$\begingroup$

Let $f,g:[0,1]\to \Bbb{R}$ be continuous functions. Show that $$||fg||\le||f||\space||g||$$ What I have got so far :

$|f(x)| \le\max|f(x)|=$ norm of $f$, $||f||$.$\forall x\in[0,1]$. (Note: I have replaced supremum with maximum.)

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

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

As I have to show that $||fg||\le||f||\space||g||$ :

Can I say $|f(x) g(x)|=||fg||$? I'm not sure about that Because $|f(x)g(x)| \le\max|f(x)g(x)|=\max|f(x)| \space \max|g(x)|$

I feel I am missing concept to prove $|f(x) g(x)|=||fg||$, through which I think I finally can prove $||fg||\le||f||\space||g||$

please If you guys could clarify.