Separate continuity and joint continuity

Let function f0,1)^n -> R be strictly increasing and separately continuous in each variable. Can we state that each plane (a solution of a system of n-2 linear equations), intersecting the set (0,1)^n, contains a point where f is continuous as a function of n variables.
Any reference will be very much appreciated.

Thanks in advance,
Mikhail

 

If I've understood correctly, "yes", because f is continuous under
your hypotheses. Briefly, if x is fixed, d>0 (i.e., d is an n-tuple
with positive entries), and we write "a < b" to mean "a_i < b_i for
i=1, ..., n", then

(i) x-d < y < x+d implies f(x-d) < f(y) < f(x+d) by monotonicity in
each variable, and

(ii) |f(x \pm d) - f(x)| is small when d is small, by continuity in
each variable.

For definiteness, consider the case n=2. If x = (x_1,x_2) and y =
(y_1,y_2) are in the unit square,

(1) f(y) - f(x) = f(y_1,y_2) - f(x_1,x_2) = [f(y_1,y_2) -
f(y_1,x_2)] + [f(y_1,x_2) - f(x_1,x_2)].

In particular, x<y implies f(x) < f(y) by separate monotonicity.

Fix (x_1,x_2) and e>0 arbitrarily and use continuity in the first
variable to choose d_1>0 such that

(2) |f(x_1 \pm d_1,x_2) - f(x_1,x_2)| < e/2.

Then use continuity in the second variable to choose d_2>0 such that

(3) |f(x_1 \pm d_1,x_2 \pm d_2)| - f(x_1 \pm d_1,x_2)| < e/2.

If (y_1,y_2) is an element of the (2d_1 x 2d_2) rectangle centered at
(x_1,x_2), namely if

x - d < y < x+d,

then (1)--(3) and monotonicity in each variable imply

- e < f(x-d) - f(x) < f(y) - f(x) < f(x+d) - f(x) < e.

Hope that's helpful,

Andy

Andrew D. Hwang
Dept of Mathematics and CS
College of the Holy Cross
Worcester, MA, 01610-2395, USA