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Jan 7, 2018 at 1:05 comment added reuns Locally $g(x) = h \circ \phi(x)$ where $\phi$ is bi-smooth and $h(x)=h(0)+ a^\top x+ \frac12 x^\top B x$ is a quadratic polynomial. Then $\nabla_u h(x) = a^\top u+ x^\top B u$ and $\nabla_v g(x) = \nabla_{\Phi(x) v}h(x) =(a^\top + x^\top B) \Phi(x)v$ where $\Phi(x)_{ij} = \partial_j\phi(x)_i$. This should impose some conditions on $\nabla g$ (together with $\Phi(x)$ inversible and $\Phi(x)\cdot dx$ being an exact form) which when not satisfied means $g(x) = h \circ \phi(x)$ cannot be true globally.
Jan 6, 2018 at 13:18 comment added Johnny T. @reuns I am confused by your question...
Jan 5, 2018 at 11:59 history asked Johnny T. CC BY-SA 3.0