Let \((X, \|\cdot\|_{X})\) and \((Y, \|\cdot\|_{Y})\) be two normed spaces. A map \(T:X \to Y\) between two normed vector spaces is linear if it satisfies
(Homogenity) \[T(\lambda x)=\lambda T(x), \quad \text{ for } \quad x \in X, \ \lambda \in \mathbb{C}.\]
(Additivity) \[T(x_1+x_2)=T(x_1)+T(x_2), \quad \text{ for } \quad x_1,x_2 \in X.\]
Moreover, a linear map \(T:X \to Y\) between two normed vector spaces is bounded if there exists \(C \geq 0\) such that \[\|Tx\|_Y \leq C\|x\|_X \quad \text{ for all } \quad x \in X.\]
Let \((X, \|\cdot\|_{X})\) and \((Y, \|\cdot\|_{Y})\) be normed vector spaces and \(T:X \to Y\) be a linear map. Then the following are equivalent:
\(T\) is continuous at a point,
\(T\) is continuous at every point,
\(T\) is bounded.
Proof. The implication (ii)\(\implies\)(i) is obvious, so it suffices to establish the implications (iii)\(\implies\)(ii) and (i)\(\implies\)(iii).
Proof (iii)\(\implies\)(ii). Let \(x \in X\) and \(x_n \ _{\overrightarrow{n\to \infty}} \ x\) in \(X\), then by linearity and boundedness we obtain \[\begin{aligned} \|Tx_n-Tx\|_Y=\|T(x_n-x)\|_Y\le C\|x_n-x\|_X \ _{\overrightarrow{n\to \infty}} \ 0, \end{aligned}\] so \(T\) is continuous at \(x \in X\). $$\tag*{$\blacksquare$}$$
Proof (i)\(\implies\)(iii). We first show that \(T\) is continuous at \(0\). Let \(x_n \ _{\overrightarrow{n\to \infty}} \ 0\) in \(X\), then \(u_n=x_n+x_0 \ _{\overrightarrow{n\to \infty}} \ x_0\) implying \[Tx_n+Tx_0=T(x_n+x_0)=Tu_n \ _{\overrightarrow{n\to \infty}} \ Tx_0,\] so \[Tx_n \ _{\overrightarrow{n\to \infty}} \ 0=T(0).\]
Suppose for a contradiction that (iii) is not true. This means that for every \(n \in \mathbb{N}\) there is \(x_n \in X\) such that \[\|Tx_n\| > n\|x_n\|.\]
In particular, \(x_n \neq 0\). Define \({\color{purple}u_n=\frac{1}{\sqrt{n}}\frac{x_n}{\|x_n\|}}\), then \(\|u_n\|=\frac{1}{\sqrt{n}}\) and \(u_n \ _{\overrightarrow{n\to \infty}} \ 0\). Thus \[\|Tu_n\|=\frac{1}{\sqrt{n}}\frac{\|Tx_n\|}{\|x_n\|}>\frac{n}{\sqrt{n}}=\sqrt{n} \ _{\overrightarrow{n\to \infty}} \ \infty,\] which is impossible, since \(\|Tu_n\| \ _{\overrightarrow{n\to \infty}} \ 0\).$$\tag*{$\blacksquare$}$$
If \((X, \|\cdot\|_{X})\) and \((Y, \|\cdot\|_{Y})\) are normd vector spaces, we denote the space of all bounded linear maps from \(X\) to \(Y\) by \(L(X,Y)\). It is easily verified that \(L(X,Y)\) is a vector space and that the function \[L(X,Y)\ni T \mapsto \|T\|_{X\to Y}\in [0, \infty)\] defined by \[\begin{aligned} \|T\|_{X\to Y}&={\color{red}\sup\{\|Tx\|_Y: \|x\|=1\}}\\ &={\color{blue}\sup\bigg\{\frac{\|Tx\|_Y}{\|x\|_X}: x \neq 0\bigg\}} \\&={\color{purple}\inf\{C \geq 0: \|Tx\|_Y \leq C\|x\|_X \ \text{ for all } \ x \in X\} } \end{aligned}\] it is a norm on \(L(X,Y)\) called the operator norm.
