Proof of The Dual of a Real Normed Space is a Real Banach Space, and the Dual Norm is the Least Bound

Each result cited is universally quantified over the data in its own statement. Elementary arithmetic and order facts in $\mathbb{R}$, the properties of $|\cdot|$, the limit laws, the order properties of limits and the uniqueness of limits are those put in force by The Real Numbers: Standing Notation and Background §background and The Real Numbers: Standing Notation and Background §sequences. For $\ell\in E^{*}$ let $B_{\ell}$ be the set of nonnegative bounds for $\ell$, as in The Dual Space of a Real Normed Space and the Dual Norm §norm. Two facts are used repeatedly. First, linear functionals are additive and homogeneous by The Dual Space of a Real Normed Space and the Dual Norm §functional, which is used below without further mention; in particular $\ell(0_{E})=0$ for every linear functional $\ell$: $0_{E}=0\,0_{E}$ by claim 3 of Elementary Identities in a Vector Space, so $\ell(0_{E})=0\,\ell(0_{E})=0$. Second, $\lVert v\rVert=0$ only if $v=0_{E}$, and $0\le\lVert v\rVert$, by Real Normed Space and Real Banach Space §norm.

Claim 1. By The Dual Space of a Real Normed Space and the Dual Norm §dual, $\ell+m$ and $\lambda\ell$ belong to $E^{*}$ for $\ell,m\in E^{*}$ and $\lambda\in\mathbb{R}$. Two elements of $E^{*}$ are equal when they take the same value at every $v\in E$, and at each $v$ the conditions 1, 2, 5, 6, 7 and 8 of Vector Space over a Field for $E^{*}$ read as the associativity and commutativity of addition, the associativity of multiplication, $1s=s$, and the two distributive laws in $\mathbb{R}$, applied, for $\ell,m,h\in E^{*}$ and scalars $\lambda,\mu$, to the real numbers $\ell(v)$, $m(v)$, $h(v)$, $\lambda$ and $\mu$. The zero functional $0^{*}:v\mapsto0$ is linear, as $0=0+0$ and $\lambda0=0$, and $0$ is a bound for it, so $0^{*}\in E^{*}$, and $(\ell+0^{*})(v)=\ell(v)$ for every $v$; this is condition 3. For $\ell\in E^{*}$, $(-1)\ell\in E^{*}$ and $(\ell+(-1)\ell)(v)=\ell(v)-\ell(v)=0$ for every $v$, which is condition 4. So $E^{*}$ is a real vector space, and by claims 1 and 2 of Elementary Identities in a Vector Space its zero vector is $0^{*}$ and the additive inverse of $\ell$ is $(-1)\ell$. In particular, for $\ell,m\in E^{*}$ the difference $\ell-m=\ell+(-1)m$ satisfies $(\ell-m)(v)=\ell(v)-m(v)$.

Claim 2. Let $\ell\in E^{*}$ and $\nu=\lVert\ell\rVert_{E^{*}}=\inf B_{\ell}$. Since $0$ is a lower bound of $B_{\ell}$ and $\nu$ is the greatest lower bound, $0\le\nu$. Let $v\in E$. If $\lVert v\rVert=0$, then $v=0_{E}$ and $|\ell(v)|=0=\nu\lVert v\rVert$. If $\lVert v\rVert\ne0$, then $0<\lVert v\rVert$ and, for every $C\in B_{\ell}$, $|\ell(v)|\le C\lVert v\rVert$ gives $|\ell(v)|/\lVert v\rVert\le C$; thus $|\ell(v)|/\lVert v\rVert$ is a lower bound of $B_{\ell}$, hence at most $\nu$, and multiplying by $\lVert v\rVert>0$ gives $|\ell(v)|\le\nu\lVert v\rVert$. So $\nu$ is a nonnegative bound for $\ell$. If $C$ is a nonnegative bound for $\ell$, then $C\in B_{\ell}$ and $\nu\le C$ because $\nu$ is a lower bound of $B_{\ell}$.

Claim 3. We check (a), (b) and (c) of Real Normed Space and Real Banach Space §norm for $\ell,m\in E^{*}$ and $\lambda\in\mathbb{R}$.

(a) $0\le\lVert\ell\rVert_{E^{*}}$ by claim 2. If $\lVert\ell\rVert_{E^{*}}=0$, claim 2 gives $|\ell(v)|\le0$, so $\ell(v)=0$, for every $v$; hence $\ell=0^{*}$, the zero vector of claim 1.

(b) By claim 2 and the computation in The Dual Space of a Real Normed Space and the Dual Norm §dual, $|\lambda|\,\lVert\ell\rVert_{E^{*}}$ is a nonnegative bound for $\lambda\ell$, so $\lVert\lambda\ell\rVert_{E^{*}}\le|\lambda|\,\lVert\ell\rVert_{E^{*}}$ by claim 2. If $\lambda=0$, then $\lambda\ell=0^{*}$, which has the bound $0$, so $\lVert\lambda\ell\rVert_{E^{*}}\le0$ by claim 2 and equality holds by (a); both sides are $0$. If $\lambda\ne0$, then $\ell=\lambda^{-1}(\lambda\ell)$, as the two agree at every $v$, so the inequality just proved, applied to $\lambda\ell$ and $\lambda^{-1}$, gives $\lVert\ell\rVert_{E^{*}}\le|\lambda|^{-1}\lVert\lambda\ell\rVert_{E^{*}}$; multiplying by $|\lambda|>0$ yields $|\lambda|\,\lVert\ell\rVert_{E^{*}}\le\lVert\lambda\ell\rVert_{E^{*}}$.

