[proofplan] The argument is a direct unwinding of the two definitions. A common pointwise bound for all functions immediately bounds each supremum norm by taking the supremum over $E$. Conversely, an upper bound for all supremum norms controls every value $|f_n(x)|$, because the supremum is an upper bound for the set of pointwise absolute values. [/proofplan] [step:Turn a common pointwise bound into an upper bound for the supremum norms] Assume that $(f_n)_{n \in \mathbb{N}}$ is uniformly bounded on $E$. By definition, there exists a real number $M \geq 0$ such that \begin{align*} |f_n(x)| \leq M \end{align*} for every $n \in \mathbb{N}$ and every $x \in E$. Fix $n \in \mathbb{N}$. Since $f_n: E \to \mathbb{R}$ is bounded and $E$ is nonempty, the set $\{|f_n(x)| : x \in E\}$ is a nonempty bounded subset of $\mathbb{R}$, so its supremum is finite. The inequality above says that $M$ is an upper bound for this set. Therefore, by the definition of supremum, \begin{align*} \|f_n\|_{\infty,E} = \sup_{x \in E} |f_n(x)| \leq M. \end{align*} Since this holds for every $n \in \mathbb{N}$, the numerical sequence $(\|f_n\|_{\infty,E})_{n \in \mathbb{N}}$ is bounded above. [guided] Assume that $(f_n)_{n \in \mathbb{N}}$ is uniformly bounded on $E$. This means that the bound is independent of both the index $n$ and the point $x$. Thus there exists a real number $M \geq 0$ such that \begin{align*} |f_n(x)| \leq M \end{align*} for every $n \in \mathbb{N}$ and every $x \in E$. Now fix one index $n \in \mathbb{N}$. The function $f_n: E \to \mathbb{R}$ is bounded by hypothesis, and $E$ is nonempty, so the set of values \begin{align*} \{|f_n(x)| : x \in E\} \end{align*} is a nonempty bounded subset of $\mathbb{R}$. Hence its supremum is a finite real number. The common pointwise estimate says exactly that $M$ is an upper bound for this set. Since the supremum is the least upper bound, it follows that \begin{align*} \|f_n\|_{\infty,E} = \sup_{x \in E} |f_n(x)| \leq M. \end{align*} Because the same $M$ works for every index $n$, the numerical sequence $(\|f_n\|_{\infty,E})_{n \in \mathbb{N}}$ is bounded above. [/guided] [/step] [step:Turn an upper bound for the supremum norms into a common pointwise bound] Conversely, assume that the numerical sequence $(\|f_n\|_{\infty,E})_{n \in \mathbb{N}}$ is bounded above. Then there exists a real number $A \in \mathbb{R}$ such that \begin{align*} \|f_n\|_{\infty,E} \leq A \end{align*} for every $n \in \mathbb{N}$. Define $C := \max\{A,0\}$, so $C \geq 0$ and $\|f_n\|_{\infty,E} \leq C$ for every $n \in \mathbb{N}$. Let $n \in \mathbb{N}$ and $x \in E$. Since $\|f_n\|_{\infty,E}$ is the supremum of the set $\{|f_n(y)| : y \in E\}$, it is an upper bound for that set. Therefore \begin{align*} |f_n(x)| \leq \|f_n\|_{\infty,E} \leq C. \end{align*} The constant $C$ is independent of both $n$ and $x$, so $(f_n)_{n \in \mathbb{N}}$ is uniformly bounded on $E$. This proves the equivalence. [/step]