Proof of The Dual of a Real Normed Space Is a Real Banach Space; Weak-Star Limits

Throughout, elements of $X^{*}$ are functions $X\to\mathbb{R}$, two of which are equal when they agree at every point; linear functionals are additive and homogeneous, which is used without further mention; and for $F\in X^{*}$, $B_{F}$ is its set of nonnegative bounds. Two facts about the dual norm are used repeatedly. First, $\lVert F\rVert_{*}\ge0$ and $\lVert F\rVert_{*}$ is itself a bound for $F$, by The Nonnegative Bounds of a Bounded Linear Functional Have an Infimum, Which Is a Bound (i) and The Nonnegative Bounds of a Bounded Linear Functional Have an Infimum, Which Is a Bound (ii):

$$|F(x)|\le\lVert F\rVert_{*}\,\lVert x\rVert\qquad(x\in X).\tag{1}$$

Second, $\lVert F\rVert_{*}\le C$ for every nonnegative bound $C$ of $F$, because the infimum $\lVert F\rVert_{*}=\inf B_{F}$ is a lower bound of $B_{F}$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv), Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (i)).

Step 1: closure. Let $F,G\in X^{*}$, with bounds $C$ and $D$, and $\lambda\in\mathbb{R}$. The pointwise sum is additive, since $(F+G)(u+v)=F(u)+F(v)+G(u)+G(v)=(F+G)(u)+(F+G)(v)$, and homogeneous in the same way; and $C+D$ is a bound for it, since $|F(x)+G(x)|\le|F(x)|+|G(x)|\le(C+D)\lVert x\rVert$. Likewise $\lambda F$ is linear with bound $|\lambda|C$, because $|\lambda F(x)|=|\lambda|\,|F(x)|$. The zero function is linear and has the bound $0$. Hence $F+G$, $\lambda F$ and $0$ lie in $X^{*}$ (Bounded Linear Functionals and the Dual Space of a Real Normed Space (iv)), and so does $-F=(-1)F$.

Step 2: the vector-space laws. The pointwise operations on real-valued functions on $X$ satisfy the commutative-ring and scalar laws carried by the setting: addition is associative and commutative, $F+0=F$ and $F+(-F)=0$, and $\lambda(\mu F)=(\lambda\mu)F$, $1F=F$, $\lambda(F+G)=\lambda F+\lambda G$ and $(\lambda+\mu)F=\lambda F+\mu F$. By Step 1 these identities hold within $X^{*}$, which gives Vector Space over a Field (i), Vector Space over a Field (ii), Vector Space over a Field (iii) (with $z=0$), Vector Space over a Field (v), Vector Space over a Field (vi), Vector Space over a Field (vii) and Vector Space over a Field (viii). For Vector Space over a Field (iv), let $z\in X^{*}$ satisfy $H+z=H$ for every $H\in X^{*}$; taking $H=0$ gives $z=z+0=0+z=0$, and then $y=-F$ satisfies $F+y=z$. So $X^{*}$ is a real vector space (Vector Space over a Field (x)). Its zero vector is $0$ and the negative of $F$ is $-F$, since these satisfy the defining properties; hence the difference $F-G=F+(-G)$ is the pointwise difference, $(F-G)(x)=F(x)-G(x)$. This proves (i). Combined with (1), it gives

$$|F(x)-G(x)|\le\lVert F-G\rVert_{*}\,\lVert x\rVert\qquad(F,G\in X^{*},\ x\in X).\tag{2}$$

Step 3: the dual norm is a norm. Let $F,G\in X^{*}$ and $\lambda\in\mathbb{R}$. Nonnegativity is the first fact above. Definiteness: if $\lVert F\rVert_{*}=0$, then (1) gives $|F(x)|\le0$, so $F(x)=0$ for every $x$, and $F=0$. Triangle inequality: by (1), $|(F+G)(x)|\le(\lVert F\rVert_{*}+\lVert G\rVert_{*})\lVert x\rVert$, so $\lVert F\rVert_{*}+\lVert G\rVert_{*}$ is a nonnegative bound of $F+G$ and therefore $\lVert F+G\rVert_{*}\le\lVert F\rVert_{*}+\lVert G\rVert_{*}$. Absolute homogeneity: in the same way $|\lambda|\,\lVert F\rVert_{*}$ is a nonnegative bound of $\lambda F$, so

$$\lVert\lambda F\rVert_{*}\le|\lambda|\,\lVert F\rVert_{*}.\tag{3}$$

If $\lambda=0$, then $\lambda F=0$, which has the bound $0$, so both sides of (3) are $0$. If $\lambda\neq0$, then $F=\lambda^{-1}(\lambda F)$, and (3) applied to $\lambda F$ and $\lambda^{-1}$ gives $\lVert F\rVert_{*}\le|\lambda|^{-1}\lVert\lambda F\rVert_{*}$, that is $|\lambda|\,\lVert F\rVert_{*}\le\lVert\lambda F\rVert_{*}$. So equality holds in (3). By Norms and Real Normed Spaces (v), $\lVert\cdot\rVert_{*}$ is a norm on the real vector space $X^{*}$, and $(X^{*},\lVert\cdot\rVert_{*})$ is a real normed space by Norms and Real Normed Spaces (vi). This proves (ii).

