Let $F: \mathbb R^n \to \mathbb R^m$ be smooth and such that for all $a \in \mathbb R$, $x \in \mathbb R^n$, $F(ax) = aF(x)$. Prove $F$ is linear

Question

Let $F: \mathbb R^n \to \mathbb R^m$ be smooth and such that for all $a \in \mathbb R$, $x \in \mathbb R^n$, $F(ax) = aF(x)$. Prove $F$ is linear.

I am sorry to say this, but I am really stuck on this problem. I tried to write $F(x + y) = 2^k F(\frac{x + y}{2^k})$ for any $k \geq 0$. Once we show the linearity for rational $n$-tuples, we would be done by smoothness. However, I am stuck on this. Can anyone help?

Answer 1

It is enough to consider the case $m = 1$. Taking gradients, we obtain $$ a(\nabla F)(ax) = a (\nabla F)(x) $$ for all $a \in \mathbb{R}$. Hence $$ (\nabla F)(ax) = (\nabla F)(x) $$ for all $a \ne 0$. Letting $a \to 0$ gives $$ (\nabla F)(0) = (\nabla F)(x), $$ i.e. the gradient of $F$ is constant. It follows that $F$ is linear.