Showing Value "a" for Continuous Function f(x) in a Closed Interval

In summary, the conversation discusses how to show the existence of a value "a" small enough such that for arbitrary x, the function f(x+a) - f(x) is less than a given lower bound e. This can be proved by using the definition of continuity and the Heine-Cantor theorem for functions on a closed interval. The existence of such a value is also connected to the convergence of Riemann sums for continuous functions on a closed interval.
  • #1
Werg22
1,431
1
This is a simple question: given a continuous function, f(x), in a closed interval, how do we show that there is value "a" small enough such as for arbitrary x:

f(x+a) - f(x) < e

Where e lower bound is 0.

?

Thxs in advance.
 
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  • #2
If [itex]x_0[/itex] is in the closed interval then the definition of "continuous" is that [itex]\lim_{x\rightarrow x_0} f(x)= f(x_0)[/itex]. That in turn means "given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x-x_0|< \delta[/itex] then [itex]|f(x)-f(x_0)|< \epsilon[/itex].
To get your result, let [itex]x_0= x[/itex] and [itex]x= x_0+ \delta[/itex]. Then your "a" is "[itex]\delta[/itex]", "e" is "[itex]\epsilon[/itex] and [itex]|f(x)- f(x_0)|<\epsilon[/itex] becomes |f(x+a)-f(x)|< e. Of course, it's still true without the absolute value since if f(x+a)- f(x) is negative it's still less than the positive e.
 
  • #3
If the function is uniformly continuous, the proof is easy. This would mean that for any e>0, there is a d>0 such that |x-y|<d implies |f(x)-f(y)|<e for all x,y in the domain. So just pick your a as d/2.

It's a little more complicated to show this from just the assumption of continuity (which, remember, is a local property, where as you're looking for an a with a ceratin global property), but the http://en.wikipedia.org/wiki/Heine-Cantor_theorem" [Broken] states that these definitions are equivalent if the domain is compact, which a closed interval is. If you can't just cite this theorem, there is a link to a proof on the wikipedia page.
 
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  • #4
Mmh, yes, the existence of the 'a' for all x is assured by definition of uniform continuity + Heine-Cantor in the case [itex](X,d(x,y))=(\mathbb{R},|x-y |)[/itex].P.S. Halls, do you know how to write in italic w/o using Latex?
 
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  • #5
just subdivide your interval into unit length subintervals, then subintervals of length 1/10, then 1/100, and so on...assuming no subdivision ever has intervals on all of which the variation is less than epsilon, resulots in the existence of an infinite decimal, i.e. a real number, in our interval where the function is not continuous.
 
  • #6
My question has been misunderstood... sorry the lack of clarity. What I meant is:

Let e and c be constants, for all x. Is there such a constant for which, regardless of x;

If within the interval [a,b]

(</= stands for inferior or equal)

f(x+e) </= a

then

f(x+e) - f(x) </= c


Additionally,

f(x+l) - f(x) ; if l < k < e

then

f(x+l) - f(x) < f(x+k) - f(k) < c


?

If there is, what is the proof?


This all has to do with a proof that all Riemann sums converge towards the same value when the span of its subdivision tends to 0. Thanks in advance.
 
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  • #7
I can't make any sense out of your last post, werg. If what you're trying to do is prove all Riemann sums converge to the same limit for continuous functions on a closed interval, you would want to start with the question we thought you asked and then answered.
 
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  • #8
Okay, so after a little thought I solved it. Sorry for the lack of clarity again, I tried my best to explain. N/P now :smile:
 

1. What does it mean to show value "a" for a continuous function f(x) in a closed interval?

Showing value "a" for a continuous function f(x) in a closed interval means that the function has a specific output, represented by "a", at every point within the interval. In other words, the function is defined and has no interruptions or gaps within the given interval.

2. How can one prove that a function has value "a" in a closed interval?

To prove that a function has value "a" in a closed interval, one can use the Intermediate Value Theorem. This theorem states that if a function is continuous on a closed interval and takes on two different values at the endpoints of the interval, then it must also take on every value between those two endpoints, including "a". Therefore, if a function is continuous and has values "a" at both endpoints of the interval, it must have value "a" at every point within the interval.

3. Can a function have more than one value "a" in a closed interval?

No, a continuous function can only have one value "a" in a closed interval. This is because the definition of a continuous function implies that there are no gaps or interruptions in the function within the given interval. Therefore, there can only be one output, or value "a", for every input within the interval.

4. What is the importance of showing value "a" for a continuous function in a closed interval?

Showing value "a" for a continuous function in a closed interval is important because it confirms that the function is well-defined and consistent within the given interval. This allows us to make meaningful conclusions and predictions about the behavior of the function within the interval, without any unexpected or undefined values.

5. Can a function be continuous without having value "a" in a closed interval?

No, a function cannot be continuous without having value "a" in a closed interval. As mentioned earlier, the definition of a continuous function includes the absence of any gaps or interruptions within the given interval. Therefore, if a function is continuous, it must have value "a" at every point within the interval.

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