(c) By claim 2 and the computation in The Dual Space of a Real Normed Space and the Dual Norm §dual, $\lVert\ell\rVert_{E^{*}}+\lVert m\rVert_{E^{*}}$ is a nonnegative bound for $\ell+m$, so $\lVert\ell+m\rVert_{E^{*}}\le\lVert\ell\rVert_{E^{*}}+\lVert m\rVert_{E^{*}}$ by claim 2.

Hence the dual norm is a norm, and $E^{*}$ with it is a real normed space, with distance $d^{*}(\ell,m)=\lVert\ell-m\rVert_{E^{*}}$ by Real Normed Space and Real Banach Space §distance. By claim 2 and claim 1,

Claim 4. Let $(\ell_{k})_{k\in\mathbb{N}}$ be a Cauchy sequence in $(E^{*},d^{*})$, as in Real Normed Space and Real Banach Space §topology.

Pointwise limit. Fix $v\in E$ and a positive $\varepsilon$. Choose $N$ with $\lVert\ell_{k}-\ell_{j}\rVert_{E^{*}}<\varepsilon/(\lVert v\rVert+1)$ for all $k,j\ge N$; by ($*$), $|\ell_{k}(v)-\ell_{j}(v)|\le\varepsilon\lVert v\rVert/(\lVert v\rVert+1)<\varepsilon$ for $k,j\ge N$. So $(\ell_{k}(v))_{k}$ is a Cauchy sequence of real numbers and converges by Every Cauchy Sequence of Real Numbers Converges; let $\ell(v)$ be its limit.

Linearity. For $u,v\in E$ and $\lambda\in\mathbb{R}$, $\ell_{k}(u+v)=\ell_{k}(u)+\ell_{k}(v)$ and $\ell_{k}(\lambda v)=\lambda\ell_{k}(v)$ for every $k$; by claims 1 and 3 of Arithmetic of Limits of Real Sequences and uniqueness of limits, $\ell(u+v)=\ell(u)+\ell(v)$ and $\ell(\lambda v)=\lambda\ell(v)$.

Uniform estimate. Let $\varepsilon$ be positive and choose $N$ with $\lVert\ell_{k}-\ell_{j}\rVert_{E^{*}}<\varepsilon/2$ for all $k,j\ge N$. Fix $k\ge N$ and $v\in E$. For every $j\ge N$, ($*$) gives $-\tfrac{\varepsilon}{2}\lVert v\rVert\le\ell_{k}(v)-\ell_{j}(v)\le\tfrac{\varepsilon}{2}\lVert v\rVert$; as $j\to\infty$ the middle term converges to $\ell_{k}(v)-\ell(v)$ by claim 3 of Arithmetic of Limits of Real Sequences, so claim 1 of Order Properties of Limits of Real Sequences, applied to the sequences indexed by $j\ge N$ (re-indexed from $1$), gives

Boundedness and convergence. Apply ($**$) with $\varepsilon=1$ and its index $N_{1}$: for every $v$, $|\ell(v)|\le|\ell_{N_{1}}(v)|+\tfrac12\lVert v\rVert\le(\lVert\ell_{N_{1}}\rVert_{E^{*}}+1)\lVert v\rVert$ by claim 2. So $\ell$ is a bounded linear functional, $\ell\in E^{*}$. For a positive $\varepsilon$ and its $N$, ($**$) and claim 1 say that $\varepsilon/2$ is a nonnegative bound for $\ell_{k}-\ell$ whenever $k\ge N$, so $d^{*}(\ell_{k},\ell)=\lVert\ell_{k}-\ell\rVert_{E^{*}}\le\varepsilon/2<\varepsilon$ by claim 2. Thus $(\ell_{k})$ converges to $\ell$ in $(E^{*},d^{*})$, and $(E^{*},d^{*})$ is complete; that is, $E^{*}$ is a real Banach space by Real Normed Space and Real Banach Space §banach.

Claim 5. If $(\ell_{m})$ converges weak-star to $\ell$ and to $\ell'$ in $E^{*}$, then for every $v\in E$ the sequence $(\ell_{m}(v))$ converges to $\ell(v)$ and to $\ell'(v)$ by Weak-Star Convergence in the Dual of a Real Normed Space §weak-star, so $\ell(v)=\ell'(v)$ by uniqueness of limits; hence $\ell=\ell'$. If $(\ell_{m})$ converges to $\ell$ in $(E^{*},d^{*})$, fix $v\in E$ and a positive $\varepsilon$, and choose $N$ with $\lVert\ell_{m}-\ell\rVert_{E^{*}}<\varepsilon/(\lVert v\rVert+1)$ for $m\ge N$; by ($*$), $|\ell_{m}(v)-\ell(v)|\le\varepsilon\lVert v\rVert/(\lVert v\rVert+1)<\varepsilon$ for $m\ge N$. So $(\ell_{m}(v))$ converges to $\ell(v)$ for every $v$, which is weak-star convergence to $\ell$.