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Why is the last step (setting $\lambda = 1$ allowed?

I have trouble accepting this because if I set $\lambda =1 $ at the very start, then:

$f(\lambda x , \lambda y)=\lambda^r f(x,y)$


$f(x,y) = f(x,y)$


so I can't prove the theorem.

Why does setting $\lambda = 1$ at the end of the proof work?


For example, let $f(x,y)=x^2y^2$

Therefore, $f(x,y)$ is homogeneous of degree 4.

Therefore, the second to last step will be:


The above equation is only true if $\lambda=1$

Why doesn't the theorem make a qualification that $\lambda$ must be equal to 1?

It seems to me that this theorem is saying that there is a special relationship between the derivatives of a homogenous function and its degree but this relationship holds only when $\lambda=1$. Please correct me if my observation is wrong.

  • 1
    $\begingroup$ $\lambda$ could be any reals. $\endgroup$
    – Mikasa
    Commented Jan 30, 2014 at 17:36
  • $\begingroup$ The author is fixing $x$ and $y$ and viewing $g(\lambda)=f(\lambda x, \lambda y)$ as a function of $\lambda$. They then calculate $g'$, and at the end, consider the value $g'(1)$. $\endgroup$
    – Jack M
    Commented Oct 23, 2016 at 17:58

6 Answers 6


You can't set $\lambda = 1$ in the line $f(\lambda x,\lambda y) = \lambda^r f(x,y)$ because the very next step is to differentiate with respect to $\lambda$ which makes no sense when the variable isn't present.

At the end of the proof, you are taking advantage of the fact that $\lambda$ is an arbitrary element of $\mathbb{R}$. So you pick an element that makes the calculations easy.

Variant of proof

Define the function $g : \mathbb{R} \rightarrow \mathbb{R}$ by $g(t) = f(tx,ty)$. Since $f$ is homogeneous, we can write $g(t) = t^r f(x,y)$. Find $g'(t)$.

Using $g(t) = t^r f(x,y)$, it is clear that $g'(t) = rt^{r-1} f(x,y)$.

Using $g(t) = f(tx,ty)$, we get that $g'(t) = \frac{\partial f}{\partial (tx)}\cdot\frac{d(tx)}{dt} + \frac{\partial f}{\partial (ty)}\cdot\frac{d(ty)}{dt} = x\frac{\partial f}{\partial (tx)}+y\frac{\partial f}{\partial (ty)}$.

So we have that for all $t$, $rt^{r-1} f(x,y) = x\frac{\partial f}{\partial (tx)} + y\frac{\partial f}{\partial (ty)}$. If we let $t=1$, then we have that $g(1) = f(x,y)$, our original function, and $rf(x,y) = x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}$, the desired result.

To address the specific example of $f(x,y) = x^2y^2$. You have the second to last step wrong on the RHS.

When doing the proof, we are working with $f(\lambda x, \lambda y)$ throughout. So when calculating $\dfrac{df}{d\lambda}$ we get:

$\begin{align} \dfrac{df}{d\lambda} &= \dfrac{\partial f}{\partial (\lambda x)}\cdot \dfrac{d (\lambda x)}{d\lambda} + \dfrac{\partial f}{\partial (\lambda y)}\cdot \dfrac{d (\lambda y)}{d\lambda} \\ &= 2(\lambda x)(\lambda y)^2\cdot x + 2(\lambda x)^2(\lambda y)\cdot y \\ &= 4\lambda^3 x^2y^2 = \dfrac{d}{d\lambda}\left(\lambda^4 f(x,y)\right) \end{align}$

So we could choose any $\lambda$ we want and it would still be a true equation. But then to get the desired result, we would have to divide that back out of both sides. Choosing $\lambda = 1$ saves a bit of algebra.

  • $\begingroup$ Yes. the differentiation would make nothing there when it became $1$ and that's why the OP said ...I can't prove.... :-) $\endgroup$
    – Mikasa
    Commented Jan 30, 2014 at 17:38
  • $\begingroup$ @John Habert But if I assume at the end that $\lambda=1$ at the end, why doesn't that immediately make $\frac{du}{d\lambda}=0$ and $\frac{dv}{d\lambda}=0$ in the preceding step? $\endgroup$
    – mauna
    Commented Jan 30, 2014 at 23:25
  • $\begingroup$ For the same reason that if you want to find the derivative of the line $y=x$ at the point $x=1$ that you don't substitute first. You take the derivative and then evaluate. I'm going to edit in a slight variant of the proof that might help. $\endgroup$ Commented Jan 31, 2014 at 4:04
  • $\begingroup$ @JohnHabert Thank you for your time and effort. Your variant of the proof does help to clear some of my doubts. I've edited my question highlight the specific area I don't understand. Would be great if you could address them in your answer too. $\endgroup$
    – mauna
    Commented Feb 2, 2014 at 0:25

You can derive Euler theorem without imposing $\lambda=1$.

