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Let $F: R^n \longrightarrow R$ be twice differentiable and $x,y \in R^n$ with $F(x)=F(y)$. Further let $\phi [0,1] \rightarrow R^n$ be a nice curve with $\phi(0)=x$ and $\phi(1)=y$. If we know that $F(\phi(t))=F(x)$ for all $t\in [0,1]$. What does this imply for the Hessians at the points $\phi(t)$ where $t \in (0,1)$? Are the equivalent to the Hessian of F at x?

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1 Answer 1

Certainly not. The conditions only determine $F$ along the curve, which might be e.g. a straight line. The function can do whatever it wants off the curve. For example, $F(x,y)=x^2y^2$ is constant on the $x$-axis, but $F_{yy}=2x^2$ isn't.

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Hmm your example $F(x,y)$ is not constant on the x-aches, but on the curve $x\cdot y=1$?! But I guess I see the problem. Thanx! –  Benedikt Dec 14 '12 at 10:11
    
@Benedikt: Assuming that you're referring to the $x$-axis: Sure it's constant there: $F(x,0)=x^2\cdot0^2=0$. There's no contradiction with also being constant on many other curves. –  joriki Dec 14 '12 at 10:23
    
Yes, sorry. I realized this too now when I had my first morning coffee.. ;-) You never should post a question and a comment without coffee –  Benedikt Dec 14 '12 at 10:52
    
But is there something that they still share? Like the Trace or one common eigenvalue? –  Benedikt Dec 14 '12 at 10:58
    
But what happens if I know that the gradient is zero all along the curve? –  Benedikt Dec 14 '12 at 19:47

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