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By definition of homogeneity, if $f(x, y)$ is homogeneous of degree $n$, then $$f(tx, ty) = t^nf(x, y)$$

Problem

Prove that if $f(x, y)$ is homogeneous of degree $1$, then $f_{xx}(x, y) f_{yy}(x, y) = (f_{xy})^2$

Attempted solution

I tried using Euler's theorem on homogeneous functions, which asserts that if $f(x, y)$ is homogeneous of degree $n$, then $$xf_x(x, y) + yf_y(x, y) = nf(x, y)$$

But this didn't lead me to the desired result:

$$xf_x(x, y) + yf_y(x, y) = nf(x, y)$$ $$\frac{xf_x(x, y)}{nf(x, y)} + \frac{yf_y(x, y)}{nf(x, y)} = 0$$ $$???$$

user1255490
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1 Answers1

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By the title, you have that $f(x,y)$ is homogeneus of degree 1, so in particular you have $$xf_{x}(x,y)+yf_{y}(x,y)=f(x,y)$$ You can try deriving that expression by $x$ and by $y$ so you get the form of $f_{xx}$ and $f_{yy}$

DanyWose
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    I figured it out. You need to differentiate it with respect to x first using the product rule for $xf_x(x, y)$, then differentiate the original expression again with respect to y, then combine the two expressions. – user1255490 Nov 21 '23 at 11:36