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There are several equivalent definitions for strongly convex.

For example, some literature said:

A function $f$ is strongly convex with modulus $c$ if either of the following holds

  1. $$f(\alpha x+(1-\alpha)x')\leq\alpha f(x)+(1-\alpha)f(x')-\frac{1}{2}c\alpha(1-\alpha)\|x-x'\|^2$$

  2. $f-\frac{c}{2}\|\cdot\|^2$ is convex.

I do not know how to prove the equivalence of the above statements. The difficulty here is that the norm is an arbitrary norm, not necessarily the $\ell_2$ norm.

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    I presume the fraction in the second definition is $\frac{c}{2}$ rather than $\frac{1}{2}$. Also, is the norm induces by an inner-product or not?2012-10-02
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    Yes, you are right. The fraction is $\frac{c}{2}$. As I said in the post, the norm is an arbitrary norm, not necessarily $\ell_2$. So the norm is not necessarily induced by the inner product.2012-10-02
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    Using each of the definitions you can write a lower bound for $\alpha f\left(x\right)+\left(1-\alpha\right)f\left(x'\right)-f\left(\alpha x+\left(1-\alpha\right)x'\right)$. To have the equivalence, I think these lower bounds must be identical. If you set $\alpha=\frac{1}{2}$ in the mentioned equation you get the parallelogram law, which implies that your space is equipped with an inner-product. Check if there's something assumed about the domain of $f$ like being a Hilbert space or an inner-product space.2012-10-02
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    The domain of $f$ is just linear space. No Hilbert space structure assumed.2012-10-03

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