converse of Euler’s homogeneous function theorem

Theorem.  If the function $f$ of the real variables $x_1,\mathrm{\dots },x_k$ satisfies the identity

$x_1{\displaystyle \frac{\partial f}{\partial x_1}}+\mathrm{\dots }+x_k{\displaystyle \frac{\partial f}{\partial x_k}}=nf,$ (1)

then $f$ is a homogeneous function of degree $n$.

Proof.  Let  $f(t{x}_1,\mathrm{\dots },t{x}_k):=\phi (t)$.  Differentiating with respect to $t$ we obtain

$${\phi }^{\prime }(t)=x_1{f}_{x_1}^{\prime }(t{x}_1,\mathrm{\dots },t{x}_k)+\mathrm{\dots }+x_k{f}_{x_k}^{\prime }(t{x}_1,\mathrm{\dots },t{x}_k)=\frac{1}{t}[t{x}_1{f}_{x_1}^{\prime }(t{x}_1,\mathrm{\dots },t{x}_k)+\mathrm{\dots }+t{x}_k{f}_{x_k}^{\prime }(t{x}_1,\mathrm{\dots },t{x}_k)],$$

which by (1) may be written

$${\phi }^{\prime }(t)=\frac{n}{t}f(t{x}_1,\mathrm{\dots },t{x}_k)=\frac{n}{t}\phi (t).$$

Accordingly,

$$\frac{{\phi }^{\prime }(t)}{\phi (t)}=\frac{n}{t},$$

which implies the integrated form

$$\ln |\phi (t)|=\ln t^n+\ln C$$

for any positive $t$.  Thus we have  $\phi (t)=C{t}^n$,  where $C$ is independent on $t$.  Choosing  $t=1$  we see that  $C=\phi (1)$,  and therefore  $\phi (t)=t^n\phi (1)$.  This last equation means that

$$f(t{x}_1,\mathrm{\dots },t{x}_k)=t^{n}f(x_1,\mathrm{\dots },x_k)$$

saying that $f$ is a (positively) homogeneous function of degree $n$.

Title	converse of Euler’s homogeneous function theorem
Canonical name	ConverseOfEulersHomogeneousFunctionTheorem
Date of creation	2013-03-22 18:07:56
Last modified on	2013-03-22 18:07:56
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	7
Author	pahio (2872)
Entry type	Theorem
Classification	msc 26B12
Classification	msc 26A06
Classification	msc 15-00
Synonym	converse of Euler’s theorem on homogeneous functions
Related topic	ConverseTheorem
Related topic	ChainRuleSeveralVariables
Related topic	Logarithm