Prove the following statements. (a) $\|fg\|_u \le \|f\|_u\|g\|_u$; (b) $\sup_{n\in N} \|f_n\|_u < \infty$ Suppose that $f_n, g_n, f, g:\mathbb R^d→\mathbb R$ are all bounded functions, and suppose further that $f_n→f$ uniformly and $g_n→g$ uniformly. Prove the following statements.
(a) $\|fg\|_u \le \|f\|_u\|g\|_u$
(b) $\sup_{n\in N} \|f_n\|_u < \infty$
Note on (b): For each $n$ we know that $w_n=\|f_n\|_u$ is a finite number, because $f_n$ is a bounded function. You have to prove that $\{w_n:n\in\mathbb N\}$ is a bounded set of numbers.
By definition, $\|f\|_u=\sup\{|f(x)|: x \in\mathbb  R^d\}$. So for (a), $\|fg\|_u=\sup\{|f(x)g(y)|:x,y \in\mathbb  R^d\}$, and I have to show it is less than or equal to $\sup\{|f(x)|\}\sup\{|g(y)|\}$. It seems really simple but I can't connect the dots. As for b), isn't it trivial that if the largest $|f_n|$ is bounded, then $w_n$ can't go to infinity?
 A: For (a) you have the bound
$$|(fg)(x)| = |f(x)||g(x)| \leq \|f\| \cdot \|g\|.$$
Taking the $\sup$ on both sides gives $\|fg\| \leq \|f\| \cdot \|g\|$.
For (b) you have that there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $\|f_n - f\| \leq 1$. By reverse triangle inequality, this means $\|f_n\| \leq \|f\| + 1$ for $n \geq N$. This implies $\sup_{n \geq N}\|f_n\| < \infty$, which implies $\sup_{n \geq 1}\|f_n\| < \infty$.
A: $$
\|f_n\|_u = \|(f_n-f) + f\|_u \le \|f_n-f\| + \|f\|_u
$$
The inequality here is an instance of the triangle inequality. The last term does not change as $n$ changes. The first term after the $\text{“}\le\text{”}$ is bounded because as a function of $n,$ it is a convergent sequence, approaching $0.$
Your question states that $\|fg\|_u = \sup\{|f(x)||g(y)| : x,y\in\mathbb R^d\},$ but according to the definition you stated, it should be $\|fg\|_u = \sup\{|f(x)g(x)| : x,y\in\mathbb R^d\}.$
You have $|f(x)g(x)| = |f(x)||g(x)|.$ The first factor $|f(x)|$ does not exceed $\|f\|_u$ and the second factor $|g(x)|$ does not exceed $\|g\|_u,$ so their product does not exceed $\|f\|_u\|g\|_u.$
