Prove f is a continuous function

Printable View

February 28th 2010, 03:58 PM
southprkfan1

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]}?
February 28th 2010, 04:15 PM
Plato

Quote:

Originally Posted by southprkfan1

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.
February 28th 2010, 05:20 PM
Drexel28

Quote:

Originally Posted by southprkfan1

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?
February 28th 2010, 06:49 PM
southprkfan1

Quote:

So, http://www.mathhelpforum.com/math-he...85e84c1b-1.gif. And http://www.mathhelpforum.com/math-he...fb9f86e5-1.gif and http://www.mathhelpforum.com/math-he...993c95e8-1.gif is continuous for each fixed http://www.mathhelpforum.com/math-he...46852fac-1.gif respectively. Then, define http://www.mathhelpforum.com/math-he...058750f1-1.gif is equicontinuous. Prove that http://www.mathhelpforum.com/math-he...09471012-1.gif is continuous?

Yes to everything to said about the functions. Here is how E is defined in the book:

E = {x http://upload.wikimedia.org/math/0/5...78bbf9bac6.png f(x,y) : y http://upload.wikimedia.org/math/7/b...e1a4c7cff0.png [0,1]}
March 1st 2010, 12:18 PM
southprkfan1

Ok I have the answer now.

Fix e>0 and (a,b) http://upload.wikimedia.org/math/7/b...e1a4c7cff0.png [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 http://upload.wikimedia.org/math/7/b...e1a4c7cff0.png [0,1]

Similarly, by continuity of f(x0, y) where x0 is constant and x0 http://upload.wikimedia.org/math/7/b...e1a4c7cff0.png [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}...