Proving Uniform Continuity of f on [1, $\infty$]

In summary: MacLeodIn summary, the problem is to prove that a continuous function f in [1,\infty) with a finite limit as x approaches infinity is also uniformly continuous in the same interval. One approach is to divide the interval into two segments, [1,x0+1] and (x0,\infty), and show that f is uniformly continuous in both segments. Since the two segments overlap, f is also uniformly continuous in the entire interval. Another approach is to consider the minimum and maximum values of f in [1,x0+1] and use the fact that f converges to a finite limit to show that the difference between these values is finite. From here, it can be shown that there exists a constant delta that
  • #1
talolard
125
0
Hello,

Homework Statement


Given that f is continuous in [tex][1,\infty)[/tex] and [tex] lim_{x->\infty}f(x) [/tex] exists and is finite, prove that f is uniformly continuous in [1,[tex]\infty)[/tex]



The Attempt at a Solution


We will mark [tex] lim_{x->\infty}f(x) = L [/tex]. So we know that there exists [tex]x_{0}[/tex] such that for every [tex] x>x_{0} |L-f(x)|<\epsilon [/tex] so f is uniformly continuos in [tex] (x_{0},\infty) [/tex]
We shall look at the segment [tex] [1,x_{0}+1] [/tex]. We know that a continuous function ina closed segment is uniformly continuous.

From here I am puzled. I have two routes.
The easy route sais that since we showed f is uniformly continuous in those two segments and they overlap then f is uniformly continuous on the whole segment. I havea feeling this is not enough.
Route 2:
We know that a continuous function ina closed segment has a minimum and maximal value there. Then we can look at the closed segment [tex] [1,x_{0}+1] [/tex] and mark [tex] m=MIN(f(x)), M= MAX(f(x)) [/tex]. But since f converges to L from [tex] x=,x_{0} [/tex] we have taken into account all of the possible values of f.
we will mark [tex] d= | |M| - |m| | [/tex] the biggest difference in value f gets.
and from here I am stuck. I am lost as to how to tie up the proof. Gudiance would be greatly apreciated.
Thanks
Tal
 
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  • #2
I think that I would try to use your "route 1", if you know that the mapping is uniformly continuous in both domains, and they overlap on [x0,x0+1]. Let's call them domains 1 and 2.

Uniform continuity in a domain demands that for any constant epsilon>0, there should be a constant delta>0 such that |f(y)-f(x)| < epsilon for all y with |y-x| < delta.

For a given epsilon, you can find such a delta for domain 1, and one for domain 2, just choose the smallest one of those, and I think is it OK, right? I guess delta also has to be less than the overlap of the two domains. You can always choose such a delta.

It's been a while since I worked on such problems, so please be very critical towards my statements!

Torquil
 

FAQ: Proving Uniform Continuity of f on [1, $\infty$]

What does it mean for a function to be uniformly continuous?

Uniform continuity is a property of a function that describes its behavior on a given interval. A function is uniformly continuous if, for every pair of points on the interval, the difference in the function values is proportional to the difference in the points. In other words, as the points get closer together, the function values also get closer together.

Why is proving uniform continuity important?

Proving uniform continuity is important because it allows us to make predictions about how the function will behave on the given interval. It ensures that the function is well-behaved and does not have any sudden jumps or discontinuities. This is particularly important in applications such as physics, engineering, and economics where precise and continuous functions are needed to make accurate predictions.

What is the difference between uniform continuity and continuity?

Continuity and uniform continuity are both properties of functions, but they describe different aspects of a function's behavior. Continuity means that a function is continuous at every point on the given interval, meaning that there are no sudden jumps or breaks in the function. Uniform continuity, on the other hand, means that the function behaves similarly at every point on the interval, with no sudden changes in behavior as the points get closer together.

How do you prove uniform continuity of a function on a given interval?

To prove uniform continuity, we typically use the epsilon-delta definition. This involves showing that for any given epsilon (a small positive number), there exists a corresponding delta (another small positive number) such that the difference in function values is less than epsilon whenever the difference in points is less than delta. This ensures that the function is well-behaved and does not have any sudden jumps or discontinuities on the given interval.

Can a function be uniformly continuous on one interval but not on another?

Yes, a function can be uniformly continuous on one interval but not on another. This is because uniform continuity depends on the behavior of the function on a specific interval, and different intervals may have different properties. For example, a function may be uniformly continuous on the interval [0,1], but not on the interval [-1,1]. This is because the function may have a different behavior or slope on the two intervals, leading to different rates of change and differences in function values.

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