Supremum Norm Identity

Question

Let $f,g : [0,1] → \mathbb{R}$ be bounded functions. Prove that $ ∥fg∥ ≤ ∥f ∥∥g ∥$

Where we define the sup norm of a bounded function $f : D → \mathbb{R}$ as $∥f∥ := sup${|f(x)| : x ∈ D}$ $.

Any ideas of where to start with this proof or theorems that are useful would be helpful.

Answer 1

$x \in [0,1];$

$|f(x)|\le \sup |f(x)|;$ $|g(x)|\le \sup |g(x)|;$

$|f(x)g(x)|=|f(x)| |g(x)|\le$

$ |f(x)| ||g|| \le ||f|| ||g||.$

$||f|| ||g||$ is an upper bound for $|f(x)g(x)|$.

$\sup |f(x)g(x)|$ is the smallest upper bound.

Hence

$||fg|| \le ||f||$ $||g||.$

Answer 2

$$\|f \cdot g\|_\infty=\sup_{x\in D} |f(x) \cdot g(x)|= \sup_{x\in D}(|f(x)|| g(x)|)\leq \sup_{x\in D}( \|f\|_\infty\cdot |g(x)|) = \|f\|_\infty \cdot \|g\|_\infty$$