The Supremum and Bounded Functions [duplicate]

Solution 1:

Put $M = \sup_x |f(x)|$. Then $$ |f(x)g(x)| \le M |g(x)| \le M \sup_x |g(x)|.$$ Since this holds for all $x$ we have $$\sup_x |f(x)g(x)|\le \sup_x |f(x)| \sup_x|g(x)|.$$

Solution 2:

This is related to this answer. This can be proven by taking the $\log$ of both sides and applying that result. Here I repeat that argument recast for products.

$$ \sup_{x\in A}(|f(x)g(x)|)\le\sup_{x\in A}|f(x)|\sup_{x\in A}|g(x)|\tag{1} $$

$(1)$ is an instance of the fact that the supremum over a set is no smaller than the supremum over a subset. The left hand side of $(1)$ is $$ \sup_{\substack{x,y\in A\\x=y}}(|f(x)|\,|f(y)|)\tag{2} $$ whereas the right hand side of $(1)$ is $$ \sup_{x,y\in A}(|f(x)|\,|f(y)|)\tag{3} $$ The set of $x$ and $y$ being considered in $(2)$ is a subset of the $x$ and $y$ being considered in $(3)$, so $(1)$ follows.