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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!

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  • $\begingroup$ One approach could be mean value theorem $\endgroup$ May 14 '19 at 8:26
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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.

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