# Prove unit ball of $L^p(\mu)$ is strictly convex, when $1<p<\infty$

When $1<p<\infty$, $||f||_p=1$,$||g||_p=1$,$f\ne g$,then $\frac{1}{2} ||f+g||_p<1$.

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 $1<p<\infty$, 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 $f\ne g$,$||f-g||^2>0$ ?I just use the definition 5.2 in rudin's book (c)$||x||=0$ implies $x=0$.

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The parallelogram law is only valid on inner product spaces. $L^p$ when $p\neq 2$ is not an inner product space, so you'll have to use another method. –  icurays1 Nov 12 '12 at 23:24

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 $0< t< 1$, $1=(1-t) + t \geq |(1-t)f|_p+|tg|_p \geq |(1-t)f+tg|_p$