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I want to ask some naive questions about the convexity of the functional. Consider $I:E \rightarrow \mathbb{R}$.

  1. The definition of the convexity: is it given by the second derivative that for every $u \in E$ $$\frac{d^2 I(u+tv)}{dt^2} \ge 0$$ at $t=0$ for any $v \in E$, or it's like the definition for the function case $$I(\alpha u + (1-\alpha)v) \le \alpha I(u)+(1-\alpha)I(v)?$$

  2. About the type of its critical point, if a functional is convex, then the critical point must be the saddle point or global minimum, right?

  3. For the simplest Poisson equation $\Delta u =f$, it’s the critical point of $I(u)=\int | \nabla u |^2 + fu dx$, it’s convex, does this mean that the critical point could only be the global minimizer or saddle? Because the other possibility (such as local minimum or maximum) will all change its convexity.

Elio Li
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    $I(x) = |x|$ is a nice convex function without second derivatives, so the second definition wins. – daw Jan 30 '24 at 09:43
  • About 3. $I(u)$ cannot be concave: $\nabla u$ is linear in $u,.$ Hence $|\nabla u|$ and $|\nabla u|^2$ are convex in $u,.$ Since $fu$ is linear in $u$ it follows that $I(u)=\int|\nabla u|^2+fu,dx$ is convex in $u,.$ – Kurt G. Jan 30 '24 at 13:33
  • @KurtG. , thanks, I carelessly made a mistake. – Elio Li Jan 30 '24 at 16:00

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