Step 4: a Cauchy sequence has a pointwise limit. Let $d_{*}(F,G)=\lVert F-G\rVert_{*}$ be the norm metric of $X^{*}$, and let $(F_{k})_{k\in\mathbb{N}}$ be a Cauchy sequence in $(X^{*},d_{*})$. Fix $x\in X$. Given $\varepsilon>0$, choose $N$ with $\lVert F_{k}-F_{j}\rVert_{*}<\varepsilon/(\lVert x\rVert+1)$ for all $k,j\ge N$; by (2), $|F_{k}(x)-F_{j}(x)|\le\varepsilon\lVert x\rVert/(\lVert x\rVert+1)<\varepsilon$ for all $k,j\ge N$. So $(F_{k}(x))_{k}$ is a Cauchy sequence of real numbers, and it converges by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences (v). Let $F(x)$ be its limit; this defines a function $F:X\to\mathbb{R}$.

Step 5: the limit is linear. For $u,v\in X$ and $\lambda\in\mathbb{R}$ we have $F_{k}(u+v)=F_{k}(u)+F_{k}(v)$ and $F_{k}(\lambda u)=\lambda F_{k}(u)$ for every $k$. The left-hand sides converge to $F(u+v)$ and $F(\lambda u)$, and the right-hand sides to $F(u)+F(v)$ and $\lambda F(u)$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi); by uniqueness of limits, Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii), $F(u+v)=F(u)+F(v)$ and $F(\lambda u)=\lambda F(u)$.

Step 6: a uniform estimate. Let $\varepsilon>0$, and choose $N$ with $\lVert F_{k}-F_{j}\rVert_{*}<\varepsilon$ for all $k,j\ge N$. Fix $k\ge N$ and $x\in X$. By (2), $|F_{k}(x)-F_{j}(x)|\le\varepsilon\lVert x\rVert$ for every $j\ge N$. As $j\to\infty$, $F_{k}(x)-F_{j}(x)\to F_{k}(x)-F(x)$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i) and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi), and then $|F_{k}(x)-F_{j}(x)|\to|F_{k}(x)-F(x)|$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (viii). Passing to the limit in the inequality, Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x) gives

$$|F_{k}(x)-F(x)|\le\varepsilon\,\lVert x\rVert\qquad(k\ge N,\ x\in X).\tag{4}$$

Step 7: the limit lies in $X^{*}$, and the sequence converges to it. Apply (4) with $\varepsilon=1$ and its index $N_{1}$: by (1), $|F(x)|\le|F_{N_{1}}(x)|+|F(x)-F_{N_{1}}(x)|\le(\lVert F_{N_{1}}\rVert_{*}+1)\lVert x\rVert$ for every $x$. So the linear functional $F$ is bounded, and $F\in X^{*}$. Now let $\varepsilon>0$ and let $N$ be the index of (4) for $\varepsilon/2$. For $k\ge N$, (4) and $(F_{k}-F)(x)=F_{k}(x)-F(x)$ say that $\varepsilon/2$ is a nonnegative bound of $F_{k}-F$, so $d_{*}(F_{k},F)=\lVert F_{k}-F\rVert_{*}\le\varepsilon/2<\varepsilon$. Thus $(F_{k})$ converges to $F$ in $(X^{*},d_{*})$. Every Cauchy sequence in $(X^{*},d_{*})$ therefore converges, so $(X^{*},d_{*})$ is complete, and $(X^{*},\lVert\cdot\rVert_{*})$ is a real Banach space by Real Banach Space (i). This proves (iii).

Step 8: weak-star limits. If $(F_{m})$ converges weak-star to $F$ and to $G$, then for every $x\in X$ the real sequence $(F_{m}(x))_{m}$ converges to $F(x)$ and to $G(x)$ by Weak-Star Convergence in the Dual of a Real Normed Space (i), so $F(x)=G(x)$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii); hence $F=G$, proving (iv). Suppose now that $\lVert F_{m}-F\rVert_{*}\to0$, and fix $x\in X$. By Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi), $\lVert x\rVert\,\lVert F_{m}-F\rVert_{*}\to\lVert x\rVert\cdot0=0$, and by (2),

$$|F_{m}(x)-F(x)|\le\lVert x\rVert\,\lVert F_{m}-F\rVert_{*}\qquad(m\in\mathbb{N}).$$

So $F_{m}(x)\to F(x)$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (xii). As $x$ was arbitrary, $(F_{m})$ converges weak-star to $F$, proving (v).