Given $U(\lambda x)=\frac{1}{\lambda}U(x)$, compute an expression for $\nabla U|_{\lambda x}$.

Question

Suppose $U:\mathbb{R}^n-\{0\}\rightarrow\mathbb{R}$ is a smooth function which satisfies for all nonzero $\lambda\in\mathbb{R}$

$U(\lambda x)=\frac{1}{\lambda}U(x)$.

Compute an expression for $\nabla U|_{\lambda x}$.

My attempt: I understand that the gradient of $U(x)$ will be some vector $(\frac{\partial U}{\partial x_1},\dots,\frac{\partial U}{\partial x_n})$, but I'm not sure how the given expression helps to find this. I think I'm mostly hung up on the fact that the function maps to $\mathbb{R}$ and not $\mathbb{R}^n$.

Any help appreciated!

Answer 1

Let $\lambda\neq 0$ be a fixed real number and $\varphi:x\mapsto \lambda x$.

$$d(U\circ \varphi)(x)(h)=dU(\varphi(x))(d\varphi(x)(h))=dU(\lambda x)(\lambda h)=\lambda dU(\lambda x)(h)=\lambda\langle\nabla U(\lambda x),h \rangle$$

Since $U\circ \varphi=\frac 1\lambda U$, we get $$\frac 1 \lambda dU(x)(h)=\lambda\langle\nabla U(\lambda x),h \rangle$$

that is $$\frac 1 \lambda \langle\nabla U( x),h \rangle = \lambda\langle\nabla U(\lambda x),h \rangle$$

$$\langle\nabla U(\lambda x),h \rangle = \langle \frac 1{\lambda^2} \nabla U( x),h \rangle$$

Hence $\displaystyle \nabla U(\lambda x)=\frac 1{\lambda^2} \nabla U( x)$