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I came across a statement which says that $$F(x+e t) = F(x) + ct $$, for all $x \in \mathbb R^m$ where $c$ is constant , and $e = (1,1,....,1)$ is equivalent to saying that $$\sum_{j=1}^m \frac {\partial^2 F}{\partial x_i \partial x_j}(x) =0 ..\forall i \in 1,...., m , x \in \mathbb R^m$$

Here $F$ is twice differentiable function .

Can someone help me to see that both of them are equivalent statements . Thanks

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up vote 1 down vote accepted

Using the chain-rule, we get $$ \sum_{j=1}^m\frac{\partial}{\partial x_j}F(x)=e\cdot\nabla F(x)=\frac{\partial}{\partial t}F(x+et)=c\tag{1} $$ Taking $\dfrac{\partial}{\partial x_i}$ of $(1)$ gives your result.

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