Prove unit ball of $L^p(\mu)$ is strictly convex, when $1<p<\infty$ When $1&lt;p&lt;\infty$, $||f||_p=1$,$||g||_p=1$,$f\ne g$,then $\frac{1}{2} ||f+g||_p&lt;1$.

I use parallelogram law

$||f+g||^2+||f-g||^2=||f||^2+||g||^2=4\\$

Since $f\ne g$,
$||f-g||^2&gt;0$

then$||f+g||_p&lt;1$

But my proof does not use $1&lt;p&lt;\infty$, and this obviously fail when p is 1 or $\infty$.
I just want to know where my proof is wrong.

I guess is this because the step Since $f\ne g$,$||f-g||^2&gt;0$ ?I just use the definition 5.2 in rudin's book
(c)$||x||=0 $ implies $x=0$.
Note that norm satisfies the triangle inequality by the definition of norm, $|f|_p+|g|_p\geq |f+g|_p$

So if $|g|_p$, $|f|_p \leq 1$, then for $0&lt; t&lt; 1$, $
    1=(1-t) + t \geq |(1-t)f|_p+|tg|_p \geq |(1-t)f+tg|_p$
Note that norm satisfies the triangle inequality by the definition of norm, $|f|_p+|g|_p\geq |f+g|_p$

So if $|g|_p$, $|f|_p \leq 1$, then for $0&lt; t&lt; 1$, $
    1=(1-t) + t \geq |(1-t)f|_p+|tg|_p \geq |(1-t)f+tg|_p$