yes but i dont know what is smaller then |x-y|
i only know that delta is bigger then |x-y|
I don't think so. Every function which is differentiable and has bounded derivative is uniformly continuous (Wikipedia), and it is possible to have a function that is above y = x and has a bounded derivative. The problem with is that its derivative is not bounded on all . In fact, the link above has an outline of the proof that is not uniformly continuous.
This is correct, but you are interested in the lower, not upper bound on : you want to show that it is .so we must prove that there is so for all
there is x,y in R which and
It's in the section "Properties," just below the sentence I quoted.
ok they input there the values and say that for sufiently bigger x the wole thing is bigger then epsilon
i choose
now i need to choose x so i will get a number bigger then 2 and without delta.
i have
i cant think of an x so delta will go away and we get number bigger then 2
i thought of
but then i cant get because i get
and i cant prove that its smaller then delta
??
yes but i cant write this explanation,its not a way of prooving uniform continuety.
i have to show bigger then 2 number in the result to finish this prove
we have to choose x and y so there subtraction is less then delta
and our result is bigger then 2 (epsilon=2)
how to choose such variables
Lemma from algebra: For all a, b and c, if a > 0, then either for all (this happens when the discriminant ) or for all , for example, for .
We need to make , i.e., . By the lemma above, if the discriminant of this polynomial is negative, you can take any x; otherwise, take x to be the rightmost root plus 1 (and the second point would be ).
i need to follow the guideline of the proof.i cant replace a step of the proof with verbal explanation
its very strict
so you take x=1
ok there subtraction is smaller then delta.
but if we input them to our expression
we get an expression which is a delta polinomial.
and we cant say that this expression is bigger then 2
??
I provided a proof of the existence of x such that (modulo the proof of the algebraic lemma). From there, it follows that for any there exists an x such that . In mathematics, it is not customary to take a proof and then do something different that may or may not work. Try to understand the proof. You may need further explanation, but ultimately either accept that the proof is correct or give a reason why it's flawed. In the original problem ("prove or desprove that 'uniformly' continuous in R) I don't see any requirement that the proof has to be of a certain kind (such as, not use any lemmas).
ok i will try to comprehend this last step.
could you guide me in my other posts,i have made some proggress i just need guidance like you did here to help me
developing the solution
1.
http://www.mathhelpforum.com/math-he...on-187134.html
2.
http://www.mathhelpforum.com/math-he...on-187133.html
3.
http://www.mathhelpforum.com/math-he...on-187135.html
thanks