Proposition. Let \((X, \|\cdot\|_{X})\) and \((Y, \|\cdot\|_{Y})\) be normd vector spaces. If \(Y\) is complete so is \(L(X,Y)\).
Proof. Let \((T_n)_{n \in \mathbb{N}}\subseteq L(X, Y)\) be a Cauchy sequence, then \[\|T_n - T_m\|_{X\to Y} \ _{\overrightarrow{m, n\to \infty}} \ 0.\] Thus for every \(x \in X\) we have \[\|T_nx-T_mx\|_Y \leq \|T_n-T_m\|_{X\to Y}\|x\|_X \ _{\overrightarrow{m, n\to \infty}} \ 0,\] hence \((Tx_n)_{n \in \mathbb{N}}\) is Cauchy in \(Y\). So it converges since \(Y\) is complete. Let \[Tx=\lim_{n \to \infty}Tx_n\in Y.\] We have to show that \(T\) is bounded and \(\|T-T_n\|_{X\to Y} \ _{\overrightarrow{n\to \infty}} \ 0\).
Fix \(\varepsilon>0\) then there is \(N_{\varepsilon} \in \mathbb{N}\) such that for every \(n,m \geq N_{\varepsilon}\) we have \(\|T_n-T_m\|_{X\to Y}<\varepsilon\). Let \(n \geq N_\varepsilon\), then \[\|(T_n-T)x\|_Y=\|T_nx -Tx\|_Y=\lim_{m \to \infty}\|T_nx-T_mx\|_Y.\]
On the other hand, if \(m \geq N_{\varepsilon}\), then \[\|T_nx-T_mx\|_Y=\|(T_n-T_m)x\|_Y \leq \|T_n-T_m\|_{X\to Y}\|x\|_X \leq \varepsilon\|x\|_X.\]
Hence, if \(n \geq N_{\varepsilon}\), then \[\|(T_n-T)x\|_Y \leq \varepsilon \|x\|_X, \quad \text{ which gives} \quad \|T_n-T\|_{X\to Y} \leq \varepsilon.\]
In particular, \(T_{N_\varepsilon}-T\) and \(T_{N_\varepsilon}\) are bounded thus \[\|T\|_{X\to Y} \leq \|T-T_{N_{\varepsilon}}\|_{X\to Y}+\|T_{N_\varepsilon}\|_{X\to Y} \leq \varepsilon+\|T_{N_\varepsilon}\|_{X\to Y}.\]
Hence we proved that \(T \in L(X,Y)\) and \(\|T_n-T\|_{X\to Y} \ _{\overrightarrow{n\to \infty}} \ 0\). $$\tag*{$\blacksquare$}$$
Let \((X, \|\cdot\|_{X})\) be a vector space over \(\mathbb K\) where \(\mathbb K=\mathbb{R}\) or \(\mathbb K=\mathbb{C}\). A linear map from \(X\) to \(\mathbb K\) is called a linear functional on \(X\).
If \((X, \|\cdot\|_{X})\) is a normed space over \(\mathbb K\), the space \(L(X,\mathbb K)\) of bounded linear functionals on \(X\) is called the dual space of \(X\) and it is denoted by \[X^*=L(X,\mathbb K).\]
Let \((X, \|\cdot\|_{X})\) be a vector space over \(\mathbb K\), where \(\mathbb K=\mathbb{R}\) or \(\mathbb K=\mathbb{C}\). Then the dual space \((X^*, \|\cdot\|_{X^*})\) is a Banach space with the operator norm, i.e. \[\|f\|_{X^*}=\|f\|_{X\to\mathbb K}\quad \text{ for every } \quad f\in X^*.\]
Example 1. Let \(K\) be a compact Hausdorff topological space. Let \((K, {\rm Bor}(K), \mu)\) be a Borel measure space. Let \(X=C(K)\) be the space of all continuous functions on \(K\). Then \[\Lambda(f)=\int_K f(x)d\mu(x), \quad \text{ for } \quad f\in X\] defies a bounded linear functional on \(X\).
