Note that in the case of the reals or complex we have the norm as the absolute value. Also we say that a linear functional is real if $F=\mathbb{R}$ and say that it is complex if $F=\mathbb{C}$

The above equates to saying that there exists some $c\in {\mathbb{R}}^{+}$ such that for all $x\in dom\left(f\right),\left|f\left(x\right)\right|\le c\Vert x\Vert$

What's interesting is that the above holds whether or not $X$ is a banach space.

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5.6 The Hahn-Banach Theorem.

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5.7 The Complex Hahn-Banach Theorem. Let $𝒳$ be a complex vector space, $p$ a seminorm on $X,ℳ$ a subspace of $𝒳$, and $f$ a complex linear functional on $ℳ$ such that $|f(x)|\le p(x)$ for $x\in ℳ$. Then there exists a complex linear functional $F$ on $X$ such that $|F(x)|\le p(x)$ for all $x\in 𝒳$ and $F\mid ℳ=f$. Proof. Let $u=Ref$. By Theorem 5.6 there is a real linear extension $U$ of $u$ to $X$ such that $|U(x)|\le p(x)$ for all $x\in X$. Let $F(x)=U(x)-iU(ix)$ as in Proposition 5.5. Then $F$ is a complex linear extension of $f$, and as in the proof of Proposition 5.5, if $\alpha =\overline{sgnF(x)}$ we have $|F(x)|=\alpha F(x)=F(\alpha x)=U(\alpha x)\le p(\alpha x)=p(x)$.

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5.8 Theorem. Let $x$ be a normed vector space. a. If $ℳ$ is a closed subspace of $\dot{X}$ and $x\in \dot{X}\backslash ℳ$, there exists $f\in X$ such that $f(x)\ne 0$ and $f\mid ℳ=0$. In fact, if $\delta ={inf}_{y\in ℳ}\Vert x-y\Vert .f$ can be taken to satisfy $\Vert f\Vert =1$ and $f(x)=\delta$. b. If $x\ne 0\in X$, there exists $f\in X^{*}$ such that $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$. c. The bounded linear functionals on $X$ separate points. d. If $x\in X$, define $\hat{x}:X^{+}\to ℂ$ by $\hat{x}(f)=f(x)$. Then the map $x\mapsto \hat{x}$ is a linear isometry from $X$ into $X^{**}$ (the dual of $X^{*}$ ). Proof. To prove (a), define $f$ on $ℳ+ℂx$ by $f(y+\lambda x)=\lambda \delta (y\in ℳ,\lambda \in ℂ)$. Then $f(x)=\delta ,f\mid ℳ=0$, and for $\lambda \ne 0,|f(y+\lambda x)|=|\lambda |\delta \le |\lambda |\Vert {\lambda }^{-1}y+x||=$ $\Vert y+\lambda x\Vert$. Thus the Hahn-Banach theorem can be applied, with $p(x)=\Vert x\Vert$ and $ℳ$ replaced by $M+ℂx$. (b) is the spectal case of (a) with $M=\{0\}$, and (c) follows immediately: if $x\ne y$, there exists $f\in X^{+}$with $f(x-y)\ne 0$, i.e., $f(x)\ne f(y)$. As for (d), obviously $\hat{x}$ is a linear functional on $X^{*}$ and the map $x\mapsto \hat{x}$ is linear. Moroever, $|\hat{x}(f)|=|f(x)|\le \Vert f\Vert \Vert x\Vert$, so $\Vert \hat{x}\Vert \le \Vert x\Vert$. On the other hand, (b) implies that $\Vert \hat{2}\Vert \ge \Vert x\Vert$. With notation as in Theorem 5.8d, let $\hat{X}=\{\hat{x}:x\in X\}$. Since $x^{*}$ is always complete, the closure $\bar{\hat{X}}$ of $\hat{X}$ in $X^{**}$ is a Banach space, and the map $x\mapsto \hat{x}$ embeds $x$ into $\overline{X}$ as a dense subspace. $\overline{x}$ is called the completion of $X$. In particular, if $X$ is itself a Banach space then $\overline{x}=\hat{x}$. If $x$ is finite-dimensional, then of course $\hat{x}=x^{**}$, since these spaces have the same dimension. For infinite-dimensional Banach spaces it may or may not happen that $\hat{x}=x·*$; if it does, $x$ is called reflexive. The examples of Banach spaces we have examined so far are not reflexive except in trivial cases where they turn out to be finite-dimensional. We shall prove some cases of this assertion and present examples of reflexive Banach spaces in later sections. Usually we shall identify $\hat{x}$ with $x$ and thus regard $x^{**}$ as a superspace of $x$; reflexivity then means that $x^{**}=x$

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We already know (Background Information): (Folland)Theorem 5.8 - Let $𝒳$ be a normed vector space. a.) If $M$ is a closed subspace of $𝒳$ and $x\in 𝒳\backslash M$, there exists $f\in {𝒳}^{*}$ such that $f(x)\ne 0$ and $f(M)=\{0\}$. In fact, if $\delta ={inf}_{y\in M}\Vert x-y\Vert ,f$ can be taken to satisfy $\Vert f\Vert =1$ and $f(x)=\delta$. Question: Let $X$ be a normed vector space. a.) If $M$ is a closed subspace and $x\in X\backslash M$ then $M+ℂx$ is closed. (Use Theorem 5.8a.)). b.) Every finite-dimensional subspace of $X$ is closed. Proof a.) - Let $M$ be a proper closed subspace of $X$ and let $x\in X\backslash M$. There exists $f\in X^{*}$ such that $f(x)\ne 0$ and $f(M)=\{0\}$. Let ${\{u_n+a_{n}x\}}_1^{\infty }$ be a sequence in $M+Kx$ that converges to $y\in X$. Then $${\displaystyle f(y)=f\left({lim}_{n\to \infty }(u_n+a_{n}x)\right)={lim}_{n\to \infty }f(u_n+a_{n}x)={lim}_{n\to \infty }{a}_{n}f(x)}$$ since $f$ is continuous, so ${\left\{a_n\right\}}_1^{\infty }$ converges to $a:=f(y)/f(x)$, which implies that ${\left\{a_{n}x\right\}}_1^{\infty }$ converges to $ax$. Therefore $${\displaystyle {\left\{u_n\right\}}_1^{\infty }={\{(u_n+a_{n}x)-a_{n}x\}}_1^{\infty }\to (y-ax)}$$ which lies in $M$ because $M$ is closed. It follows that $y\in M+Kx$, which shows that $M+Kx$ is closed. Proof b.) - Let $Y$ be a finite dimensional subspace of $X$. Let $\{e_1,\dots ,e_n\}$ be a basis of $Y$. Now, note that $M=\{0\}$ is a closed subspace of $X$. So, by item a.), $M+C{e}_1$ is a closed subspace of $X$. Again, by item a.) we have that $M+ℂe_1+\mathbb{C}\mathbb{e}{e}_2$ is a closed subspace of $X$. Repeating the argument enough times, we have that $M+\mathbb{C}\mathbb{e}{e}_1+\dots +ℂe_n$ is a closed subspace of $X$. But $M+\mathbb{C}\mathbb{e}{e}_1+\dots +ℂe_n=Y$. So we have proved that $Y$ is a closed subspace of $X$.

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