Starting from $f(\lambda x, \lambda y) = \lambda^n \times f(x,y)$, one can write the differentials of the LHS and RHS of this equation:

  • LHS

$df(\lambda x, \lambda y) = \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} \times d(\lambda x) + \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} \times d(\lambda y)$

One can then expand and collect the $d(\lambda x)$ as $xd\lambda + \lambda dx$ and $d(\lambda y)$ as $yd\lambda + \lambda dy$ and achieve the following relation:

$df(\lambda x, \lambda y) = \lambda \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} \times dx + \lambda \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} \times dy + \left[x\left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x}\right] \times d\lambda$

  • RHS

$d\left[\lambda^n f(x,y)\right] = n\lambda^{n-1}f(x,y) \times d\lambda + \lambda^n \left[\left(\frac{\partial f}{\partial x}\right)_y \times dx + \left(\frac{\partial f}{\partial y}\right)_x \times dy\right]$

  • RHS = LHS

Both differentials must be equal. Thus, terms in front of $dx$, $dy$ and $d\lambda$ must be equal in both expressions. This leads to the following system:

$\begin{cases} \lambda \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} = \lambda^n \times \left(\frac{\partial f}{\partial x}\right)_y\\ \lambda \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} = \lambda^n \times \left(\frac{\partial f}{\partial y}\right)_x\\ x\left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} = n \lambda^{n-1} \times f(x,y) \end{cases}$

From this point, the classical next step is to take the third equation, set $\lambda=1$ (although you want the identity to be true for all $\lambda$) and conclude that the identity is proven. Clearly this assumption is not required as you can inject the first and second equations into the third and obtain the required identity:

$\lambda^{n-1} x\left(\frac{\partial f}{\partial x}\right)_{y} + \lambda^{n-1} y\left(\frac{\partial f}{\partial y}\right)_{x} = n \lambda^{n-1} \times f(x,y)$

You can see that all $\lambda^{n-1}$ cancel and lead to the expected relation:

$x\left(\frac{\partial f}{\partial x}\right)_{y} + y\left(\frac{\partial f}{\partial y}\right)_{x} = n \times f(x,y)$

  • $\begingroup$ beautiful! thanks. $\endgroup$
    – agks mehx
    Commented May 8 at 18:50
  • 1
    $\begingroup$ also, i have corrected what seemed like a typographical error; please review my edit $\endgroup$
    – agks mehx
    Commented May 8 at 19:07

Regarding your example: In the second to last step where you compute $$ \frac{\partial f}{\partial u} \frac{\mathrm du}{\mathrm d\lambda} + \frac{\partial f}{\partial v} \frac{\mathrm dv}{\mathrm d\lambda} = r\lambda^{r-1}f(x,y) $$ In your example, the left-hand side of that is computed as follows: We differentiate $f(\lambda x, \lambda y)$ with respect to $\lambda$. To do that, we use the chain rule: The outer derivative is the derivative of $f$ evaluated at $(\lambda x, \lambda y)$, the inner derivative is just $(x,y)$. So we need to compute (I'm writing this as a product of vectors here) $$ Df(\lambda x, \lambda y) \cdot (x,y) $$ Since $Df(x,y) = (2xy^2, 2x^2y)$, this gives $$ Df(\lambda x, \lambda y) \cdot (x,y) = (2(\lambda x)(\lambda y)^2, 2(\lambda x)^2(\lambda y)) \cdot (x,y) = (2\lambda^3 xy^2, 2\lambda^3 x^2y) \cdot (x,y) = 2 \lambda^3 x^2y^2 + 2 \lambda^3 x^2y^2 = 4 \lambda^3 x^2y^2$$ as desired.


I can say that the general property which must be satisfied for all $\lambda>0$ is : $$x\frac{\partial f}{\partial x}(\lambda x,\lambda y)+y\frac{\partial f}{\partial y}(\lambda x,\lambda y)=k\lambda^{k-1}f(x,y).$$ From which we deduce the desired Euler Theorem by setting $\lambda=1$.


Since $f(x,y)=x^2y^2$, therefore, it can be written as $$f(x,y)=x^2\left(\frac{y}{x}\right)\times x^2=x^4\left(\frac{y}{x}\right).$$ This shows that $f$ is a homogeneous function of degree $4$. Hence, by Euler's theorem, we have $$x\frac{\partial f}{\partial x} + x\frac{\partial f}{\partial x}=4f.$$


Consider a function $f$ that is homogeneous of degree $r$, i.e.,

$f(tx, ty) = t^rf(x, y)$ ......$(1)$

Differentiate both sides of equation (1) with respect to $t$. Then by Chain Rule:

$\begin{equation} \frac{\partial f(tx, ty)}{\partial(tx)}x + \frac{\partial f(tx, ty)}{\partial(ty)}y = rt^{r-1}f(x, y) \end{equation}$ ......$(2)$

Differentiate both sides of equation $(1)$ with respect to $x$. Then by chain rule:

$\begin{equation} \frac{\partial f(tx, ty)}{\partial(tx)}t = t^r\frac{\partial f(x, y)}{\partial(x)} \end{equation}$

$\implies$$\begin{equation} \frac{\partial f(tx, ty)}{\partial(tx)} = t^{r-1}\frac{\partial f(x, y)}{\partial x} \end{equation}$ ......$(3)$


$\begin{equation} \frac{\partial f(tx, ty)}{\partial(ty)} = t^{r-1}\frac{\partial f(x, y)}{\partial y} \end{equation}$ ......$(4)$

Put $(3)$ and $(4)$ in $(2)$ to get the result:

$\begin{equation} \frac{\partial f(x, y)}{\partial x}x + \frac{\partial f(x, y)}{\partial y}y = rf(x, y) \end{equation}$


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