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Feb 16, 2018 at 21:50 vote accept CommunityBot
Feb 16, 2018 at 8:47 comment added RaphaelB4 (a=4*3=12.) Yes, with positive probabily there exists $\lambda<0$, this particular example don't work otherwise. Insteed of $\|x\|^4$, one could have also taken $f(x)=F(\| x \|)$, with $F$ has a surjective seconde derivative.
Feb 15, 2018 at 21:42 comment added lcv What is $a$? (this comment was otherwise too short)
Feb 15, 2018 at 17:59 comment added user31317 Hi, what I don't understand is that you assume $\lambda, \alpha$ have opposite signs with positive probability? $t$ is a real so $\frac{-\lambda}{\alpha ||v||^2}>0$.
Feb 15, 2018 at 16:12 history edited RaphaelB4 CC BY-SA 3.0
added 100 characters in body
Feb 15, 2018 at 16:02 history edited RaphaelB4 CC BY-SA 3.0
added 100 characters in body
Feb 12, 2018 at 8:37 history answered RaphaelB4 CC BY-SA 3.0