# converse of Euler’s homogeneous function theorem

Theorem.  If the function $f$ of the real variables $x_{1},\,\ldots,\,x_{k}$ satisfies the identity

 $\displaystyle x_{1}\frac{\partial f}{\partial x_{1}}+\ldots+x_{k}\frac{% \partial f}{\partial x_{k}}=nf,$ (1)

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

Proof.  Let  $f(tx_{1},\,\ldots,\,tx_{k}):=\varphi(t)$.  Differentiating with respect to $t$ we obtain

 $\varphi^{\prime}(t)=x_{1}f^{\prime}_{x_{1}}(tx_{1},\,\ldots,\,tx_{k})+\ldots+x% _{k}f^{\prime}_{x_{k}}(tx_{1},\,\ldots,\,tx_{k})=\frac{1}{t}[tx_{1}f^{\prime}_% {x_{1}}(tx_{1},\,\ldots,\,tx_{k})+\ldots+tx_{k}f^{\prime}_{x_{k}}(tx_{1},\,% \ldots,\,tx_{k})],$

which by (1) may be written

 $\varphi^{\prime}(t)=\frac{n}{t}f(tx_{1},\,\ldots,\,tx_{k})=\frac{n}{t}\varphi(% t).$

Accordingly,

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

which implies the integrated form

 $\ln|\varphi(t)|=\ln{t^{n}}+\ln{C}$

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

 $f(tx_{1},\,\ldots,\,tx_{k})=t^{n}f(x_{1},\,\ldots,\,x_{k})$

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

## References

• 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset II.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1932).
 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