[proofplan] The proof is obtained by expanding the two definitions. [Uniform convergence](/page/Uniform%20Convergence) says that the pointwise error $|f_n(x)-f(x)|$ is eventually small uniformly in $x \in E$, while convergence in the [uniform norm](/page/Uniform%20Norm) says that the supremum of the same error is eventually small. The only technical point is to pass between pointwise uniform bounds and bounds on the supremum, using $\varepsilon/2$ in the forward direction to obtain a strict final inequality. [/proofplan] [step:Handle the empty set by the convention for the uniform norm] If $E=\varnothing$, then every quantified condition over $x \in E$ is vacuous, so $f_n \to f$ uniformly on $E$. Also, by the standard convention in $B(E)$, the uniform norm of the zero function on the empty set is $0$, and hence $\|f_n-f\|_\infty=0$ for every $n \in \mathbb{N}$. Therefore $\|f_n-f\|_\infty \to 0$. For the rest of the proof, assume $E \neq \varnothing$. [/step] [step:Convert uniform convergence into convergence of the uniform norm] Assume that $f_n \to f$ uniformly on $E$. For each $n \in \mathbb{N}$, define the bounded function \begin{align*} h_n: E \to \mathbb{R}, \qquad h_n(x)=f_n(x)-f(x). \end{align*} Let $\varepsilon>0$. By uniform convergence, applied with $\varepsilon/2>0$, there exists $N \in \mathbb{N}$ such that for every $n \geq N$ and every $x \in E$, \begin{align*} |h_n(x)| < \frac{\varepsilon}{2}. \end{align*} Taking the supremum over $x \in E$ gives, for every $n \geq N$, \begin{align*} \|f_n-f\|_\infty = \|h_n\|_\infty = \sup_{x \in E}|h_n(x)| \leq \frac{\varepsilon}{2} < \varepsilon. \end{align*} Thus $\|f_n-f\|_\infty \to 0$. [guided] Assume that $f_n \to f$ uniformly on $E$. We want to prove convergence of the real sequence whose $n$th term is the size of the error in the uniform norm. For each $n \in \mathbb{N}$, define the error function \begin{align*} h_n: E \to \mathbb{R}, \qquad h_n(x)=f_n(x)-f(x). \end{align*} Because $f_n \in B(E)$ and $f \in B(E)$, the difference $h_n=f_n-f$ is also bounded, so its uniform norm is finite. Let $\varepsilon>0$. Uniform convergence of $f_n$ to $f$ on $E$ means that the same index $N$ works for every point $x \in E$. Applying this definition with the positive number $\varepsilon/2$, there exists $N \in \mathbb{N}$ such that whenever $n \geq N$ and $x \in E$, we have \begin{align*} |f_n(x)-f(x)| < \frac{\varepsilon}{2}. \end{align*} Equivalently, for every $n \geq N$ and every $x \in E$, \begin{align*} |h_n(x)| < \frac{\varepsilon}{2}. \end{align*} The uniform norm is the supremum of these pointwise absolute errors. Therefore, for every $n \geq N$, \begin{align*} \|f_n-f\|_\infty = \|h_n\|_\infty = \sup_{x \in E}|h_n(x)| \leq \frac{\varepsilon}{2}. \end{align*} Since $\varepsilon/2<\varepsilon$, we obtain \begin{align*} \|f_n-f\|_\infty < \varepsilon \end{align*} for every $n \geq N$. This is exactly the definition of $\|f_n-f\|_\infty \to 0$. [/guided] [/step] [step:Convert convergence of the uniform norm into uniform convergence] Assume that $\|f_n-f\|_\infty \to 0$. Let $\varepsilon>0$. By convergence of the real sequence $\|f_n-f\|_\infty$ to $0$, there exists $N \in \mathbb{N}$ such that for every $n \geq N$, \begin{align*} \|f_n-f\|_\infty < \varepsilon. \end{align*} For every $n \geq N$ and every $x \in E$, the definition of the supremum norm gives \begin{align*} |f_n(x)-f(x)| \leq \sup_{y \in E}|f_n(y)-f(y)| = \|f_n-f\|_\infty < \varepsilon. \end{align*} Thus, for every $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that for every $n \geq N$ and every $x \in E$, \begin{align*} |f_n(x)-f(x)| < \varepsilon. \end{align*} Hence $f_n \to f$ uniformly on $E$. [/step]