Why is the supremum of two functions smaller than the individual suprema

Question

Let $X$ denote a topological space, and write $C^b(X)$ for the algebra of bounded, continuous functions on $X$. Why does it then hold that $\left|fg\right|_{X} \leq \left|f\right|_{X} \ \left|g\right|_{X}$, where $\left|\ \cdot \ \right|_{X} := \text{sup}\{ |f(x)| \ : \ x\in X \}$.

EDIT: I actually doubt that this holds, but how can $C^b(X)$ be an algebra then? Counter example: $X := [0, 10], f(x) := \sin(x), g(x):= x^2$. I then used a graphics program (namely, GeoGebra) to see that $\text{sup}\{|\sin(x)\cdot x^2| : x\in X\} = 63.63$, $\text{sup}\{|\sin(x)| : x\in X\} = 1, \text{sup}\{|x^2|: x\in X\}= 100$. I am confused ...

Kind regards, MathIsFun

Answer 1

Just note that $|f(x)g(x)|=|f(x)||g(x)| \leq |f|_X|g|_X$ for all $x \in X$. So $|f|_X|g|_X$ is an upper bound for the set $\{|f(x)g(x)| \mid x \in X\}$. This means that $|fg|_X \leq |f|_X|g|_X$.