Example 2. Let \((X,\mathcal M,\mu)\) be a \(\sigma\)-finite measure space, and let \(g \in L^q(X)\), with \(1 \leq q \leq \infty\). Then \[\Lambda(f)=\int_X f(x)\overline{g(x)}d\mu(x)\] defines a bounded linear functional on \(L^p(X)\), where \(\frac{1}{p}+\frac{1}{q}=1\).
Suppose \(X\) is a vector space. A map \(p : X \to \mathbb R\) is called a sublinear functional if it satisfies
(positive homogeneity) \(p(\alpha x) = \alpha p(x)\) for all \(\alpha\ge 0\) and \(x \in X\),
(subadditivity) \(p(x + y) \le p(x) + p(y)\) for all \(x, y \in X\).
Condition (a) implies that \(p(0) = 0\). Condition (b) is also called the triangle inequality.
Examples.
Any linear functional is a sublinear functional.
Every seminorm is a sublinear functional.
Proposition. A sublinear functional \(p\) on a real vector space \(X\) is linear if and only if \(p(-x) = -p(x)\) for all \(x \in X\).
Proof. Certainly, if \(p\) is linear, then the conclusion follows. Suppose now that \(p(-x) = -p(x)\) for all \(x \in X\). Then, by the triangle inequality, \[p(x + y) = -p(-x - y) \ge - (p(-x) + p(-y)) = p(x) + p(y).\] The reverse inequality is simply subadditivity of \(p\). $$\tag*{$\blacksquare$}$$
Let \(\mathfrak B_X\) be the collection of all sublinear functionals on a real vector space \(X\). We define an order on the set \(\mathfrak B_X\) by saying \(p \le q\) whenever \[p(x) \le q(x)\quad \text{ for all } \quad x\in X.\]
Let \(X\) be a real vector space and let \(p\) be a sublinear functional on \(X\). If \(V\) is a subspace of \(X\) and if \(f : V \to \mathbb R\) is a linear functional such that \[f (x) \le p(x) \quad \text{ for } \quad x \in V,\] then there is an extension \(\hat{f} : X \to \mathbb R\) of \(f\) that is linear and satisfies \[\hat{f} (x) \le p(x) \quad \text{ for } \quad x \in X,\]
Proof. We begin by showing that if \(x \in X\setminus V\), then \(f\) can be extended to a linear functional \(g\) on \(V + \mathbb{R}x\) satisfying \(g(y) \leq p(y)\) for \(y\in V + \mathbb{R}x\). If \(y_1, y_2 \in V\), we have \[f(y_1) + f(y_2) = f(y_1+y_2) \leq p(y_1+y_2) \leq p(y_1-x) + p(x+y_2).\]
Equivalently, for every \(y_1, y_2 \in V\) we have that \[f(y_1) - p(y_1-x) \leq p(x+y_2) - f(y_2).\]
Hence \[\sup \{ f(y) - p(y-x) : y \in V \} \leq \inf \{ p(x+y) - f(y) : y \in V \}.\]
Choose \(\alpha\in \mathbb R\) such that \[\sup \{ f(y) - p(y-x) : y \in V \} \leq \alpha \leq \inf \{ p(x+y) - f(y) : y \in V \},\] and define \(g: V + \mathbb{R}x \to \mathbb R\) by setting \[g(y+\lambda x) = f(y) + \lambda \alpha.\]
Then, \(g\) is clearly linear, and \(g_{|V} = f\), so \(g(y) \leq p(y)\) for \(y \in V\).
If \(\lambda > 0\) and \(y \in V\), then \[\begin{aligned} g(y+\lambda x) = \lambda [f(y/\lambda) + \alpha] &\leq \lambda [f(y/\lambda) + p(x+(y/\lambda)) - f(y/\lambda)]\\ &= p(y+\lambda x). \end{aligned}\]
If \(\lambda = -\mu < 0\), then \[\begin{aligned} g(y+\lambda x) = \mu [f(y/\mu) - \alpha] &\leq \mu [f(y/\mu) - f(y/\mu) + p((y/\mu) - x)] \\ &= p(y+\lambda x). \end{aligned}\]
Thus \(g(y) \leq p(y)\) for \(y \in V+\mathbb Rx\).
Evidently, the same reasoning can be applied to any linear extension \(F\) of \(f\) satisfying \(F \leq p\) on its domain and it shows that the domain of a maximal linear extension satisfying \(F \leq p\) must be the whole space \(X\).
