(x)$ for all $x\in Z$. By Hahn-Banach, there exists some linear extension $\tilde{f}$ on $X$ such that $\tilde{f}(x)\le p(x)$ for all $x\in X$. This immediately implies $|\tilde{f}(x)|\le p(x)$ as $-\tilde{f}(x)=\tilde{f}(-x)\le p(-x)=p(x)$

We now focus on the case where $X$ is a complex vector space. Note that for any arbitrary linear functional $g$ over a complex vector space, we can represent $g$ as $g=g_1+i{g}_2$ where $g_1$ and $g_2$ are real-valued (non-linear) functionals. By linearity, $g_1$ and $g_2$ must satisfy $i{g}_1(x)-g_2(x)=ig(x)=g(ix)=g_1(ix)+i{g}_2(ix)\Rightarrow g_2(x)=-g_1(ix)$. Thus, $g$ can be represented as $g=g_1-i{g}_1(ix)$ for some real-valued functional $g_1$.

We use this decomposition to write $f$ as $f=f_1-i{f}_1(ix)$ over $Z$. Consider the vector fields $Z_r$ and $X_r$ generated by limiting scalar multiplication in the complex fields $Z$ and $X$ to the real numbers (this immediately defines a new set of real vector fields containing the same vectors). The functional $f_1$ is linear in $Z_r$; as $|f_1(x)|\le |f(x)|\le p(x)$ in $Z_r$, there is some linear extension $\widetilde{f_1}$ from $Z_r$ to $X_r$ of $f_1$ such that $|\widetilde{f_1}|\le p(x)$ in $X_r$, by the real case proven above. We prove the corresponding extension $\tilde{f}(x)=\widetilde{f_1}(x)-i\widetilde{f_1}(ix)$ of $f$ from $Z$ to $X$ is our desired linear extension.

First, we show $\tilde{f}$ is linear on $X$. As we already know $\tilde{f}$ is linear on $X_r$, it suffices to show linearity under complex scalar multiplication. For any $a+ib\in \mathbb{C}$,

$$\tilde{f}((a+ib)x)=\widetilde{f_1}((a+ib)x)+-i\widetilde{f_1}(i(a+ib)x)$$ $$=a{\tilde{f}}_1(x)+b{\tilde{f}}_1(ix)-ai{\tilde{f}}_1(ix)+bi{\tilde{f}}_1(x)=(a+ib)({\tilde{f}}_1(x)-i{\tilde{f}}_1(ix))=(a+ib)\tilde{f}(x)$$

It now suffices to show that $|\tilde{f}|$ is majorized by $p$ in $X$. As $p(x)\ge 0$ for any $x\in X$, for points $x$ where $\tilde{f}(x)=0$ the inequality $|\tilde{f}(x)|\le p(x)$ trivially holds. Otherwise, one can decompose $\tilde{f}(x)$ as $\tilde{f}(x)=|\tilde{f}(x)|e^{i\theta }$, and thus $|\tilde{f}(x)|=\tilde{f}(x)e^{-i\theta }=\tilde{f}(x{e}^{-i\theta })$. We finish by noticing that

$$|\tilde{f}(x)|=\tilde{f}(x{e}^{-i\theta })={\tilde{f}}_1(x{e}^{-i\theta })\le p(x{e}^{-i\theta })=p(x)|e^{-i\theta }|=p(x)$$