[proofplan]
The proof is a direct comparison of pointwise bounds. Since $f$ and $g$ are bounded, their uniform norms are finite upper bounds for $|f(x)|$ and $|g(x)|$ at every point $x \in E$. Multiplying these pointwise inequalities gives a uniform bound for the pointwise product $fg$, which proves both boundedness of $fg$ and the stated supremum estimate.
[/proofplan]
[step:Use the uniform norms as pointwise upper bounds]
Define the pointwise product function $h: E \to \mathbb{R}$ by $h(x) := f(x)g(x)$.
Since $f \in B(E)$ and $g \in B(E)$, the quantities $M_f := \|f\|_\infty$ and $M_g := \|g\|_\infty$ are finite nonnegative [real numbers](/page/Real%20Numbers). By the defining upper-bound property of the supremum, for every $x \in E$, $|f(x)| \le M_f$ and $|g(x)| \le M_g$.
[guided]
We first translate the statement from norm language into pointwise language. Define the pointwise product function $h: E \to \mathbb{R}$ by $h(x) := f(x)g(x)$. This is a function from $E$ to $\mathbb{R}$ because $f(x)$ and $g(x)$ are real numbers for each $x \in E$, and the product of two real numbers is real.
Now define the two constants $M_f := \|f\|_\infty$ and $M_g := \|g\|_\infty$. Because $f,g \in B(E)$, both $f$ and $g$ are bounded functions, so $M_f$ and $M_g$ are finite nonnegative real numbers. The definition of the [uniform norm](/page/Uniform%20Norm) says that $M_f$ is the supremum of the set of values $|f(x)|$ with $x \in E$, and $M_g$ is the supremum of the set of values $|g(x)|$ with $x \in E$. Therefore each supremum is an upper bound for its corresponding set. Hence, for every $x \in E$, $|f(x)| \le M_f$ and $|g(x)| \le M_g$.
This is the only place where boundedness of $f$ and $g$ is used: it gives finite constants that control the two functions uniformly over the whole set $E$.
[/guided]
[/step]
[step:Multiply the pointwise bounds to bound the product uniformly]
For every $x \in E$, the absolute value is multiplicative on $\mathbb{R}$, and the constants $M_f$ and $M_g$ are nonnegative. Thus
\begin{align*}
|h(x)| = |f(x)g(x)| = |f(x)|\,|g(x)| \le M_f M_g.
\end{align*}
Hence $h$ is bounded on $E$, so $h \in B(E)$. Since $h=fg$ by definition, this proves $fg \in B(E)$.
[guided]
Fix an arbitrary point $x \in E$. The absolute value on $\mathbb{R}$ is multiplicative, so applying it to the real numbers $f(x)$ and $g(x)$ gives
\begin{align*}
|h(x)| = |f(x)g(x)| = |f(x)|\,|g(x)|.
\end{align*}
From the preceding step, $|f(x)| \le M_f$ and $|g(x)| \le M_g$. Since $M_f$ and $M_g$ are nonnegative real numbers, multiplication preserves the inequality, and therefore
\begin{align*}
|h(x)| \le M_f M_g.
\end{align*}
This bound holds for every $x \in E$, and the right-hand side does not depend on $x$. Thus $h: E \to \mathbb{R}$ is bounded. By the definition of $B(E)$ as the [vector space](/page/Vector%20Space) of bounded functions from $E$ to $\mathbb{R}$, this proves $h \in B(E)$. Since $h$ was defined by $h(x)=f(x)g(x)$ for every $x \in E$, it is exactly the pointwise product $fg$, so $fg \in B(E)$.
[/guided]
[/step]
[step:Take the supremum to obtain the norm inequality]
The preceding step shows that $M_fM_g$ is an upper bound for the set
\begin{align*}
\{|h(x)| : x \in E\}.
\end{align*}
Therefore the least-upper-bound property of the supremum gives
\begin{align*}
\|fg\|_\infty = \|h\|_\infty = \sup_{x \in E} |h(x)| \le M_fM_g = \|f\|_\infty \|g\|_\infty.
\end{align*}
If $E=\varnothing$, the same conclusion follows from the stated convention that the supremum of the empty set is $0$, so $\|fg\|_\infty=0$ and the displayed inequality remains valid. This completes the proof.
[guided]
The pointwise estimate from the preceding step says that $|h(x)| \le M_fM_g$ for every $x \in E$. Equivalently, $M_fM_g$ is an upper bound for the set of all absolute values of $h$:
\begin{align*}
\{|h(x)| : x \in E\}.
\end{align*}
The uniform norm of $h$ is the least upper bound of this set, with the convention from the theorem statement when $E$ is empty. Since the supremum is the least upper bound, it is no larger than any upper bound. Therefore
\begin{align*}
\|h\|_\infty = \sup_{x \in E} |h(x)| \le M_fM_g.
\end{align*}
Finally, $h=fg$ by the definition of $h$, while $M_f=\|f\|_\infty$ and $M_g=\|g\|_\infty$ by definition. Substituting these identities gives
\begin{align*}
\|fg\|_\infty = \|h\|_\infty \le M_fM_g = \|f\|_\infty\|g\|_\infty.
\end{align*}
If $E=\varnothing$, the convention in the theorem statement gives $\|fg\|_\infty=0$, so the same inequality holds. This proves the asserted norm estimate and completes the proof.
[/guided]
[/step]
