In the French Wikipedia page for strictly convex spaces it is stated that a normed space $(X, || \cdot ||)$ is strictly convex if and only if $|| \cdot ||^p$ is a strictly convex function for every $p>1$. However, isn't that always true?

Given $x, y \in X$ with $x \not = y$ and $\lambda \in (0,1)$ we have that for any $p>1$:

  • If $x=0$ then $$||\lambda x + (1-\lambda)y||^p = (1-\lambda)^p ||y||^p < (1-\lambda) ||y||^p = \lambda ||x||^p +(1-\lambda) ||y||^p$$
  • If $y=0$ then $$||\lambda x + (1-\lambda)y||^p = \lambda^p ||x||^p < \lambda ||x||^p = \lambda ||x||^p +(1-\lambda) ||y||^p$$
  • If $x, y \not =0$ then $$||\lambda x + (1-\lambda)y||^p \leq \left( \lambda ||x|| + (1-\lambda) ||y|| \right)^p < \lambda ||x||^p + (1-\lambda) ||y||^p$$ where the first inequeality is true because $f(t)=t^p$ is increasing in $[0,+\infty)$ and the second one because $f$ is strictly convex in $(0, +\infty)$.

Have I made any silly mistake in my proof or am I right?

EDITED: The problem with the proof is that in the last case it could happen that $\lambda ||x|| = (1-\lambda) ||y||$, so the strict convexity of the norm is needed to accomplish a strict inequality for the triangle inequality.

  • 2
    $\begingroup$ It is not true for all norms. For example, consider $\|(x,y)\| = |x| + |y|$ and take your two points in your convex combination to be $(1,0)$ and $(0,1)$. $\endgroup$ May 15 at 13:01
  • $\begingroup$ Oh! Now I see the problem. Thanks! $\endgroup$
    – Eparoh
    May 15 at 14:08


You must log in to answer this question.

Browse other questions tagged .