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Math Help - Prove f is a continuous function

  1. #1
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    Prove f is a continuous function

    Let F: [0,1]X[0,1] --> R

    where F is such st both Fy(x) and Fx(y) are continuous. [That is, Fy(x) is the function holding y constant]

    Furthermore, assume that the set of functions, E, is equicontinuous. Where E

    E = {maps that take x-->f(x,y) : y is in [0,1]}


    Prove that f is continuous.


    I guess my biggest problem is that I'm not entirely clear what the set E represents. Is E the set of all {Fy(x) for all y in [0,1]}?
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  2. #2
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    Quote Originally Posted by southprkfan1 View Post
    Let F: [0,1]X[0,1] --> R where F is such st both Fy(x) and Fx(y) are continuous. [That is, Fy(x) is the function holding y constant]
    Furthermore, assume that the set of functions, E, is equicontinuous. Where E
    E = {maps that take x-->f(x,y) : y is in [0,1]}
    Prove that f is continuous.
    I guess my biggest problem is that I'm not entirely clear what the set E represents. Is E the set of all {Fy(x) for all y in [0,1]}?
    Why don't you learn to use LaTeX?
    As is, I for one, have no idea about what you have posted.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Let F: [0,1]X[0,1] --> R

    where F is such st both Fy(x) and Fx(y) are continuous. [That is, Fy(x) is the function holding y constant]

    Furthermore, assume that the set of functions, E, is equicontinuous. Where E

    E = {maps that take x-->f(x,y) : y is in [0,1]}


    Prove that f is continuous.


    I guess my biggest problem is that I'm not entirely clear what the set E represents. Is E the set of all {Fy(x) for all y in [0,1]}?
    So, F:[0,1]\times[0,1]\mapsto\mathbb{R}. And F(x,y_0) and F(x_0,y) is continuous for each fixed x_0,y_0\in [0,1] respectively. Then, define E=\left\{f_{y_0}:[0,1]\mapsto\mathbb{R}:\text{ }f_{y_0}(x)=F(x,y_0),\text{ }y_0\in[0,1]\right\} is equicontinuous. Prove that F is continuous?
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  4. #4
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    So, . And and is continuous for each fixed respectively. Then, define is equicontinuous. Prove that is continuous?
    Yes to everything to said about the functions. Here is how E is defined in the book:

    E = {x f(x,y) : y [0,1]}
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  5. #5
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    Ok I have the answer now.

    Fix e>0 and (a,b) [0,1] X [0,1]. We want to find a d such that:

    l(x,y) - (a,b)l < d implies lf(x,y) - f(a,b)l < e

    By equicontinuity of E, there is a d1 where

    l x - a l < d1 implies lf(x,y) - f(a,y)l < e/2 for all y [0,1]

    Similarly, by continuity of f(x0, y) where x0 is constant and x0 [0,1], there is a d2 where:

    l y - b l < d2 implies lf(a, y) - f(a, b)l < e/2.

    Let d =min{d1, d2}...
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