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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 $f(0)=h(x,y)=0,\ \forall x,y \in \Bbb{R}^2$.