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A function $f:\mathbb{R}^n →\mathbb{R}$ is said to be homogeneous of degree $k$ ($k\in \mathbb{R},k>0$) if $f (t\mathbf{x}) = t^k f (\mathbf{x})$ for every $t ∈ \mathbb{R}, \mathbf{x} ∈ \mathbb{R}^n$. Show that if $f$ is homogeneous of degree $k$, then $\langle \nabla f(\mathbf{x}),\mathbf{x}\rangle = k f(\mathbf{x})$ for all $x ∈ \mathbb{R}^n$.

I tried that $\partial_t f(t\mathbf{x})=\partial_t t^kf(\mathbf{x})=kf(\mathbf{x})t^{k-1}=\langle\nabla f(t\mathbf{x}),t\rangle$, but how can I start from here to $\langle \nabla f(\mathbf{x}),\mathbf{x}\rangle=kf(\mathbf{x})$?

Thank you all.

2 Answers 2

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Hint: By chain rule: $$\frac{\partial f}{\partial t}(tx)=\left\langle \nabla f(tx), x\right\rangle.$$ Moreover $$\frac{\partial}{\partial t}(t^kf(x))=kt^{k-1}f(x).$$ Then set $t=1$ gives you the result.

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Starting with $f(tx)=t^kf(x)$ and differentiating both sides with respect to $t$, we get $$ \sum_{i=1}^n\frac{\partial f(tx)}{\partial x_i}x_i=kt^{k-1}f(x) $$ by the chain rule, and then if we set $t=1$ then the above becomes $$ \sum_{i=1}^n\frac{\partial f(x)}{\partial x_i}x_i=kf(x)$$ which is the desired result.

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    but why can we set $t=1$? by the definition of homogeneous, $t$ should be any real number. – 2017-02-08
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    The first equation is true for all $t$, which means that it remains true if we plug in a particular value of $t$. – 2017-02-08
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    Also for this function, do it has to be differentiable for us to differentiate it? – 2017-02-08
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    Since the problem involves the gradient of $f$, it seems safe to assume that $f$ is differentiable. – 2017-02-08