[proofplan] The implication "norm-bounded implies weakly (resp. $w^*$) bounded" follows from $|f(x)| \leq \|f\| \|x\|$, valid for all $x \in X$ and $f \in X^*$. The reverse implications are applications of the **Principle of Uniform Boundedness** (Banach-Steinhaus): a family of bounded linear operators from a Banach space to a normed space that is pointwise bounded is uniformly bounded in operator norm. For (i), we view $A$ inside the bidual via the canonical isometric embedding $\hat{x}(f) = f(x)$; weak boundedness of $A$ is exactly pointwise boundedness of $\hat{A} \subseteq \mathcal{L}(X^*, \mathbb{R})$ on the Banach space $X^*$, so PUB gives uniform operator-norm boundedness of $\hat{A}$, and the isometry $\|\hat{x}\| = \|x\|$ transfers this to norm-boundedness of $A$. For (ii), the Banach space hypothesis on $X$ allows us to apply PUB directly to $B \subseteq \mathcal{L}(X, \mathbb{R})$ on the Banach space $X$, with $w^*$-boundedness providing pointwise boundedness on $X$. [/proofplan] [step:Recall the Principle of Uniform Boundedness] **Principle of Uniform Boundedness (PUB / Banach-Steinhaus).** Let $E$ be a Banach space, $F$ a normed space, and $(T_\alpha)_{\alpha \in I}$ a family in $\mathcal{L}(E, F)$. If \begin{align*} \sup_{\alpha \in I} \|T_\alpha(x)\|_F < \infty \quad \text{for every } x \in E, \end{align*} then $\sup_{\alpha \in I} \|T_\alpha\|_{\mathcal{L}(E, F)} < \infty$. The hypothesis is that the family is **pointwise bounded** on $E$; the conclusion is that the family is **uniformly bounded in operator norm**. The completeness of $E$ is essential. This is the tool used in the converse implications below. [/step] [step:Recall the canonical embedding $J : X \to X^{**}$ as an isometry] Define $J : X \to X^{**}$ by $J(x) = \hat{x}$, where $\hat{x}(f) := f(x)$ for $f \in X^*$. Each $\hat{x}$ is linear in $f$ with $|\hat{x}(f)| \leq \|f\|\|x\|$, so $\hat{x} \in X^{**} = \mathcal{L}(X^*, \mathbb{R})$ with $\|\hat{x}\| \leq \|x\|$. The Hahn-Banach norming-functional theorem yields the reverse inequality: for any $x \neq 0$, there exists $f_0 \in X^*$ with $\|f_0\| = 1$ and $f_0(x) = \|x\|$, so $\|\hat{x}\| \geq |\hat{x}(f_0)|/\|f_0\| = \|x\|$. Hence \begin{align*} \|\hat{x}\|_{X^{**}} = \|x\|_X \quad \text{for all } x \in X. \end{align*} Thus $J$ is an isometric linear embedding. [/step] [step:Forward direction of (i): norm-bounded implies weakly bounded] Suppose $A \subseteq X$ is norm-bounded, with $M := \sup_{x \in A} \|x\| < \infty$. For any $f \in X^*$, \begin{align*} \sup_{x \in A} |f(x)| \leq \sup_{x \in A} \|f\|\|x\| = \|f\| M < \infty, \end{align*} so $\{f(x) : x \in A\}$ is bounded for every $f \in X^*$, i.e. $A$ is weakly bounded. [/step] [step:Converse of (i): weakly bounded implies norm-bounded via PUB on $X^*$] Suppose $A \subseteq X$ is weakly bounded: for every $f \in X^*$, \begin{align*} \sup_{x \in A} |f(x)| < \infty. \end{align*} Consider the family $\hat{A} = \{\hat{x} : x \in A\} \subseteq X^{**} = \mathcal{L}(X^*, \mathbb{R})$. For any $f \in X^*$, \begin{align*} \sup_{x \in A} |\hat{x}(f)| = \sup_{x \in A} |f(x)| < \infty \end{align*} by hypothesis. Thus $\hat{A}$ is pointwise bounded on $X^*$. The space $X^*$ is a Banach space (the dual of any normed space is complete; this is the standard fact that $\mathcal{L}(X, \mathbb{F})$ is a Banach space because $\mathbb{F}$ is). So we apply the **Principle of Uniform Boundedness** with $E = X^*$ (Banach), $F = \mathbb{R}$, and the family $\hat{A} \subseteq \mathcal{L}(X^*, \mathbb{R})$ (pointwise bounded on $E = X^*$ by the calculation above). PUB gives \begin{align*} \sup_{x \in A} \|\hat{x}\|_{X^{**}} < \infty. \end{align*} By the isometry of Step 2, $\|\hat{x}\| = \|x\|$, so \begin{align*} \sup_{x \in A} \|x\| = \sup_{x \in A} \|\hat{x}\| < \infty, \end{align*} i.e. $A$ is norm-bounded. [guided] Suppose $A \subseteq X$ is weakly bounded: for every $f \in X^*$, \begin{align*} \sup_{x \in A} |f(x)| < \infty. \tag{$*$} \end{align*} We must show $\sup_{x \in A} \|x\| < \infty$. The strategy is to interpret weak boundedness as pointwise boundedness of a family of bounded linear operators acting on $X^*$, then apply PUB. But here is the subtlety: the elements of $A$ live in $X$, not in any space of operators. The trick is to use the canonical embedding $J : X \to X^{**}$ to transfer $A$ to the bidual, where it does become a family of operators on $X^*$. *Setting up the operator family.* Define $\hat{x}(f) := f(x)$ for $x \in X, f \in X^*$. Each $\hat{x}$ is a bounded linear functional on $X^*$, i.e. $\hat{x} \in \mathcal{L}(X^*, \mathbb{R}) = X^{**}$. Form the family \begin{align*} \hat{A} := \{\hat{x} : x \in A\} \subseteq \mathcal{L}(X^*, \mathbb{R}). \end{align*} *Verifying pointwise boundedness on $X^*$.* For any $f \in X^*$, \begin{align*} \sup_{x \in A} |\hat{x}(f)| = \sup_{x \in A} |f(x)| < \infty, \end{align*} where the equality is the definition of $\hat{x}$ and the finiteness is the weak boundedness hypothesis $(*)$. So $\hat{A}$ is pointwise bounded on $X^*$. *Applying PUB.* To apply PUB we need the *domain* of the operators to be Banach. Here the operators $\hat{x}: X^* \to \mathbb{R}$ have domain $X^*$, which is a Banach space (the dual of any normed space is complete: a Cauchy sequence $f_n$ in $X^*$ has $f_n(x)$ Cauchy in $\mathbb{R}$ for each $x$, the pointwise limit $f$ is linear and bounded by the same uniform constant, and $f_n \to f$ in operator norm by a standard $\varepsilon/2$ argument). Note: completeness of $X$ itself is **not** required here — even when $X$ is incomplete, $X^*$ is automatically complete, and that is what PUB consumes. Apply PUB with $E = X^*$, $F = \mathbb{R}$, family $\hat{A}$ to conclude \begin{align*} \sup_{x \in A} \|\hat{x}\|_{X^{**}} < \infty. \end{align*} *Transferring back to $X$.* By the isometry $J$ of Step 2 (which used the Hahn-Banach norming functional), $\|\hat{x}\|_{X^{**}} = \|x\|_X$. Therefore $\sup_{x \in A} \|x\| < \infty$, i.e. $A$ is norm-bounded. [/guided] [/step] [step:Forward direction of (ii): norm-bounded implies $w^*$-bounded] Suppose $B \subseteq X^*$ is norm-bounded, with $M := \sup_{f \in B} \|f\| < \infty$. For any $x \in X$, \begin{align*} \sup_{f \in B} |f(x)| \leq \sup_{f \in B} \|f\|\|x\| = M\|x\| < \infty, \end{align*} i.e. $\{\hat{x}(f) : f \in B\} = \{f(x) : f \in B\}$ is bounded for every $x \in X$. So $B$ is $w^*$-bounded. [/step] [step:Converse of (ii): $w^*$-bounded implies norm-bounded via PUB on $X$] Suppose $B \subseteq X^*$ is $w^*$-bounded: for every $x \in X$, \begin{align*} \sup_{f \in B} |f(x)| = \sup_{f \in B} |\hat{x}(f)| < \infty. \end{align*} This says exactly that the family $B \subseteq \mathcal{L}(X, \mathbb{R})$ is pointwise bounded on $X$. By hypothesis, $X$ is a Banach space. Apply the **Principle of Uniform Boundedness** with $E = X$ (Banach), $F = \mathbb{R}$, family $B \subseteq \mathcal{L}(X, \mathbb{R})$ (pointwise bounded on $X$). PUB gives \begin{align*} \sup_{f \in B} \|f\|_{X^*} < \infty, \end{align*} i.e. $B$ is norm-bounded. [/step] [step:Conclude] Steps 3-4 establish (i): $A \subseteq X$ is weakly bounded if and only if $A$ is norm-bounded. Steps 5-6 establish (ii): for $X$ a Banach space, $B \subseteq X^*$ is $w^*$-bounded if and only if $B$ is norm-bounded. The proof is complete. The asymmetry between (i) and (ii) — that (ii) requires $X$ to be Banach while (i) does not — comes from where PUB is invoked: in (i), the relevant Banach space is $X^*$ (always complete, irrespective of completeness of $X$), while in (ii) the relevant Banach space is $X$ itself, requiring the explicit completeness hypothesis. [/step]