# Prove f is a continuous function

• Feb 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]}?
• Feb 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.
• Feb 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, $\displaystyle F:[0,1]\times[0,1]\mapsto\mathbb{R}$. And $\displaystyle F(x,y_0)$ and $\displaystyle F(x_0,y)$ is continuous for each fixed $\displaystyle x_0,y_0\in [0,1]$ respectively. Then, define $\displaystyle E=\left\{f_{y_0}:[0,1]\mapsto\mathbb{R}:\text{ }f_{y_0}(x)=F(x,y_0),\text{ }y_0\in[0,1]\right\}$ is equicontinuous. Prove that $\displaystyle F$ is continuous?
• Feb 28th 2010, 06:49 PM
southprkfan1
• Mar 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}...