Euler's Theorem for homogeneous functions states that $f: \mathbb{R}^n \to \mathbb{R}$ is homogeneous of degree $k$ ($f(cx) = c^k f(x)$ for all $c > 0$, $x \in \mathbb{R}^n$), if and only if the partial derivatives of $f$ satisfy $$ k f(x) = \sum_{i=1}^n x_i \frac{\partial f(x)}{\partial x_i} $$ Clearly this theorem requires differentiability of $f$. However, it seems to often be stated with the requirement of continuous differentiability, e.g., here. On the other hand, it is also sometimes stated without this requirement, and I don't see how any of the proofs actually use continuity of partial derivatives (e.g., the proof here).

I would greatly appreciate if anyone could clear this up!


1 Answer 1


No, you don't need to assume continuous differentiability. In fact, the result can be proved point-wise, as follows.

Fix a point $x$ where $f$ is differentiable. Consider the function $g(c)=f(cx)$.

Homogeneity implies that if $f$ is differentiable at $x$, then it is differentiable at $tx$ for every $t\neq 0$, because \begin{eqnarray*} f(tx+h)-f(tx)&=&t^k(f(x+h/t)-f(x))\\ &=&t^{k}(f'(x)(h/t)+o(\|h\|_2/t))\\ &=&t^{k-1}f'(x)(h)+t^ko(\|h\|_2/t)\\ \end{eqnarray*} so that $f'(tx)=t^{k-1}f'(x)$ for every $t\neq 0$. The notation $f'(x)(w)$ in this context means the dot product between the gradient of $f$ at $x$ and the vector $w$.

It follows that $g$ is differentiable at every $c\neq 0$, and by the chain rule, $$g'(c)=f'(cx)(x)=c^{k-1}f'(x)(x)$$ On the other hand, since $g(c)=f(cx)=c^{k}f(x)$, direct differentiation gives $g'(c)=kc^{k-1}f(x)$, and we deduce the announced formula, $kf(x)=f'(x)(x)$.

All we needed was the chain-rule, which does not require continuous derivatives.


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