# Uniformly convex implies strictly convex [closed]

How to prove Uniformly convex implies strictly convex

## closed as off-topic by RRL, Kavi Rama Murthy, Tianlalu, Alex Provost, Adrian KeisterApr 4 at 13:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Kavi Rama Murthy, Tianlalu, Alex Provost, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.

• What have you done so far? If you cannot start the proof you can look at any book that discusses these concepts. – Kavi Rama Murthy Apr 4 at 5:38
• @KaviRamaMurthy.. i seen but there are different definitions are there so im confusing –  SURYA Apr 4 at 5:48
• Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Alex Provost Apr 4 at 13:23

Since $$\delta > 0$$ in the definition of uniform convexity, it follows that $$\left\|\frac{1}{2}x + \frac{1}{2}y\right\| < 1$$ whenever $$\|x\| = \|y\| = 1$$, which is the definition of strict convexity in the special case for $$\lambda = \frac{1}{2}$$. As it turns out, verifying strict convexity for $$\lambda = \frac{1}{2}$$ (or indeed, for any one fixed value of $$\lambda \in (0, 1)$$) is sufficient to prove strict convexity in general.
To prove this, we start by assuming $$\lambda \in \left(0, \frac{1}{2}\right)$$. We have that $$\lambda x + (1 - \lambda) y = 2\lambda\left(\frac{x}{2} + \frac{y}{2}\right) + (1 - 2\lambda)y,$$ hence \begin{align*} \|\lambda x + (1 - \lambda) y\| &= \left\|2\lambda\left(\frac{x}{2} + \frac{y}{2}\right) + (1 - 2\lambda)y\right\| \\ &\le 2\lambda\left\|\frac{x}{2} + \frac{y}{2}\right\| + (1 - 2\lambda)\|y\| \\ &< 2\lambda + 1 - 2\lambda = 1. \end{align*} If $$\lambda \in \left(\frac{1}{2}, 1\right)$$ instead, a similar approach will work. Or, indeed, interchange the roles of $$x$$ and $$y$$, and replace $$\lambda$$ with $$1 - \lambda$$. So, in any case, we have $$\|\lambda x + (1 - \lambda)y\| < 1$$ for $$\lambda \in (0, 1)$$.