Question

Apply the Young's theorem to justify that:

fxy(1,1)= fyx(1,1) for the function f : R^2→R , defined by

f(x,y)= |x+y|

Answer 1

Function is given as

f(x,y)= |x+y|

This is piecewise function due to mod

Thus function is f(x,y)=+( x+y), if x+y>0

f(x,y)=-(x+y), if x+y<0

f(x,y)=0, if x+y=0

Let's take f(x,y)=x+y

$\partial f(x,y)/\partial x=1$

Similarly partial derivatives of function wrt y is also 1

Partial derivative of (partial derivatives of function wrt x) with respect to y =0 i.e fxy=0

Partial derivative of (partial derivatives of function wrt y) with respect to x =0 i.e fyx=0

fxy(1,1)=0

And

fyx(1,1)=0

Thus we can say

fxy(1,1)= fyx(1,1)