# Property of norm

Let $X$ be a compact Hausdorff space and let $C(X)$ denote the set of continuous complex valued functions on $X$. Define $$\|f\|:=\sup\{|f(x)|:x\in X\},$$

then prove that $\|fg\|\leq \|f\|\|g\|$.

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What are you struggling with? Do you understand the definition of the norm? – Michael Albanese Jul 7 '14 at 8:04
yes i know the definition but i do not know why it is inequality,it should be equality.But now i understand that how this is inequality.Thanks to everyone. – Nannes Jul 7 '14 at 9:06
In future, you should include such concerns in the post so people can address them. – Michael Albanese Jul 7 '14 at 9:10

For every $x \in X$ we have $|g(x)| \leq \sup_{z \in X} |g(z)|$ by definition of the supremum, so for every $x \in X$ we may observe that $$|f(x)g(x)| = |f(x)||g(x)|\leq |f(x)|\left(\sup_{z \in X}|g(z)|\right) =|f(x)|\|g\|,$$ Since this is true for every $x\in X$ we may take the supremum on both sides of the equation to get $$\|fg\| = \sup_{x \in X}|f(x)g(x)| \leq \sup_{x \in X} |f(x)|\|g\| = \|f\|\|g\|$$

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yes perfect.Thanx – Nannes Jul 7 '14 at 9:08

Alternatively, as $\{(x,x)| x\in X\} \subset \{(x,y)| x,y\in X\}$: $$\sup_{x\in X} |f(x)g(x)| \le \sup_{x,y\in X} |f(x)g(y)| = \sup_{x\in X}|f(x)| \sup_{y\in X}|g(y)|$$

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How do you justify $\sup_{x \in X} |f(x)g(x)| \leq \sup_{x,y \in X} |f(x)g(y)|$ ? – Surb Jul 7 '14 at 8:27
this is because $\{(x,x) | x\in X\}\subset X\times X$. – mookid Jul 7 '14 at 8:49
ok, it is really alternative then :). – Surb Jul 7 '14 at 8:52

For all $x \in X$ we have $$|f(x)| \leq \|f\| ,$$ as well as $$|g(x)| \leq \|g\|.$$ Multiplying the two inequalities gives

$$|f(x)g(x)| \leq \|f\|\|g\| \; \forall x \in X .$$

Thus $\|f\| \|g\|$ is an upper bound of the set $\{|f(x)g(x)|:x \in X\}$, hence the least upper bound can't be larger.

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