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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)|? $

Thanks.

  • 0
    Ok, thanks for the tip2012-03-05

2 Answers 2

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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.