Do not understand setting lambda = 1 in proof of Euler's homogeneous function theorem

[clustro]

As the title suggests, I do not understand why people set lambda = 1 in proofs of Euler's homogeneous function theorem.

Euler's homogeneous function theorem is:

i. Definition of homogeneity.

Given a differentiable function, [itex]f(\vec x)[/itex], that function is said to be "homogeneous of degree k" if:

[itex]f(\alpha\vec x) = \alpha^kf(\vec x)[/itex].

ii. The theorem:

Given [itex]f(\vec x)[/itex], iff it satisfies (i), then:

[tex]kf(\vec x) = \vec x \cdot \nabla f(\alpha \vec x)[/tex]

I am studying this because it is used in thermodynamics. The use there is relatively simple. Any thermodynamics function is homogeneous of degree 1 with respect to its extensive variables; e.g. if I increase the mass by a factor of 2, then my energy has increased by a factor of 2 (keeping the intensive variables constant).

The proof in this pdf is representative of the proofs I found (its on first page): http://tinyurl.com/3ky2ud5


I do not understand why they are arbitrarily setting [itex]\lambda = 1[/itex]. I understand and see that it gives the correct results, but I should be able to scale my variables by any factor of [itex]\lambda[/itex] I want.

I guess my question is really, "Why is this a proof of a general theorem, and not a proof for a singular case where [itex]\lambda = 1[/itex]."

 

[clustro]

Oh, gee, sorry.

I kind of switched greek letters on you.

Wherever you see [itex]\alpha[/itex] in the first post, change it to [itex]\lambda[/itex].

 

[lurflurf]

That should be (without alpha)
k f=x.grad(f)
"Why is this a proof of a general theorem, and not a proof for a singular case where ."
You can use any scale you like, but it will make no difference in the end.
a=1 arises because in a derivative we consider
[f(x)-f(y)]/(x-y) with x close to y
a x is close to x when a is close to 1

x.grad(f)=lim_{a->1} [f(a x)-f(x)]/(a-1)
if f is homogeneous we have
lim_{a->0}[f(a x)-f(x)]/(a-1)=[lim_{a->1}(a^k-1^k)/(a-1)] f(x)=k f(x)