[proofplan]
We verify the three norm axioms directly from the definition of the [uniform norm](/page/Uniform%20Norm). Boundedness ensures that the supremum defining $\|f\|_\infty$ is finite, while nonnegativity follows from the nonnegativity of absolute value. Definiteness follows by comparing each point value to the supremum, homogeneity follows from $|\lambda f(x)| = |\lambda||f(x)|$, and the triangle inequality follows from the pointwise triangle inequality before taking suprema.
[/proofplan]
[step:Check that the uniform norm is finite and nonnegative]
Let $f \in B(E)$. If $E$ is empty, then $\|f\|_\infty = 0$ by definition, so $\|f\|_\infty \in [0,\infty)$.
Assume now that $E$ is nonempty. Since $f: E \to \mathbb{R}$ is bounded, there exists a number $M \in [0,\infty)$ such that $|f(x)| \le M$ for every $x \in E$. The set
\begin{align*}
A_f := \{|f(x)| : x \in E\}
\end{align*}
is therefore a nonempty subset of $[0,\infty)$ bounded above by $M$. Hence $\sup A_f$ exists in $[0,\infty)$, and by definition
\begin{align*}
\|f\|_\infty = \sup A_f \in [0,\infty).
\end{align*}
[/step]
[step:Prove definiteness by comparing point values to the supremum]
Let $0_E: E \to \mathbb{R}$ denote the zero function, defined by $0_E(x)=0$ for every $x \in E$.
If $f = 0_E$, then $|f(x)| = 0$ for every $x \in E$, and the definition of $\|\cdot\|_\infty$ gives $\|f\|_\infty = 0$.
Conversely, suppose $\|f\|_\infty = 0$. If $E$ is empty, then $f = 0_E$ because there is only one function from $\varnothing$ to $\mathbb{R}$. If $E$ is nonempty, then for every $x \in E$,
\begin{align*}
0 \le |f(x)| \le \sup_{y \in E} |f(y)| = \|f\|_\infty = 0.
\end{align*}
Thus $|f(x)|=0$ for every $x \in E$, so $f(x)=0$ for every $x \in E$. Therefore $f=0_E$.
[/step]
[step:Prove absolute homogeneity from the pointwise absolute value identity]
Let $f \in B(E)$ and let $\lambda \in \mathbb{R}$.
If $E$ is empty, then both $\|\lambda f\|_\infty$ and $\|f\|_\infty$ are equal to $0$, so
\begin{align*}
\|\lambda f\|_\infty = |\lambda|\|f\|_\infty.
\end{align*}
Assume now that $E$ is nonempty. For every $x \in E$, the absolute value product rule gives
\begin{align*}
|(\lambda f)(x)| = |\lambda f(x)| = |\lambda||f(x)|.
\end{align*}
If $\lambda = 0$, then $|(\lambda f)(x)|=0$ for every $x \in E$, and hence $\|\lambda f\|_\infty = 0 = |\lambda|\|f\|_\infty$.
If $\lambda \ne 0$, then $|\lambda|>0$. Multiplication by the positive number $|\lambda|$ preserves suprema of bounded nonempty subsets of $\mathbb{R}$, so
\begin{align*}
\|\lambda f\|_\infty = \sup_{x \in E} |\lambda||f(x)| = |\lambda|\sup_{x \in E}|f(x)| = |\lambda|\|f\|_\infty.
\end{align*}
[/step]
[step:Prove the triangle inequality from the pointwise triangle inequality]
Let $f,g \in B(E)$.
If $E$ is empty, then $\|f+g\|_\infty=0$, $\|f\|_\infty=0$, and $\|g\|_\infty=0$, so the desired inequality holds.
Assume now that $E$ is nonempty. For every $x \in E$, the pointwise triangle inequality in $\mathbb{R}$ gives
\begin{align*}
|(f+g)(x)| = |f(x)+g(x)| \le |f(x)|+|g(x)|.
\end{align*}
By the definition of the supremum, for every $x \in E$,
\begin{align*}
|f(x)| \le \|f\|_\infty
\end{align*}
and
\begin{align*}
|g(x)| \le \|g\|_\infty.
\end{align*}
Combining these inequalities gives, for every $x \in E$,
\begin{align*}
|(f+g)(x)| \le \|f\|_\infty + \|g\|_\infty.
\end{align*}
Thus $\|f\|_\infty + \|g\|_\infty$ is an upper bound for the set $\{|(f+g)(x)| : x \in E\}$. Taking the least upper bound yields
\begin{align*}
\|f+g\|_\infty \le \|f\|_\infty + \|g\|_\infty.
\end{align*}
[guided]
The triangle inequality for the uniform norm is obtained by first proving the corresponding estimate at each point of $E$, and only then taking the supremum. Let $f,g \in B(E)$.
If $E$ is empty, the convention in the statement gives
\begin{align*}
\|f+g\|_\infty = \|f\|_\infty = \|g\|_\infty = 0,
\end{align*}
so
\begin{align*}
\|f+g\|_\infty \le \|f\|_\infty + \|g\|_\infty.
\end{align*}
Now suppose $E$ is nonempty. Fix an arbitrary point $x \in E$. Since addition in $B(E)$ is pointwise, we have
\begin{align*}
(f+g)(x)=f(x)+g(x).
\end{align*}
Applying the triangle inequality for the absolute value on $\mathbb{R}$ gives
\begin{align*}
|(f+g)(x)| = |f(x)+g(x)| \le |f(x)|+|g(x)|.
\end{align*}
The purpose of the supremum is to provide a single bound valid for every point. By definition,
\begin{align*}
\|f\|_\infty = \sup_{y \in E}|f(y)|
\end{align*}
and
\begin{align*}
\|g\|_\infty = \sup_{y \in E}|g(y)|.
\end{align*}
Therefore the supremum property gives
\begin{align*}
|f(x)| \le \|f\|_\infty
\end{align*}
and
\begin{align*}
|g(x)| \le \|g\|_\infty.
\end{align*}
Substituting these two bounds into the pointwise triangle inequality gives
\begin{align*}
|(f+g)(x)| \le \|f\|_\infty + \|g\|_\infty.
\end{align*}
Because the point $x \in E$ was arbitrary, the number $\|f\|_\infty+\|g\|_\infty$ is an upper bound for every element of the set
\begin{align*}
\{|(f+g)(x)| : x \in E\}.
\end{align*}
The supremum is the least upper bound, so it is no larger than any particular upper bound. Hence
\begin{align*}
\|f+g\|_\infty = \sup_{x \in E}|(f+g)(x)| \le \|f\|_\infty + \|g\|_\infty.
\end{align*}
[/guided]
[/step]
[step:Conclude that the uniform norm is a norm on $B(E)$]
The previous steps show that $\|\cdot\|_\infty$ maps $B(E)$ into $[0,\infty)$, is definite, is absolutely homogeneous under real scalar multiplication, and satisfies the triangle inequality. Therefore $\|\cdot\|_\infty$ is a norm on the real [vector space](/page/Vector%20Space) $B(E)$.
[/step]