Let \(\mathcal F\) be the family of linear extensions \((F, M)\) of \((f, V)\), where \(M\) is a linear subspace of \(X\) containing \(V\) and \(F\) is a linear functional on \(M\) such that \(F(x)\le p(x)\) for all \(x\in M\), and \(F(x)=f(x)\) for all \(x\in V\).
The family \(\mathcal F\) is partially ordered by the relation \[(f_1, M_1)\prec (f_2, M_2) \ \iff \ M_1\subset M_2 \ \text{ and } \ f_1=f_2 \ \text{ on } \ M_1.\]
Since the union of any increasing family of subspaces of \(X\) is again a subspace, one easily sees that the union of any linearly ordered subfamily of \(\mathcal{F}\) lies in \(\mathcal{F}\).
The proof is therefore completed by invoking Kuratowski–Zorn’s lemma, as the maximal element \((\hat{f}, M)\in \mathcal F\) is the extension with the desired properties. We must have \(M=X\), otherwise we could find an extension of \(\hat{f}\), but this would contradict maximality of \((\hat{f}, M)\). $$\tag*{$\blacksquare$}$$
Let \(X\) be a real vector space and let \(p\) be a sublinear functional on \(X\). Let \(f : V \to \mathbb R\) be a linear functional on a linear subspace \(V\) of \(X\) such that \(f \le p\) on \(V\). Suppose that \(\mathcal T\) is a commutative collection of linear maps on \(X\) (i.e. \(ST = TS\) for all \(S,T \in \mathcal T\)) such that
\(T[V]\subseteq V\) for any \(T\in \mathcal T\),
\(p(Tx)\le p(x)\) for any \(T\in \mathcal T\) and \(x\in X\),
\(f(Tx)=f(x)\) for any \(T\in \mathcal T\) and \(x\in V\).
Then there is an extension \(\hat{f} : X \to \mathbb R\) of \(f\) that is linear and satisfies
\(\hat{f} (x)=f(x)\) for \(x\in V\).
\(\hat{f} (x) \le p(x)\) for any \(x\in X\).
\(\hat{f} (Tx)=\hat{f} (x)\) for any \(T\in \mathcal T\) and \(x\in X\).
Let \(\mathfrak B_X\) be the collection of all sublinear functionals on a real vector space \(X\).
Proof. Consider the collection \[\mathcal C=\{q\in \mathfrak B_X: q\le p \ \text{ and } \ f\le q \text{ on} \ V \text{ and } q\circ T \le q \text{ for all } T \in \mathcal T\}.\] We will use Zorn’s Lemma to show there exists a minimal element \(q \in \mathcal C\).
Suppose \(\mathcal L=\{q_i: i\in I\}\) is a chain in \(\mathcal C\). Note that \(q = \inf_{i\in I} q_i\) is a sublinear functional on \(X\). Why? Justify this! Then \[q(T x) \le q_i (T x)\le q_i (x), \quad \text{ for all } \quad x \in X, \ T \in \mathcal T, \ i\in I,\] since \(q_i\in\mathcal C\) for all \(i\in I\). Hence \(q(T x) \le q (x)\) for all \(T \in \mathcal T\) and \(x \in X\) and consequently \(q\in \mathcal C\) is a lower bound for the chain \(\mathcal L\).
Therefore, \(\mathcal C\) contains a minimal element, by Zorn’s Lemma. Now let \(q\) be a minimal element of \(\mathcal C\) and let \(T\in \mathcal T\), \(n \in \mathbb N\) and define \[q_n(x)=q\bigg(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}x\bigg), \quad x\in X.\]
Note that \(q_n\) is sublinear since \(q\) is sublinear and \(T\) is linear.
Moreover, \(q_n\in \mathcal C\), since for any \(S\in \mathcal T\) we have \[q_n(Sx)=q\bigg(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}Sx\bigg)=q\bigg(S\bigg(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}x\bigg)\bigg)\le q_n(x),\] and \(q_n(x)=q\big(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}x\big)\ge f\big(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}x\big)=f(x)\) for \(x\in V\).
