When $1
I use parallelogram law
$||f+g||^2+||f-g||^2=||f||^2+||g||^2=4\\$
Since $f\ne g$, $||f-g||^2>0$
then$||f+g||_p<1$
But my proof does not use $1
I guess is this because the step Since $f\ne g$,$||f-g||^2>0$ ?I just use the definition 5.2 in rudin's book (c)$||x||=0 $ implies $x=0$.