If f is continuous on [a,b] it is uniformly continuous on [a,b].
f is continuous at x if derivative exists at x.
So try calculating derivative.
So we have learned Uniform Continuity recently, and I must say it's really confusing. I get the idea, however it's hard to prove that something is uniform continuous or not. I posted a picture, I have spent a lot of time on it, and I am pretty sure that what I did is not right/enough, because at the end I found a delta that is dependent on x and y , and from what I understand delta is dependent on nothing, but I would like to get feedback, I would like to get feedback, and if I am in the wrong way ,can someone give me a hint? Thank you.
Its derivative definitely exists on $[1,\infty)$, but the function is not uniformly continuous.
I have little time now, but I'll give one remark about the attached proof. The fact that $f(x)<F(x)$ (or even $|f(x)|<F(x)$) and $g(x)<G(x)$ does not imply that $|f(x)-g(x)|<|F(x)-G(x)|$. But you are right to see the problem that $\delta$ depends on $x$ and $y$.
The OP is applying the definition of continuity:
f is uniformly continuous on an interval A if for every ε>0 there is some δ>0 st for all x and y in A,
if |x-y|<δ, then |f(x)-f(y)|<ε
The procedure in OP is straight forward leading to: δ=ε/(x+y)
That is wrong, because uniform continuity requires that δ be independent of x,y.
However, if you can show |x+y|<constant over entire interval , the proof is ok.
EDIT: |x+y| obviously less than constant, so with that modification proof is ok.
Now you can concentrate on just following the algebra of inequalities. I assume that's correct.
I didn’t follow the algebra. I assumed it was correct. Will check.
1/x ≤ 1 for x ≥ 1
|sin(x^{3})| ≤ 1 for all x
Whoops. x -> infinity. OK, going too fast. Will have to go through proof.
Almost certain that this proof is true.Happy if you can check it, Thanks for the feedback guys!
Note: I wrote this very fast so there are some mistakes like the ] sign where I wrote infinity.
|f(x)-f(y)| ≤ ||f(x)| - |f(y)|| ≤ |1/|x| - 1/|y| | ≤ | x-y|/|xy| ≤ | x-y| < | x-y|/2
δ = 2ε
EDIT: OK, a little concise. Note:
if |a| ≤ |p|, and |b| ≤ |q|
Then ||a| - |b|| ≤ ||p| - |q||
This reminds me the book "Moscow 2042" by Vladimir Voinovich, where he describes how by 2042 "angry KGB generals" have built communism in a single city: Moscow. The slogan of the USSR-style socialism was "From everybody according to their abilities, to everybody according to their labor", in other words, "He who does not work, does not eat either". In contrast, the slogan of communism was supposed to be, "From everybody according to their abilities, to everybody according to their needs". That is, it does not matter what you do, your needs will be supplied for. This communism never came in real life, but the book has a dialog between the protagonist and another man. This man assures that communism has arrived and everybody can get what they want according to their needs. The protagonist asks who determines those needs: people themselves? But it turns out that individual's needs are determined by special committees consisting of party workers, security officials and so on.
Back to calculus: if the definition says that something holds "for all $\epsilon$", it won't do to ensure that it holds only for $\epsilon$ having the form 2 + ... .
I don't agree with these two inequalities, but it is true thatOriginally Posted by Hartlw
$$\left|\frac{\sin(x^3)}{x}-\frac{\sin(y^3)}{y}\right|\le\frac{1}{x}+\frac{1}{ y}\le \frac{2}{\min(x,y)}.$$Then we can find an $x_0$ such that $2/x<\epsilon$ for all $x> x_0$. Thus, if $x,y>x_0$, then $|f(x)-f(y)|<\epsilon$ regardless of how far $x$ is from $y$. Concerning the closed segment $[1,x_0]$, the function is continuous and therefore uniformly continuous there, so for the given $\epsilon$ there exists a $\delta$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. Therefore, this $\delta$ will work on the whole interval $[1,\infty)$. (To be strict, one also has to take care of the case $x<x_0<y$.)
So yes, I was wrong in post #3, and the function is UC. I was thinking that it is not Lipschitz continuous (a stronger condition that implies UC) because the derivative grows indefinitely.
This does not hold for a = 2, b = 1 and p = q = 3.Originally Posted by Hartlw
By the left side, do you mean $1/x+1/y$? Yes, if $\epsilon$ is small, say, $\epsilon=0.01$, and x = 2, then $1/x+1/y>\epsilon$. That's why I consider two cases: when x and y are small and when they are large. In the first case, $|f(x)-f(y)|$ is small because $x$ is close to $y$. This can be achieved for any continuous function on a closed segment. And when $x$ and $y$ are large, then $|f(x)-f(y)|$ is small just because both $|f(x)|$ and $|f(y)|$ are small, i.e., less than $\epsilon/2$ and $|f(x)-f(y)|\le|f(x)|+|f(y)|<\epsilon$. The segment for the first case is well-defined because I first determine when $|f(x)|$ becomes $<\epsilon/2$ and then use this point as the right border of the segment.
I am not sure I understand this.
You are right... If I define epsilon to be 2 + delta than it is true for every epsilon greater than 2... So it's not for all epsilon as I was trying to show... The thing is it's easy to prove this with the statements that we proved in class, however I want to prove this without lipschnitz or with the fact that it is a closed interval from the left ([1, ) and in infinity the function converges to a certain point so it is uniformly continuous.. Anyway I will try to prove this again then and send a new proof. Thanks for the feedback guys! I really appreciate it