# Equivalences on the definition of strictly convex normed space

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.

• 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 $(0,1)$. May 15 at 13:01
• Oh! Now I see the problem. Thanks! May 15 at 14:08