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?
Hessian equivalence
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real-analysis
differential-geometry
derivatives
1 Answers
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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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0But what happens if I know that the g$r$adient i$s$ zero all along $t$he curve? – 2012-12-14