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Putnam 2010 A3

Suppose that the function h \to \Bbb{R}^2 \to \Bbb{R} has continuous partial derivatives and satisfies the equation \displaystyle h(x,y)=a \frac{\partial h}{\partial x}(x,y)+b\frac{\partial h}{\partial y}(x,y) for some constants a,b. Prove that if there is a constant M such that |h(x,y) | \leq M for all (x,y) \in \Bbb{R}^2, then h is identically zero.

Putnam 2010 A3

Solution: If a=b=0 we have nothing to prove. Suppose at least one of them is not zero, and define f: \Bbb{R} \to \Bbb{R},\ f(t)=h(x+at,y+bt). Then it is easy to see that f^\prime(t)=\displaystyle a \frac{\partial h}{\partial x}(x+at,y+at)+b\frac{\partial h}{\partial y}(x+at,y+at)=h(x+at,y+at)=f(t). Since f^\prime (t)=f(t),\ \forall t \in \Bbb{R} and f is bounded, it must be the constant zero function. Hence f(0)=h(x,y)=0,\ \forall x,y \in \Bbb{R}^2.

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