By sublinearity and the fact that \(q(T^{n-1}x)\le q(x)\) for \(x\in X\), we have \[q_n(x)=q\bigg(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}x\bigg)\le q\bigg(\frac{x}{n}\bigg)+\ldots+q\bigg(\frac{x}{n}\bigg)=q(x),\] and by minimality of \(q\) we obtain that \(q=q_n\) for all \(n\in \mathbb N\). Hence \[q(x)=q\bigg(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}x\bigg), \quad x\in X.\]
We have proved that \(f(x)\le q(x)\le p(x)\) for all \(x\in V\). Thus by the Hahn–Banach theorem there exists an extension \(\hat{f}:X\to \mathbb R\) of \(f\) such that \[\hat{f} (x) \le q(x)\le p(x) \quad \text{ for } \quad x \in X.\]
We have to prove that \(\hat{f} (Tx)=\hat{f} (x)\) for any \(T\in \mathcal T\) and \(x\in X\). For this purpose note that for any \(x\in X\) and \(n\in \mathbb N\) we have \[\begin{aligned} q(x-Tx)&=q\bigg(\frac{1}{n}\sum_{j=0}^{n-1}T^{j}(x-Tx)\bigg)=q\bigg(\frac{1}{n}(x-T^nx)\bigg)\\ &\le \frac{1}{n}q(x)+\frac{1}{n}q(-T^nx)\le \frac{1}{n}(q(x)+q(-x)) \ _{\overrightarrow{n\to \infty}} \ 0. \end{aligned}\]
Thus \(\hat{f}(x)-\hat{f}(Tx)=\hat{f} (x-Tx)\le q(x-Tx)\le 0\), and a similar argument shows that \(\hat{f}(Tx)-\hat{f}(x)\le q(-x-T(-x))\le 0\), yielding the desired conclusion.$$\tag*{$\blacksquare$}$$
There exits a nonnegative set function \(\mu\) defined for all bounded subsets of \(\mathbb R\) such that
\(\mu(A\cup B)=\mu(A)+\mu(B)\) if \(A\cap B=\varnothing\).
\(\mu(x+A)=\mu(A)\) for every \(A\subseteq \mathbb R\) and every \(x\in \mathbb R\).
\(\mu(A)=\lambda(A)\) for every bounded \(A\in \mathcal L(\mathbb R)\).
In other words, Lebesgue measure \(\lambda\) on \(\mathbb R\) can be extended to an additive translation-invariant set function defined on all subsets of \(\mathbb R\).
Proof. Let \(B_{fin}(\mathbb R)\) be the set of all bounded functions \(f:\mathbb R\to \mathbb R\) such that \(\{x\in\mathbb R: f(x)\neq0\}\subseteq [a, b]\) for some compact interval \([a, b]\).
Let \(BL_{fin}(\mathbb R)\) be the set of all Lebesgue measurable functions in \(B_{fin}(\mathbb R)\).
Define \(p: B_{fin}(\mathbb R)\to\mathbb R\) by \[p(f)=\inf\Big\{\int_{\mathbb R}g(x)dx: g\in BL_{fin}(\mathbb R) \text{ and } |f|\le g \Big\}.\]
Note that \(p(f)\) is finite. Indeed, if \(f\in B_{fin}(\mathbb R)\) then \(|f(x)|\le m\) for some \(m\ge0\), so we may take \(g=m\mathbf{1}_{{[a, b]}}\). Then \(|f(x)|\le g(x)\) and \(p(f)\le m(b-a)<\infty\).
It is not difficult to see that \(p\) is a sublinear functional.
Define a translation operator on \(BL_{fin}(\mathbb R)\) by \[A_tf(x)=f(x+t) \quad \text{ for } \quad x, t\in \mathbb R.\]
Then we see that \(A_tA_s=A_sA_t\) for all \(s, t\in \mathbb R\) and \(A_0f=f\) for \(f \in BL_{fin}(\mathbb R)\) and \(A_t\big[BL_{fin}(\mathbb R)\big]\subseteq BL_{fin}(\mathbb R)\) for \(t\in \mathbb R\).
