Uniform Continuity on Closed and Bounded Intervals

In summary, the conversation is discussing how to prove that a continuous function f on the interval [0, infinity) is uniformly continuous given that f approaches a limit L as x approaches infinity. Relevant theorems and the definition of uniform continuity are mentioned. The Archimedean Property is also brought up, but there is a question about whether the interval is bounded above. The expert summarizer suggests that the interval is not bounded above and that it does not make sense to say that infinity is bounded above.
  • #1
bluskies
11
0

Homework Statement



Suppose that [tex]f: [0, \infty) \rightarrow \mathbb{R}[/tex] is continuous and that there is an [tex]L \in \mathbb{R}[/tex] such that [tex]f(x) \rightarrow L[/tex] as [tex]x \rightarrow \infty[/tex]. Prove that f is uniformly continuous on [tex][0,\infty)[/tex].

2. Relevant theorems

If [tex] f:I \rightarrow \mathbb{R}[/tex] is continuous on I, where I is a closed, bounded interval, then f is uniformly continuous on I.

Or, the definition of uniform continuity:

A function [tex]f:E \rightarrow \mathbb{R}[/tex] is uniformly continuous on E iff for every [tex]\epsilon >0[/tex], there is a [tex]\delta >0[/tex] such that [tex]|x-a|<\delta[/tex] and [tex]x,a \in E[/tex] imply [tex]|f(x)-f(a)|<\epsilon[/tex].

Archimedean Property: For every [tex]x \in \bb{R},[/tex] there exists an [tex]n_x[/tex] such that [tex]x<n_x[/tex].

The Attempt at a Solution



I was trying to use the theorem I listed, since we already know that f is continuous on the interval and that the interval [tex][0, \infty)[/tex] is closed, as well as bounded below. My main question is whether the interval is bounded above. From the Archimedean Property, we can see that the real line is bounded above. Does the idea that infinity is bounded above follow directly, or is there something that I'm missing?Any help would be very much appreciated. Thank you!Edit: I'm sorry about the latex attempt, I'm still trying to figure out how to use it on the forums...the R is supposed to be the real numbers.
 
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  • #2
bluskies said:

Homework Statement



Suppose that [tex]f: [0, \infty) \rightarrow \bb{R}[/tex] is continuous and that there is an $L \in \bb{R}$ such that $f(x) \rightarrow L$ as $x \rightarrow \infty$. Prove that $f$ is uniformly continuous on $[0,\infty)$.

2. Relevant theorems

If $f:I \rightarrow \bb{R}$ is continuous on $I$, where $I$ is a closed, bounded interval, then $f$ is uniformly continuous on $I$.

Or, the definition of uniform continuity:

A function $f:E \right \bb{R}$ is uniformly continuous on $E$ iff for every $\epsilon >0$, there is a $\delta >0$ such that $|x-a|<\delta$ and $x,a \in E$ imply $|f(x)-f(a)|<\epsilon$.

Archimedean Property: For every $x \in \bb{R},$ there exists an $n_x$ such that $x<n_x$.

The Attempt at a Solution



I was trying to use the theorem I listed, since we already know that f is continuous on the interval and that the interval $[0, \infty)$ is closed, as well as bounded below. My main question is whether the interval is bounded above. From the Archimedean Property, we can see that the real line is bounded above. Does the idea that infinity it bounded above follow directly, or is there something that I'm missing?

Any help would be very much appreciated. Thank you!

Edit: I'm sorry about the latex attempt, I'm still trying to figure out how to use it on the forums...

Surround your expressions with [ tex] and [ /tex] tags, without the leading spaces. For example,
[tex]f: [0, \infty) \rightarrow \bb{R}[/tex]
 
  • #3
Thank you for your help - I think I've fixed the Latex problems. Do you think I'm going in the right direction on the homework problem?
 
  • #4
The interval [0, infinity) is NOT bounded above. It makes no sense to say that infinity is bounded above.
 

What does uniform continuity mean?

Uniform continuity is a mathematical concept that describes the behavior of a function. A function is said to be uniformly continuous if it maintains a consistent rate of change throughout its entire domain, meaning that the gap between the function values at any two points is always within a certain tolerance. This means that the function does not have any sudden jumps or breaks, and it can be continuously graphed without any interruptions.

How is uniform continuity different from continuity?

While continuity refers to the smoothness and connectedness of a function, uniform continuity specifically refers to how the function behaves at all points in its domain. A function can be continuous but not uniformly continuous if it has sudden jumps or breaks in its behavior, whereas a uniformly continuous function must have a consistent rate of change throughout its domain.

Why is uniform continuity important in mathematics?

Uniform continuity is important because it helps us understand the behavior of a function as a whole, rather than just at specific points. It allows us to make predictions about the function's behavior and properties, and it is essential in many mathematical concepts and proofs, such as the Intermediate Value Theorem and the Fundamental Theorem of Calculus.

Can all functions be uniformly continuous?

No, not all functions can be uniformly continuous. For a function to be uniformly continuous, it must meet certain criteria, such as having a bounded derivative or being defined on a closed interval. Functions that have sudden jumps or breaks, or that have an unbounded derivative, are not considered uniformly continuous.

How is uniform continuity tested or proven?

The most common way to test for uniform continuity is by using the epsilon-delta definition. This involves setting a tolerance (epsilon) and finding a corresponding interval (delta) such that the function values within that interval are within the specified tolerance. If such an interval can be found for any point in the function's domain, then the function is considered uniformly continuous. Other methods, such as the Mean Value Theorem, can also be used to prove uniform continuity.

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