Continunity In Function Spaces

[jdcasey9]

Homework Statement

Let fn:R->R be continuous for each n, and suppose that fn->f uniformly on each closed,bounded interval [a,b]. Show that f is continuous on R.

Homework Equations

The definition of continuity that I will be using:

f:M->N
For any xcM, if xn->x in M, then f(xn)->f(x) in N.

Definition of uniform continuity:

f:M->N
fn->f uniformly if for all E>0 there exists an N(E) and for all xcM, all n>N, such that p(fn(x),f(x)) < E.

The Attempt at a Solution

Let E>0 and take xcR. Since fn is continuous and fn->f is uniformly continuous on a closed, bounded interval, we can find xn such that xn->x, where xc[a,b] for large enough n. By the definition of uniform continuity we have p(fn(x),f(x)) = lfn(x)-f(x)l < E. So as E -> 0, fn(x)-> f(x).
Therefore, f is continuous on a R.

 

[LCKurtz]

Homework Statement

Let fn:R->R be continuous for each n, and suppose that fn->f uniformly on each closed,bounded interval [a,b]. Show that f is continuous on R.

Homework Equations

The definition of continuity that I will be using:

f:M->N
For any xcM, if xn->x in M, then f(xn)->f(x) in N.

Definition of uniform continuity:

f:M->N
fn->f uniformly if for all E>0 there exists an N(E) and for all xcM, all n>N, such that p(fn(x),f(x)) < E.

The Attempt at a Solution

Let E>0 and take xcR. Since fn is continuous and fn->f is uniformly continuous

"fn->f is uniformly continuous" doesn't mean anything. Do you mean fn->f uniformly?

on a closed, bounded interval, we can find xn such that xn->x, where xc[a,b] for large enough n.

That is a confused statement. You can always find x_n converging to x; that doesn't have anything to do with your functions.

By the definition of uniform continuity we have p(fn(x),f(x)) = lfn(x)-f(x)l < E. So as E -> 0, fn(x)-> f(x).
Therefore, f is continuous on a R.

Back to the drawing board. Pick any x and consider the interval I = [x-1,x+1]. You need to show that given any e > 0 you can find a d such that if |x - y| < d then |f(x) -f(y)| < e.

Think about using the triangle inequality by adding and subtracting terms using the sequence and what is given.

 

[jdcasey9]

Let E>0 and E = d + 2ly-f(y)l + 2lf(x)-f(y)l. Consider the interval I = [x-1, x+1] with x,ycR. If lx-yl < d, then lx-yl <= lx-f(x)l + lf(x)-yl = lx-f(x)l + ly-f(x)l <= lx-f(y)l + lf(y)-f(x)l +ly-f(y)l + lf(y)-f(x)l = 2lf(y)-f(x)l + lx-f(y)l + ly-f(y)l <= 2lf(y)-f(x)l + lx-yl + ly-f(y)l + ly-f(y)l = 2lf(y)-f(x)l + lx-yl + 2ly-f(y)l < d + 2ly-f(y)l + 2lf(y)-f(x)l = E.
So, lf(y)-f(x)l < E.

 

[jdcasey9]

Does that seem about right?

 

[LCKurtz]

No, not even close. You haven't used anything about the f_n in your argument so how could it be right?

Look, here's the general idea. You want to get f(x) close to f(y). So think about doing it by getting f_n(x) close to f(x), f_n(x) close to f_n(y), and f_n(y) close to f(y). The devil is in the details.

 

[jdcasey9]

Ok, ok, I'm understanding.

Fix E>0 and let xc[x-1,x+1]. Since fn -> f uniformly on [x-1,x+1], for some n, p(fn(z), f(z)) < E/3 for all zc[x-1,x+1]. This fn is continuous at x, so for some delta>0, d(x,y)<delta implies p(fn(x),fn(y))<E/3. Then for such y, d(x,y)<delta, p(f(x), f(y))<=p(f(x),fn(x)) + p(fn(x),fn(y)) + p(fn(y), f(y)) < E/3 + E/3 + E/3 = E.

Therefore, f is cts on [x-1,x+1].

Does it logically follow that it is continuous on R? Or is there more to be shown?

 

[LCKurtz]

Ok, ok, I'm understanding.

Fix E>0 and let xc[x-1,x+1].

You mean suppose E > 0. (You don't need to fix it unless it is broken ). Then you don't say "let x ε [x-1,x+1]". That is always true. You are trying to show f is continuous on R so you say pick x ε R and consider the interval I = [x-1,x+1]. Then if you show f is continuous at x you will know it is continuous on R because x could have been any real number.

Since fn -> f uniformly on [x-1,x+1], for some n, p(fn(z), f(z)) < E/3 for all zc[x-1,x+1]. This fn is continuous at x, so for some delta>0, d(x,y)<delta implies p(fn(x),fn(y))<E/3.

You need to also require delta < 1 to stay on your interval.

Then for such y, d(x,y)<delta, p(f(x), f(y))<=p(f(x),fn(x)) + p(fn(x),fn(y)) + p(fn(y), f(y)) < E/3 + E/3 + E/3 = E.

Therefore, f is cts on [x-1,x+1].

Does it logically follow that it is continuous on R? Or is there more to be shown?

Yes. See above.

 

[jdcasey9]

Ok, thanks. I really appreciate it.