# Thread: Prove f is a continuous function

1. ## 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]}?

2. 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.

3. 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?

4. 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]}

5. 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}...