# Continuous functions

1. Aug 12, 2006

### Werg22

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.

???

2. Aug 12, 2006

### HallsofIvy

If $x_0$ is in the closed interval then the definition of "continuous" is that $\lim_{x\rightarrow x_0} f(x)= f(x_0)$. That in turn means "given $\epsilon> 0$ there exist $\delta> 0$ such that if $|x-x_0|< \delta$ then $|f(x)-f(x_0)|< \epsilon$.
To get your result, let $x_0= x$ and $x= x_0+ \delta$. Then your "a" is "$\delta$", "e" is "$\epsilon$ and $|f(x)- f(x_0)|<\epsilon$ 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. Aug 12, 2006

### StatusX

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.

Last edited by a moderator: May 2, 2017
4. Aug 12, 2006

### quasar987

Mmh, yes, the existence of the 'a' for all x is assured by definition of uniform continuity + Heine-Cantor in the case $(X,d(x,y))=(\mathbb{R},|x-y |)$.

P.S. Halls, do you know how to write in italic w/o using Latex?

Last edited: Aug 12, 2006
5. Aug 12, 2006

### mathwonk

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. Aug 16, 2006

### Werg22

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

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.

Last edited: Aug 16, 2006
7. Aug 17, 2006

### StatusX

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.

Last edited: Aug 17, 2006
8. Aug 17, 2006

### Werg22

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