Calc Proof(s): Uniform Continuity

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SUMMARY

The discussion focuses on two proofs related to uniform continuity. The first proof demonstrates that the function f: [0, ∞) → R defined by f(x) = 1/x is not uniformly continuous on (0, ∞) by leveraging the definition of uniform continuity and its negation. The second proof shows that the function g: [0, ∞) → R defined by g(x) = (4x)/(x+2) is uniformly continuous on [0, ∞) by manipulating the expression |g(x) - g(y)| < ε. Key insights include the importance of avoiding the point at infinity in the first proof and the necessity of establishing a delta for the second proof.

PREREQUISITES
  • Understanding of uniform continuity and its definition
  • Familiarity with limits and continuous functions
  • Basic knowledge of mathematical proofs and proof techniques
  • Experience with epsilon-delta arguments in calculus
NEXT STEPS
  • Study the formal definition of uniform continuity and its implications
  • Learn how to construct epsilon-delta proofs for various functions
  • Explore proof by contradiction techniques in mathematical analysis
  • Examine examples of uniformly continuous and non-uniformly continuous functions
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Mathematics students, particularly those studying analysis, educators teaching calculus concepts, and anyone interested in deepening their understanding of continuity and proof techniques.

mtvateallmybrains
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There are two proofs which I have attempted to work on that have beein somewhat trifling. The first of which is :

prove that the function f: [0,infin) -> R defined by f(x) = 1/x is not uniformly continuous on (0,infin).
im thinking that the way in which i should probably attempt to solve this problem is by using a theorem which states:

for a function f: D -> R and a point x_0 in its domain D, the following two assertions are equiv:
i.) the function f: D -> R is contin. on x_0
ii.) for each epsilon > 0, there exists a delta > 0 such that |f(x) - f(y)| < epsilon for all points x in D such that | x - x_0 | < delta

I suppose that if i were to do that then all i would have to do is let x_0 be a point at infinity (in the domain) and allow x to converge to x_0? then somewhere, somehow, something equals zero which is not strictly greater than epsilon (or something along these lines maybe?) also, would it perhaps be easier to solve this proof by contradiction and if so, how would one venture into doing so?

=======================
second proof: prove that g: [0, infin) -> R defined by g(x) = (4x) / (x+2) is uniformly continuous on [0, infin).

after finding a common denom. for | g(x) - g(y) | < epsilon, this resolves to ( 8 |x-y| ) / ( |x+2| |y+2|) the |x-y| in the numerator can be swapped out for a delta, but I am kind of stuck after this. what should i do next and what steps am i leaving out?


While i understand pretty much everything i read it's only a small portion of what's required for a math analysis course (as most of you know). But I am very weak in the area of actually preforming a proof - once i see the direction it's taking, however, i can usually do okay, but the initial steps always give me trouble. I've had the usual basic introduction to continuous functions and limits as well as sequences - everything you usually learn in your first math/calculus analysis course. While i understand the principle of proofs (and even forward-backward techniques) I am still just very weak at actually doing them (for lack of experience). so with all of that having been said, any help would be greatly appreciated.

thank you !
 
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The definition of uniform continuity is:
A function f : D -> R is called uniformly continuous if for every real number
ε>0 there exists a number δ>0 such that for all x, y in D with |x-y|< δ, we have |f(x)-f(y)|<ε.

So a function is NOT uniformly continous
when there exists a real number ε>0, such that for all δ>0
there exists a pair x,y in D with |x-y|< δ and |f(x)-f(y)|>=ε

That's what you need to show for f(x)=1/x.
 
infinity isn't in the domain, so you can't use that as an input.

You have the definition of uniform continuity. What is its negation? Show that there is a delta such that for all ..., which is reasonbly straightforward as long as you don't look anywhere near infinity, since that isn't the reason why it isn't uniformly continuous.
 

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