Moreover, \(p(A_tf)\le p(f)\) for any \(f \in BL_{fin}(\mathbb R)\). Indeed, if \(|f|\le g\), then \(|A_tf|\le A_tg\) and \[p(A_tf)\le \int_{\mathbb R}A_tg(x)dx=\int_{\mathbb R}g(x+t)dx=\int_{\mathbb R}g(x)dx,\] giving the claim \(p(A_tf)\le p(f)\).
We define a functional \[\pi(f)=\int_{\mathbb R}f(x)dx \quad \text{ for } \quad f \in BL_{fin}(\mathbb R).\]
Then we see that \(\pi(f)\le p(f)\), since \[\pi(f)=\int_{\mathbb R}f(x)dx\le \int_{\mathbb R}|f(x)|dx=p(f),\] and we also have \(\pi(A_tf)=\pi(f)\) for any \(f \in BL_{fin}(\mathbb R)\) and \(t\in \mathbb R\).
By the invariant Hahn–Banach theorem there exists an extension \(\Pi:B_{fin}(\mathbb R)\to \mathbb R\) of \(\pi\) such that
\(\Pi(f)=\pi(f)\) for all \(f\in BL_{fin}(\mathbb R)\).
\(\Pi(f)\le p(f)\) for all \(f\in B_{fin}(\mathbb R)\).
\(\Pi(A_tf)=\Pi(f)\) for all \(f\in B_{fin}(\mathbb R)\) and \(t\in \mathbb R\).
We show that \(\Pi(f)\ge0\) if \(f\ge0\) and \(f\in B_{fin}(\mathbb R)\). Suppose that \(f\le m\mathbf{1}_{{[a, b]}}\) for some \(m\ge 0\). Then \(0\le m\mathbf{1}_{{[a, b]}}-f\le m\mathbf{1}_{{[a, b]}}\), and \[\Pi(m\mathbf{1}_{{[a, b]}}-f)\le p(m\mathbf{1}_{{[a, b]}}-f)\le \int_{\mathbb R}m\mathbf{1}_{{[a, b]}}(x)dx=m(b-a).\]
Hence \(m(b-a)-\Pi(f)=\pi(m\mathbf{1}_{{[a, b]}})-\Pi(f)\le m(b-a)\) and consequently \(\Pi(f)\ge0\), since \(\pi(m\mathbf{1}_{{[a, b]}})=m(b-a)\).
Now for a bounded set \(A\subset \mathbb R\) we define \(\mu(A)=\Pi(\mathbf{1}_{{A}})\). This set functions has all desired properties:
\(\mu(A)\ge0\) for all bounded \(A\subset \mathbb R\).
For any \(A, B\subset \mathbb R\) such that \(A\cap B=\varnothing\) we have \[\mu(A\cup B)=\Pi(\mathbf{1}_{{A\cup B}})=\Pi(\mathbf{1}_{{A}})+\Pi(\mathbf{1}_{{B}})=\mu(A)+\mu(B).\]
For a bounded set \(A\in \mathcal L(\mathbb R)\) we also have \[\qquad\qquad\mu(A)=\Pi(\mathbf{1}_{{A}})=\int_{\mathbb R}\mathbf{1}_{{A}}(x)dx=\lambda(A). \qquad\qquad \tag*{$\blacksquare$}\]
By a similar argument Lebesgue measure on \(\mathbb R^2\) can be extended to a finitely additive measure on \(\mathcal P(\mathbb R^2)\) that is invariant under translations and rotations. The Banach–Tarski paradox prevents this result from being extended to higher dimensions.
The Banach–Tarski paradox asserts that if \(U, V\subset \mathbb R^d\) are arbitrary bounded open sets, \(d\geq 3\), then there exists \(k\in{\mathbb N}\) and subsets \(E_1,\ldots,E_k,F_1,\ldots,F_k\) of \({\mathbb R}^d\) such that
\(E_i\cap E_j =\varnothing\) whenever \(i\neq j\), and \(\bigcup_{j=1}^k E_j = U\);
\(F_i\cap F_j =\varnothing\) whenever \(i\neq j\), and \(\bigcup_{j=1}^k F_j = V\);
\(E_j \sim F_j\) for all \(j=1,\ldots,k\).
