Expert: Paul Klarreich - 4/30/2009

Question
Hi Paul,

I am really struggling with uniform continous and uniform bounded functions, so was hoping for some guidance with the question below, please.

For uniformly continuous f: R -> R, let f_n: R-. R be f_n(x) = f(x + 1/n).

i.   How does f’s graph relate to that of f_1(x), f_2(x),….,f_n(x) ??

ii.   Prove the sequence [f_n(x)} converges uniformly to f(x) on R.
Hint: Use the expression |f_n(x) – f(x)| = |f(x + 1/n) – f(x)|.

iii.   Can ii. Be extended to all continuous functions f: R -> R on R?  Answer with a counter example or a proof in a restricted case.


Answer
Questioner:   Steve
Country:  Australia
Category:  Advanced Math
Private:  No
 
Subject:  Algebra - uniform continous functions
Question:  Hi Paul,

I am really struggling with uniform continous and uniform bounded functions,

>> Yes, so did I at your age.

so was hoping for some guidance with the question below, please.

For uniformly continuous f: R -> R, let f_n: R-. R be f_n(x) = f(x + 1/n).

i. How does f’s graph relate to that of f_1(x), f_2(x),….,f_n(x) ??

ii. Prove the sequence [f_n(x)} converges uniformly to f(x) on R.
Hint: Use the expression |f_n(x) – f(x)| = |f(x + 1/n) – f(x)|.

iii. Can ii. Be extended to all continuous functions f: R -> R on R?  Answer with a counter example or a proof in a restricted case.
...............................................
Hi, Steve,

I will take a small shot at this, but I don't know how far I will get.

You want to prove that for all  x in your interval, [I guess here your interval is all R]

|f_n(x) – f(x)| = |f(x + 1/n) – f(x)| < e,
for sufficiently large n.

I think you use the U.C. idea.  You will say that since f is U.C., then for all x0, we can find d (d = delta, which I can't make) such that
whenever  |x - x0| < d,  | f(x) - f(x0)| < e.

Now lets rename things a bit.

|f_n(x0) – f(x0)| = |f(x0 + 1/n) – f(x0)|.

Now can we make that < e?  Yes, by just taking n sufficiently large, meaning > some N.

What N will do it?  Take N > 1/d, and we have  

|f(x) – f(x0)|, where  x = x0 + d. so  | x - x0 | < d.

And we know that if that is true, then

|f(x) – f(x0)| < e.

Does that make sense?  I am really saying that if you want

f[n](x0) to be close to  f(x0), it will do to take f[n](x0) to be a value of f(some x close to x0).

...........................................
About the last one, you have substituted continuity for uniform continuity, and you don't say it is bounded.  That means that if you have some e fixed, there could be some place where f(x) varies so dramatically that you need a smaller d at that point.  

Obviously, any polynomial over R will do. [But not over a closed interval -- there is this theorem:

If f(x) is continuous on [a,b] -- a closed interval, then it is uniformly continous on [a,b].  Something having to do with the Heine-Borel Theorem, but don't get me started.
..........................................
More on a counterexample, to finish up.

Uniform convergence means that for some N,

| fn(x) - f(x) | <  e

for all n > N and for all x.

Let's take  f(x) = x^2.

and take  fn(x) = (x + 1/n)^2.

Now compute:

| fn(x) - f(x) | =

| (x + 1/n)^2 - x^2 | =

| x^2 + 2x/n + 1/n^2 - x^2 | =

| 2x/n + 1/n^2 | =

for large n, we can ignore the 1/n^2.

| 2x/n | =

assume x is positive, and that is:

2x/n

That is going to be bigger than your e, because no matter what n we choose, there exists x > en/2, so

2x/n > 2(en/2)/n = e

So no N, no matter how big, works for all values of x.  And that is what Uniform Convergence is not. (What awful language!  Sorry.)
.........................
P.S. You can also construct an example on an open interval like, say (0,10), where the function is something like  1/(x - 10).  It is continuous but not uniformly, so you won't have U.C. of your fn's.