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Does it hold that for every function $a \in C^\infty(\mathbb{R}^2,\mathbb{R})$ we have $a(x,y) = \int_0^1 \left(a(t\cdot (x,y)) + xt \frac{\partial a}{\partial x}(t \cdot(x,y)) + yt \frac{\partial a}{\partial y}(t \cdot(x,y)) \right) dt$

I am asking this because in an exercise I have tried to solve I managed to get to the right hand side, but I need to reach the left hand side.

1 Answers 1

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It is correct and $a\in C^1(\mathbb{R}^2,\mathbb{R})$ is enough. Let $f(t)=t\times a(t\cdot(x,y))$. Then $f$ is $C^1$ and $f'(t)=a(t\cdot(x,y))+xt\frac{\partial a}{\partial x}(t\cdot(x,y))+yt\frac{\partial a}{\partial y}(t\cdot(x,y)).$

Therefore, the right hand side of the equality to prove is $\int_0^1 f'(t)dt$, which equals to $f(1)-f(0)=a(x,y)$.