partial differentiation?

Question

Show that if f is a real valued function on a connected open subset of E^n and f'_1 = f'_2 = ... = f'_n = 0 then f is constant.

thank you,
so is there any special reason for the subset to
be "connected" ?

Answer 1

Let S be this subset.

Suppose that there exist points x1, x2 in S such that f(x1) does not equal f(x2). Let v = x1 - x2.

Then, the mean value theorem implies that

f(x1)-f(x2) = (grad f(c) . u)v

where u = v/||v||, and c is some point along the line segment connecting x1 to x2. grad f . u is the directional derivative of f in the direction u. If this line segment does not lie entirely in S, then, because it's connected, you can always form a chain of line segments lying entirely within S to connect x1 to x2, and f must have different values at the endpoints of at least one of these segments, or else the assumption f(x1) not equal f(x2) is contradicted.

Because grad f = 0 in the subset, (grad f (c). u) = 0, for every point c along the segment, so we must have f(x2)=f(x1), contradicting our assumption that they were not equal. Therefore f must be constant in S.

Answer 2

let f is constant and assume that f_i != 0 for some point a in E^n

by definition lim x->a f_i(a) - f_i(x)/(x-a) != 0
since f is contstant f_i(a) = f_i(x) for all x in E^n.
and thus f_i(a) - f_i(x)/(x-a) != 0 is in contradiction with f_i(x) = f_i(a)