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I have the following problem where I want to show the local strong convexity based on some assumptions. Let $l(\theta, X)$ be a function with $\theta \in \mathbb{R}^d$ and $X$ a real-valued random variable. Define $M = E[l(\theta, X)]$. Let $\theta_0$ be the minimizer of $M$. Further suppose:

  1. $l(\theta, X)$ is three times continuously differentiable.
  2. $l(\theta, X)$ is $L(X)$-Lipschitz in a neighborhood of $\theta_0$.
  3. $EL(X)^2<\infty$.

I want to show that in the neighborhood of $\theta_0$ we have that $$M(\theta) \geq M(\theta_0) + \lambda ||\theta- \theta_0||^2$$ for some $\lambda$.

I think the question is related to the following previous posts:

  1. Is a smooth function convex near a local minimum?
  2. Details on equivalent strong convexity
Renat Sergazinov
  • 322
  • Are you satisfied with $\lambda = 0$, when the result is obvious from the definition of $\theta_{\circ}$? If you want $\lambda > 0$, then $l(\theta, X) = 0$ satisfies all your conditions regardless of $X$ and the resulting $M$ is not locally strongly convex. – tal Apr 15 '21 at 09:58
  • I think I can see where you are pointing. If the $M(\theta)$ is constant, then the only $\lambda$ we can choose is $0$. However, what if we assume that $M(\theta) > M(\theta_0)$ in the neighborhood of $\theta_0$? – Renat Sergazinov Apr 15 '21 at 15:28
  • That would also not be sufficient. For example, if $M(\theta) = \Vert \theta\Vert_2^4$. You more or less need the hessian of $M$ at $\theta_0$ to be positive definite and then you can use Taylor's theorem. – tal Apr 16 '21 at 12:47
  • That makes sense. After playing with counterexamples for a day, I think requiring p.d. of the hessian at the minimizer is the only way to go. – Renat Sergazinov Apr 16 '21 at 16:41

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