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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.

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1 Answer 1

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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 $\lambda \in (0,1)$.

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  • $\begingroup$ What is an example of a pair of functions where this does not hold in $L^1$ and $L^{\infty}$? $\endgroup$ Commented May 2, 2014 at 0:07
  • $\begingroup$ @JohnnyApple $L^1$: Simply consider any two positive functions $f,g$. $L^{\infty}$: Consider two positive functions $f,g$ such that the $\|f\|_{\infty}$, $\|g\|_{\infty}$ is attained at some point $x \in \Omega$. $\endgroup$
    – saz
    Commented May 2, 2014 at 5:34

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