# Minimizing sequence $\implies$ Palais–Smale sequence

Set $$F:\mathbb{R}^n\rightarrow \mathbb{R}$$ a $$C^2$$-function that is bounded from below. Set $$x_n$$ a minimizing sequence, i.e., $$F(x_n)\to \alpha = \inf F$$. I want to prove that under the assumption of $$\lVert D^2 F\rVert$$ is bounded, $$x_n$$ is a Palais–Smale sequence, i.e., $$F(x_n)\to \alpha = \inf F$$ and $$\lVert DF(x_n)\rVert\to 0$$.

I have come across the variational Ekeland principle, with which I showed that for every minimizing sequence $$x_n$$ there exists another minimizing sequence $$y_n$$ close to $$x_n$$ in the sense that $$\lVert y_n-x_n\rVert\to 0$$ and $$\lVert DF(y_n)\rVert\rightarrow 0$$. It seems that this could be useful, but I am stuck. Any help is welcome!

• One approach could be mean value theorem May 14 '19 at 8:26

In fact it's just the MVT to $$DF$$: $$\|DF(x_n)-DF(y_n)\|\le \|x_n-y_n\|\sup\|D^2F\|=O(\|x_n-y_n\|)=o(1),$$ so $$\|DF(x_n)\|=o(1)$$ too.
$$*$$ Note that if $$D^2f$$ is not bounded, it is not true, and (among other reasons) that's why Ekeland's principle is useful. Take $$n=1$$ and e.g. $$f(x)= \frac{2+\sin x^2}{1+x^2}$$. Then $$f(x)>0$$ for all $$x$$, $$\inf f=0$$, and the minimizing sequences are exactly the diverging sequences $$x_n\to+\infty$$; however the sequence $$f'(x_n)$$ may have any limit or diverge.