Is there an easy way to show that if f(x,y) is twice-differentiable on [0,1]x[0,1] with continuous second-order partial derivatives, then there must exist a point at which (d/dx)(d/dy) f(x,y) equals f(0,0)-f(0,1)-f(1,0)+f(1,1)?
Let g = (d/dx)(d/dy) f(x,y).
From the Fundamental Theorem of Calculus,
df/dy (x,y) = int_0^x g dx + df/dy (0,y) Again, from the Fundamental Theorem of Calculus, f(x,y) = int_0^y df/dy dx + f(x,0) = =int_0^y (int_0^x g dx) dy + f(0,y) - f (0,0) + f(x,0) .
From here,
f(1,1)-f(0,1)-f(1,0)+f(0,0) = int_0^1 (int_0^1 g dx) dy Now, if g < the l.h.s. on the unit square, the double integral would be less than the l.h.s. Similarly, if g > the l.h.s on the unit square, the double integral would be greater than the l.h.s. Thus, from the continuity of g, there is a point at which g = the l.h.s. Alec Mihailovs