Final answer:
The question involves proving that the norm of the product of two bounded functions is less than or equal to the product of their norms. By using the definition of norms and properties of absolute values, it is demonstrated that |fg(x)| ≤ |f(x)||g(x)| ≤ ‖f‖‖g‖ for all x in the domain, leading to the conclusion that ‖fg‖ ≤ ‖f‖‖g‖.
Step-by-step explanation:
You are asked to prove that ‖fg‖ ≤ ‖f‖‖g‖ for bounded functions f and g. To start the proof, we define the norms (or supremum of absolute values) of the functions: ‖f‖ = sup and ‖g‖ = sup. These norms represent the greatest absolute value each function can attain within its domain. Now, considering the product fg, we see that at any point x in the domain D, the absolute value of the product |fg(x)| is less than or equal to |f(x)||g(x)| because the absolute value of a product is less than or equal to the product of the absolute values. Hence, we can say |fg(x)| ≤ |f(x)||g(x)| ≤ ‖f‖‖g‖. Since this holds for any x ∈ D, taking the supremum of the left-hand side over D gives us ‖fg‖ ≤ ‖f‖‖g‖. This completes the proof and ensures that the product of two bounded functions is also a bounded function, and its norm is less than or equal to the product of the individual norms.