I need to show that if $1<p<\infty$, then the unit ball is strictly convex in $L^p$, that is,
$||\lambda x+(1-\lambda)y|| < 1$ whenever $||f|| = ||g||=1$ and $\lambda \in (0,1)$.
I tried Minkwoski's inequality, but that only yields convexity, not strict. I also never used the fact that $p$ cannot be $1$ or $\infty$.
Speaking of which, why is it not true for those values (unless the spaces are singletons of course)?
EDIT: I know this is a duplicate, but the other post contains 2 incorrect answers only.
Best Answer
For $p \in (1,\infty)$ equality in Minkowski's inequality
$$\|f+g\|_p \leq \|f\|_p+\|g\|_p$$
holds if, and only if, $f = \alpha \cdot g$ for some constant $\alpha \in \mathbb{R}$; see this question. This means that equality in
$$\|\lambda f+ (1-\lambda)g\|_p \leq \lambda \|f\|_p+(1-\lambda)\|g\|_p = 1$$
holds if, and only if, $f = \alpha \cdot g$ for some constant $\alpha \in \mathbb{R}$. As $\|f\|_p = \|g\|_p=1$, we see that $|\alpha|=1$. Therefore, we find that
$$\|\lambda f+(1-\lambda) g\|_p < 1$$
whenenver $f \neq g$, $\|g\|_p=\|f\|_p=1$ and $\lambda \in (0,1)$.