Here \(A\sim B\) means that \(A\) can be mapped into \(B\) by a combination of a translation, a rotation, and a reflection.
Thus one can cut up a ball the size of a pea into a finite number of pieces and rearrange them to form a ball the size of the earth! Needless to say, the sets \(E_j\) and \(F_j\) are very bizarre.
Proposition. Let \(X\) be a vector space over \(\mathbb C\). If \(f:X\to\mathbb C\) is a linear functional and \(u = {\rm Re}f\), then \(u\) is a real linear functional, and \(f (x) = u (x)- i u ( ix )\) for all \(x \in X\). Conversely, if \(u\) is a real linear functional on \(X\) and \(f:X\to\mathbb C\) is defined by \(f(x) = u(x) - iu(ix)\), then \(f\) is complex linear. In this case, if \(X\) is normed, we have \(\|f\|=\|u\|\).
Proof. If \(f\) is complex linear and \(u = {\rm Re}f\), then \(u\) is clearly real linear and \({\rm Im} [f(x)] = - {\rm Re}[if(x)] = -u(ix)\), so \(f(x) = u(x) - iu(ix)\).
On the other hand, if \(u\) is real linear and \(f(x) = u(x) - iu(ix)\), then \(f\) is clearly linear over \(\mathbb R\), and \(f(ix) = u(ix) - iu( -x) = u(ix) + iu(x) = if(x)\), so \(f\) is also linear over \(\mathbb C\).
Finally, if \(X\) is normed, then \(\|u\|\le \|f\|\), since \(|u(x)| = |{\rm Re} f(x)| \le |f(x)|\). Also if \(f(x) \neq 0\), let \(\alpha = \overline{{\rm sgn} f(x)}\). Then \(|f(x)|= \alpha f(x) = f(\alpha x) = u(\alpha x)\) (since \(f(\alpha x)\) is real), so \(|f(x)|\le \|u\| \|\alpha x\|\), whence \(\|f\|\le \|u\|\).$$\tag*{$\blacksquare$}$$
Let \(E\) be a complex vector space and let \(p\) be a seminorm on \(E\). If \(V\) is a subspace of \(E\) and if \(f : V \to \mathbb C\) is a linear functional such that \[|f (x)| \le p(x) \quad \text{ for } \quad x \in V,\] then there is an extension \(\hat{f} : E \to \mathbb C\) of \(f\) that is linear and satisfies \[|\hat{f} (x)| \le p(x) \quad \text{ for } \quad x \in E,\]
Proof. Let \(u = {\rm Re} f\). By the Hahn–Banach theorem for real normed vector spaces there is a real linear extension \(\hat{u}\) of \(u\) to \(X\) such that \(\hat{u}(x)\le p(x)\) for all \(x \in X\). Let \(\hat{f}(x) = \hat{u}(x) - i\hat{u}(ix)\) as in the previous proposition. Then \(\hat{f}\) is a complex linear extension of \(f\), and if \(\alpha = \overline{{\rm sgn} \hat{f}(x)}\) we have \(|\hat{f}(x)|=\alpha \hat{f}(x)=\hat{f}(\alpha x)=\hat{u}(\alpha x)\le p(\alpha x)=p(x)\). $$\tag*{$\blacksquare$}$$
Let \((X, \|\cdot\|)\) be a normed vector space.
If \(M\) is a subspace of \(X\) and \(f\in M^*\), then there exists \(\hat{f}\in X^*\) an extension of \(f\) such that \(\hat{f}=f\) on \(M\) and \(\|\hat{f}\|_{X^*}=\|f\|_{M^*}\).
If \(M\) is a closed subspace of \(X\) and \(x \in X \setminus M\), there exists \(f \in X^{*}\) such that \(f(x) \not = 0\) and \(f(y)= 0\) for every \(y\in M\). In fact, \(f\) can be taken to satisfy \(\|f\|_{X^*} = 1\) and \(f(x) = \delta\), where \(\delta = \inf_{y \in M} \|x-y\|\).
If \(0\neq x \in X\), then there exists \(f \in X^{*}\) such that \(\|f\|_{X^*} = 1\) and \(f(x) = \|x\|\).
The bounded linear functionals on \(X\) separate points.
