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I'm trying to show that this is true:

Let $X$ be a set and suppose $f$ and $g$ are bounded (real-valued) functions defined on $X$. Then,

$$ \sup_{x \in X}|f(x)g(x)| \leq \sup_{x \in X}|f(x)|\sup_{x \in X}|g(x)| $$

I think I'm pretty close but I'm not sure about the last step. First, since $f$ and $g$ are bounded, all involved suprema exist and are finite. If $a = \sup|f(x)|$ and $b = \sup|f(x)|$ then it is true that $$ a \geq |f(x)| \;\;\;\;\;\; b \geq |g(x)| $$ for every $x \in X$. Since none of the quantities involved are negative, this implies $$ a b \geq |f(x)|\cdot |g(x)| \implies \sup|f(x)|\sup|g(x)| \geq |f(x)|\cdot |g(x)| $$

Can I say now that since this last inequality holds for all $x$ that $$ \sup_{x \in X}|f(x)g(x)| = \sup_{x \in X} \left(|f(x)|\cdot |g(x)|\right) \leq \sup_{x \in X}|f(x)|\sup_{x \in X}|g(x)|? $$


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Yes: note that for every $x$ you have $|f(x)g(x)|\leq ab$; since $ab$ is an upper bound, the supremum, which is the least upper bound, must be less than or equal to $ab$. –  Arturo Magidin Mar 5 '12 at 2:03
Ok, thanks for the tip –  AFX Mar 5 '12 at 2:12
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marked as duplicate by vonbrand, Davide Giraudo, Yiorgos S. Smyrlis, Daniel Rust, user127.0.0.1 Mar 7 at 17:06

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2 Answers

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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 $$\sup_x |f(x)g(x)|\le \sup_x |f(x)| \sup_x|g(x)|.$$

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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.

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