# Integral of open set

1. Dec 3, 2012

### cummings12332

1. The problem statement, all variables and given/known data

3. The attempt at a solution
for part A i assume that f is in norm space C[0,1],||.|| , then choose a sequence fn in C[o,1] s.t fn->f then for 0<fn<1 so 0<f<1 i.e. A is closed i am not sure my answer here

for part B i assume the anti-derivatice of f(t) to be K(t)+c therefore, by F(f)=2K(1/2)+1/2-K(1)-K(0) then how should i prove it is cts ???

2. Dec 3, 2012

### micromass

Staff Emeritus
I don't see how that proves anything. You need to prove that if $f_n$ is a convergent sequence in A such that $f_n\rightarrow f$, that then $f$ is an element of A.

OK, but that doesn't really help. What definition of continuous would you like to use here? Can you state it?

Also, it would be a great help for us if you would type your posts in LaTeX: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

3. Dec 3, 2012

### cummings12332

δ
if $f_n$ is a convergent sequence in A such that $f_n\rightarrow f$, then for $f_n$ is in A , A is a subset of C[0,1] so fn is bounded by 0 and 1 and continuous on [0,1] so fn is uniformly continuous to f ,then f is bouded by 0,1 which is in A????

and secound part if i use the definition of continuous , then exists a δ,s.t. |f(x)-f(y)|<δ implies that |F(f(x))-F(g(x))|<esillope . and just fixed the esillope . and i can solve it out now many thanks

Last edited: Dec 3, 2012
4. Dec 3, 2012

### micromass

Staff Emeritus
Sure. But can you actually prove that f is bounded by 0 and 1? Once you proven that, I agree that the above shows that A is closed.

OK, so you you need to make $|F(f)-F(g)|$ small somehow by making $\|f-g\|_\infty$ small. So, let us start by writing

$$|F(f)-F(g)|=\left|\int_0^{1/2}f(t)dt + \int_{1/2}^1 (1-f(t))dt - \int_0^{1/2} g(t)dt - \int_{1/2}^1 (1-g(t))dt\right|$$

Try to find a good estimation for that.

5. Dec 3, 2012

### cummings12332

i not sure how to prove that 0<f<1 ,which way should i begin with??

6. Dec 3, 2012

### micromass

Staff Emeritus
You should not prove 0<f<1, you should prove $0\leq f\leq 1$.

Start by fixing an $x\in [0,1]$. Then try to deduce from $f_n(x)\rightarrow f(x)$ that $0\leq f(x)\leq 1$.

7. Dec 3, 2012

### cummings12332

for me ,it is obviouse. $0\leq f_n\leq 1$. then max fn =1, min fn =0 and limfn =f so by sanwich rule $0\leq f(x)\leq 1$ i just dont know how should i prove that f is bounded by [0,1] here

8. Dec 3, 2012

### micromass

Staff Emeritus
Why should $\max f_n=1$ and $\min f_n=0$?? That's really weird.
For example, take the map

$$f_n:[0,1]\rightarrow \mathbb{R}:x\rightarrow x\rightarrow \frac{1}{n}$$

then the $\max f_n=\min f_n=\frac{1}{n}$.

You're really making it harder than it is. I'm just asking you that if $(y_n)_n$ is a convergent sequence with $y_n\rightarrow y$, and if $0\leq y_n\leq 1$, then $0\leq y\leq 1$. It doesn't even have anything to do with functions.

9. Dec 3, 2012

### cummings12332

omg! thats what i thought at the beginning!!! many thanks !!!