Differentiable function satisfying $x \cdot \nabla f(x) = 0$

Question

We have a differentiable function $f: \mathbb{R}^n \setminus \{0\} \rightarrow \mathbb{R}$ that satisfies $$ x \cdot \nabla f(x) = 0 \quad \forall x \in \mathbb{R}^n \setminus \{0\} $$

I am trying to prove two things

1. By fixing $x_0 \in \mathbb{R}^n$ show that $f(tx_0)=f(x_0) \quad \forall t \gt 0$
2. Show that if $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}^n$, differentiable and satisfies our dot product, then $f$ is constant on $\mathbb{R}^n$

I would normally say what I have tried but I don't know where to start. I tried expanding $x \cdot \nabla f(x) = 0$ using f's partial derivatives but to no luck. Any hints or help would be much appreciated.

Answer 1

Hint.

For the first thing: consider the function $g(t)=f(tx_0)$, and show that $g$ is a constant (prove that $g'\equiv0$).

For the second thing: take the limit $t\to0+$ in the equality $f(tx_0)=f(x_0)$.

The details of the first thing can be find in this post.

The details of the second thing:

Since $f:\mathbb R^n\to\mathbb R$ is continuous, we have $$\lim_{t\to0+}f(tx_0)=f(0)$$ for all $x_0\in \mathbb R^n$. Taking the limit $t\to0+$ in the equality $f(tx_0)=f(x_0)$ gives that $f(0)=f(x_0)$ for all $x_0$, hence $f$ is a constant.