If \(x \in X\), define \(\hat{x}: {X}^{*} \to \mathbb{C}\) 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 of (a). Define \(p(x) = \|f\|_{M^*}\|x\|\) for \(x \in X\). Then \(|f| \le p\) on \(M\). (Note that \(f\) is defined only on \(M\).) By the Hahn–Banach extension theorem, there is \(\hat{f} \in X^{*}\) such that \(|\hat{f}| \le p\) on \(X\) and \(\hat{f} = f\) on \(M\). It follows directly that \(\|\hat{f}\|_{X^*}=\|f\|_{M^*}\). Indeed, we have \(\|\hat{f}\|_{X^*}\le \|f\|_{M^*}\), and also \(\|f\|_{M^*}\le \|\hat{f}\|_{X^*}\), since \(\hat{f}\) is an extension of \(f\). $$\tag*{$\blacksquare$}$$
Proof of (b). Define \(f\) on \(M + \mathbb{C}x\) by setting \(f(y+\lambda x) = \lambda \delta\) for \(y \in M\) and \(\lambda \in \mathbb C\). Then \(f(x) = \delta\), and \(f= 0\) on \(M\), and for \(\lambda \not = 0\), we have \(|f(y+\lambda x) | = |\lambda| \delta \leq |\lambda| \| \lambda^{-1} y + x \| = \|y+\lambda x\|\). Thus, the Hahn–Banach theorem can be applied with \(p(x) = \|x\|\) and \(M\) replaced by \(M+\mathbb{C}x\). $$\tag*{$\blacksquare$}$$
Proof of (c). It is the special case of (b) with \(M = \{ 0 \}\). $$\tag*{$\blacksquare$}$$
Proof of (d). If \(x \not = y\), then \(f(x) \not = f(y)\), since by (c) there exists \(f \in {X}^*\) with \(f(x-y)=\|x-y\| \not = 0\).$$\tag*{$\blacksquare$}$$
Proof of (e). Obviously \(\hat{x}\) is a linear functional on \({X}^*\) and the map \(x \mapsto \hat{x}\) is linear. Moreover, \(|\hat{x}(f)| = |f(x)| \leq \|f\|_{X^*} \|x\|\), so \(\|\hat{x}\|_{X^*} \leq \|x\|\). On the other hand, by (c) there exists \(f \in X^{*}\) such that \(f(x) = \|x\|\) thus \(\|x\|=|f(x)|\le \|\hat{x}\|_{X^*}\). Hence \(\|\hat{x}\|_{X^*} = \|x\|\) as desired. $$\tag*{$\blacksquare$}$$
With notation as in (e) above, let \(\hat{{X}} = \{ \hat{x} : x \in {X} \}\). Since \({X}^{**}\) is always complete, the closure \(\overline{\hat{X}}\) is a Banach space and the map \(x \mapsto \hat{x}\) embeds \({X}\) into \(\overline{\hat{X}}\) as a dense subspace. \(\overline{\hat{X}}\) is called the completion of \({X}\). In particular, if \(X\) is itself a Banach space, then \(\overline{\hat{X}}={X}\).
A Banach space \({X}\) is reflexive if \({X} = {X}^{**}\). For finite dimensional \(X\) we always have \({X} = {X}^{**}\) since these spaces have the same dimension. For infinite dimensional Banach spaces we may have \({X} \neq {X}^{**}\).
Let \((X,\mathcal M,\mu)\) be a \(\sigma\)-finite measure space. If \(1<p<\infty\), then we show that \((L^p(X))^*=L^q(X)\), where \(\frac{1}{p}+\frac{1}{q}=1\). In particular, \(L^p(X)\) is reflexive for all \(1<p<\infty\). We also show that \((L^1(X))^{*}=L^{\infty}(X)\), but \({\color{red}(L^\infty(X))^* \neq L^1(X)}\), when \(X\) is infinite.
Let \(K\) be a compact Hausdorff topological space and \(C(K)\) be the space of all continuous functions on \(K\). It follows from Riesz theorem that \((C(K))^{*}=\mathcal M(K)\), where \(\mathcal M(K)\) is the space of all complex Radon measures on \(K\). It will be proved in 502!