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I was wondering if $|f(x)g(x)| = |f(x)| |(g(x)|$ is true all the time as in the case of real numbers.

I was not convinced enough that that was true.

But I can't think of any counterexample.

Thank you.

Rio Alvarado
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user256732
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  • I also believe it is a duplicate in the end, but one comment I think is in order is that $f(x)$ and $g(x)$ are, as I'm sure you're assuming, real numbers, so as you even pointed out in your question statement we have $|f(x)g(x)|=|f(x)|\cdot |g(x)|$. – Hayden Jul 25 '15 at 02:23
  • @Hayden: The wording "as in the case of real numbers" suggests the OP is asking if the equality also holds in a more general setting ("all the time"). It does hold in more general settings (such as in the case of complex numbers), and it is a property that we require of norms generally in settings like Banach algebras. Strictly speaking we need some context to say whether the equality is true or false. – hardmath Jul 25 '15 at 02:41
  • @hardmath Yes, you may be right. More context is definitely necessary. – Hayden Jul 25 '15 at 02:44

2 Answers2

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Hint: Its is enough to prove it for the squares of the absolute values. And $\lvert z\rvert^2=z\bar z$.

Bernard
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For each $x$, $f(x)$ and $g(x)$ are real (or complex) numbers. So the equality is precisely that of real (or complex) numbers.

Martin Argerami
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