Recent content by Stardust*

Now your idea is clearer to me, but I still do not have a good idea to use g(x)... Instead I think I have found something useful starting from your initial suggestion (maybe it's what you told me till this point, but unfortunately I have not fully understood). The definition of uniform...

But we know nothing about f(x) from g(x)... Anyway you are adding a linear piece to the value of the original function in 0: this does not mean that g(x)>=f(x) always, unless you thought of g(x) as: g(x)=x/\delta +f(x) In this case I do agree g(x) must be higher than or equal to f(x) itself...

In a previuos post you suggested to use g(x)=x/d+f(0). I think you meant that even g(x) is uniformly continuous. But I fear the presence of f(0) in the formula for g(x) does not affect things so much. You are just saying: take the value of f(0) and use it to build a convenient line crossing...

A correction to my previous post. Extract the highest terms from the denominator as suggested: (the lower part of the fraction is:) \sqrt{x^2-3x}+x=(x^2-3x)^{1/2}+x=[x^{2}(1-\frac{3x}{x^2})]^{1/2}+x=|x|(1-\frac{3}{x})^{1/2}+x \sqrt{x^2-3x}+x=x^{2/2}...

Are you suggesting to use g(x) instead of f(x)? Surely g(x) is uniformly continuous, but I find no clues about how it is useful... Could you tell sth more?

ACHTUNG: This post contains some wrong parts in LaTeX I am not able to correct. Please jump it... When you use the conjugate, the objective is to eliminate all the disturbing x's floating around. If I have understood your initial steps, the limit should look like this (after using the...

f uniformly continuous --> finite slope towards infinity Homework Statement Given f:R \rightarrow R uniformly continuous. Show that \limsup_{x\rightarrow \infty} \displaystyle|f(x)|/x<\infty i.e. \exists C \in R: \, |f(x)|\leq C|x| as x \rightarrow \pm \infty. Homework Equations The...

I'm sorry, I did not want to annoy anyone, I was just trying to see if I could give a hand and if I could solve such exercises in a more or less correct way (these are the first ones of this kind I do, so this is good practice for me, too). I'll be more careful in future... Yes, I guess so...

For what I know, this fact for a row directly derives from the properties of the determinant. It is defined: a) to be a linear application in the rows, b) to give 0 if two rows are equal, c) to give 1 if you calculate det(I_n), with I_n the identity matrix. In particular the answer comes...

The last part of my previous post is: Looking at y=x^2, it is clear it is strictly increasing in the interval [0, \infty]; y=arctg(x)=h(x) is strictly increasing too. So the quantity in (*) is always positive since all the pieces of the sum are positive. This means we can always take x_2 big...

Now, I think there's something wrong in what I said for 4). I'll try again. So we have: f(x)=x^2 arctg(x)=x^2 \cdot h(x) Let's consider two points: x_1,x_2 \in [0,\infty]\, , \,x_2<x_1\, , \, x_1=x_2+\omega, where \omega>0. We get to: \left|(x_2+\omega)^2...

Now a proposal of solution for 4: Let's keep \varepsilon\geq 0 fixed and choose x_1,x_2 \in R so that: 0<x_1<x_2=x_1+\delta /2 The variation in the 'height' of the function between the two points is: |f(x_1)-f(x_2)|=|x_1^2arctg(x_1)-x_2^2arctg(x_2)|\leq...

3) f(x)=x \,arctg(x) Well, I think the work is more or less the same... Let's take x_1,x_2 \in R so that |x_1\,arctg(x_1)-x_2\, arctg(x_2)|\leq \pi|x_1-x_2|\leq \varepsilon So we can take whatever x1,x2 , have the inequality above for \varepsilon \geq \pi |x_1-x_2| and take our |x_1-x_2|\leq...

The property of uniform continuity is: A function f :A \rightarrow R is uniformly continuous is \forall \varepsilon >0, \exists \delta >0: \forall x_1,x_2 \in A, |x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|<\varepsilon This means that for every interval of length 2\varepsilon in the image of...