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Is it possible to use Hölder's inequality prove that $|x|^\alpha$ $(\alpha>1)$ is strictly convex? Here $x\in\mathbb{R}^n$ and $|\cdot|$ is the Euclidean norm.

Strictly convex means $$|\lambda_1x_1+\lambda_2x_2|^\alpha<\lambda_1|x_1|^\alpha+\lambda_2|x_2|^\alpha,\forall x_1,x_2\in\mathbb{R}\text{ and }\lambda_1+\lambda_2=1$$

I understand how to use Hölder's inequality to prove that $|x|^\alpha$ $(\alpha>1)$ is convex, but not sure how to use it to establish the strict convexity. I guess it's because I don't fully understand the equality condition, i.e. when Hölder's equality holds. Here is the Hölder's inequality,

$$\sum_{i=1}^n |x_iy_i|\leq\left(\sum_{i=1}^n|x_i|^p\right)^{1/p}\left(\sum_{i=1}^n|y_i|^q\right)^{1/q},$$ where $1/p+1/q=1$ and $x_i,y_i\in\mathbb{R}$.

Or not using Hölder, how to show the strict convexity? I appreciate your help.

Tan
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    Note that the strict inequality for strict convexity should only be for all $x_1 \ne x_2$. Also, it seems that the linked answer has been updated to show strict convexity. The key in the proof is that Hölder's inequality is an equality if and only if $\alpha \lambda_i |x_i|^p = \beta \lambda_i$ for some $\alpha, \beta$ not both zero. (That is, $|\lambda^{1/p} x|^p$ and $|\lambda^{1/q}|^q$ are linearly dependent.) This is equivalent to requiring $|x| = |y|$. When combined with the triangle inequality step, they must have the same sign. – JKL Jan 27 '21 at 06:42

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