Separate continuity and joint continuity

Discussion in 'Math Research' started by Mikhail V. Sokolov, Sep 10, 2007.

  1. Let function f:(0,1)^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
     
    Mikhail V. Sokolov, Sep 10, 2007
    #1
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  2. 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
     
    Andrew D. Hwang, Sep 11, 2007
    #2
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