So basically I read someone else's answer to a question regarding Euler Homogeneous function theorem
source: https://quant.stackexchange.com/questions/8911/what-is-exactly-eulers-decomposition Also here https://planetmath.org/converseofeulershomogeneousfunctiontheorem
I tried to do the proof by hand and this is what I get using product rules etc.
Given $$\mathbf{x}\cdot \nabla{f(\mathbf{x})}\equiv kf(\mathbf{x})$$
Replace x by $\lambda x$ (The same as the letting $\phi(\lambda)$ step) $$\lambda \mathbf{x}\cdot \nabla{f(\lambda \mathbf{x})}\equiv kf(\lambda \mathbf{x})$$ Differentiating both sides wrt $\lambda$ $$\frac{d}{d\lambda}(\lambda \mathbf{x}\cdot \nabla{f(\lambda \mathbf{x})})\equiv \frac{d}{d\lambda}(kf(\lambda \mathbf{x}))$$ $$\mathbf{x}\cdot \nabla{f(\lambda \mathbf{x})}+\lambda \mathbf{x}\cdot \frac{d}{d\lambda}(\nabla{f(\mathbf{\lambda x})})\equiv k\frac{d}{d\lambda}(f(\lambda \mathbf{x}))$$
So basically the proof is identical except for this extra term $$\lambda \mathbf{x}\cdot \frac{d}{d\lambda}(\nabla{f(\mathbf{\lambda x})})$$ As we know the proof is true, what is the rationale/reasoning or constraint that forces this term to go to zero?
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Edit: Further attempt to simplify the problematic term using index notation:
$$P=\lambda \mathbf{x}\cdot \frac{d}{d\lambda}(\nabla{f(\mathbf{\lambda x})})$$ This can be rewritten using Einstein Notation (and for convenience and clarity, using $d_{\nu}$ to denote $\frac{d}{d\nu}$ and $\lambda_i$ to denote $\lambda x^i$ when it appeared as an index. Also $\lambda$ is NOT an index in this evaluation) as $$=\lambda x^id_{\lambda}(\partial_i f(\lambda x^j))$$ Using chain rule $$=\lambda x^id_{\lambda}(\lambda\partial_{\lambda_i} f(\lambda x^j))$$ $$=\lambda x^id_{\lambda}(\lambda\partial_{\lambda_i} f(\lambda x^j))$$ Now letting $y^{i'}=\lambda x^i$ and $z^{j'}=\lambda x^j$ then
$$=y^{i'}d_{\lambda}(\lambda\partial_{i'} f(z^{j'}))$$ $$=y^{i'}(\partial_{i'} f(z^{j'})+\lambda d_{\lambda}(\partial_{i'} f(z^{j'}))$$ Using chain rule again on $d_\lambda$ $$=y^{i'}(\partial_{i'} f(z^{j'})+\lambda (x^{k}\partial_{k'}\partial_{i'} f(z^{j'})+(\partial_{\lambda}x^k)\partial_{i'} f(z^{j'}))$$ Unpacking the expression by reverting to normal indices $$=\lambda x^i(\frac{1}{\lambda}\partial_{i} f(\lambda x^j)+\lambda (x^{k}\frac{1}{\lambda^{(2)}}\partial_{k}\partial_{i} f(\lambda x^j)+(\partial_{\lambda}x^k)\frac{1}{\lambda}\partial_{i} f(\lambda x^j))$$ $$= x^i\partial_{i} f(\lambda x^j)+ x^ix^{k}\partial_{k}\partial_{i} f(\lambda x^j)+\lambda x^i(\partial_{\lambda}x^k)\partial_{i} f(\lambda x^j)$$ Unpacking the expression by reverting to del operators $$= \mathbf{x}\cdot\nabla f(\lambda \mathbf{x})+ \mathbf{x}\cdot ( (\mathbf{x}\cdot\nabla)\otimes(\nabla f(\lambda \mathbf{x}))+\lambda \frac{\partial \mathbf{x}}{\partial\lambda}\otimes(\mathbf{x}\cdot\nabla) f(\lambda \mathbf{x})$$ The partial derivative is zero since $\mathbf{x}$ is independent of $\lambda$
Thus finally $$= \mathbf{x}\cdot\nabla f(\lambda \mathbf{x})+ \mathbf{x}\cdot ( (\mathbf{x}\cdot\nabla)\otimes(\nabla f(\lambda \mathbf{x}))$$
But this term is not necessary zero for all $\mathbf{x}$, so how does the proof of the converse of the Euler Homogeneous function theorem in the pics above got rid of it?
The desired equation is not true in general for homogeneous functions. As a counter-example, take $f(\mathbf{x})=|\mathbf{x}|^2$. Then $$\lambda \mathbf{x}\cdot\frac{d}{d\lambda}\nabla f(\lambda \mathbf{x})=\lambda \mathbf{x}\cdot\nabla |\mathbf{x}|^2 \frac{d(\lambda^2)}{d\lambda} =\lambda \mathbf{x}\cdot4\lambda \mathbf{x}=4\lambda^2 |\mathbf{x}|^2\neq 0$$ except when $\mathbf{x}=0$.