F uniformly continuous -> finite slope towards infinity

In summary, the conversation discusses the concept of uniform continuity and its application to proving that the limit supremum of the absolute value of a uniformly continuous function is finite. The idea is to use the definition of uniform continuity to show that, for any given epsilon, there exists a delta such that the function's graph can be contained in small boxes of side delta and epsilon. Then, by considering a series of smaller boxes moving towards 0, the function can be shown to be bounded by a linear function as x approaches infinity. This ultimately leads to the conclusion that the limit supremum of the absolute value of the function is finite.
  • #1
Stardust*
14
0
f uniformly continuous --> finite slope towards infinity

Homework Statement


Given [tex]f:R \rightarrow R[/tex] uniformly continuous. Show that [tex]\limsup_{x\rightarrow \infty} \displaystyle|f(x)|/x<\infty[/tex] i.e.
[tex]\exists C \in R: \, |f(x)|\leq C|x| [/tex] as [tex]x \rightarrow \pm \infty[/tex].


Homework Equations


The idea of uniform continuity:
[tex]\forall \varepsilon >0, \, \exists \delta>0:\, \forall x,y \in R, |x-y|<\delta \, \Rightarrow \, |f(x)-f(y)|<\varepsilon [/tex].




The Attempt at a Solution


If f were a linear function, it could be quite easy...
But f is unknown, so I tried sth else:
let's say uniform continuity gives the certainty we can always find a box in which the studied part of the function's graph is contained properly.
In this case we need a corner of the box to be in 0 (as the denominator simply has |x-0|=|x|) and let the other corner move as far as possible towards infinity, along the x-axis. Moreover in vertical we have to forget the the typical |f(x)-f(y)|, leaving |f(x)| only.
This suggests that the contribution of f(y) must somehow be negligible, I mean we can find a larger side of length f(x) for the box to keep the function inside it, without caring about f(y).

These are my thoughts till now. The first problem is: are these ideas right?
The second one is: how to express them in a formal, mathematical language?
 
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  • #2


so how about picking e = 1, then there exists d>0 such that |f(x+d) -f(x)| < 1 for all x, now consider the function g(x) = x/d + f(0)?
 
  • #3


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?
 
  • #4


what can you say about g(x) vs f(x)?

(may neeed to be careful with axis crossings so maybe coonsidering g(x)-g(0) and f(x)-f(0) would help)
 
  • #5


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 the y-axis at f(0). But it is well known that a straight line is uniformly continuous, and the connection between g and f is quite weak for me.
Now some thoughts of mine:
chosen [tex]\varepsilon[/tex], I find my d such that [tex]y=x+d=d[/tex] (for x=0). But if [tex]\varepsilon[/tex] is constant, it binds the range in which d can run, preventing y from 'reaching' infinity, while our limit should get to infinity.
I hope somehow there exists a greater value for [tex]\varepsilon[/tex], maybe the [tex]\psi=\sup_{i \in I}\{\varepsilon_i\}[/tex], where I is an appropriate set of index. But nothing suggests [tex]\psi [/tex] cannot be infinity...

Anyway I would like to do something like this:
[tex]\lim_{y\rightarrow \infty, x=0}=\displaystyle \left| \frac{f(y)-f(x)}{y-x}\right|=\lim_{y\rightarrow \infty}\left|\frac{f(y)-f(0)/f(y)}{y}\right|[/tex]
where the fraction f(0)/f(y)n should be negligible. But this cannot be said so easily at the moment. Moreover we want the whole fraction to be finite (at least minor than a certain C), which means the speed of f(y) and y can be compared.
Am I on the right way to the solution (right way... there exist many, I think, but I'm still looking for one :blushing:)?
 
  • #6


could you convince yourself that g(x)>=f(x) for all x=nd, where n is a natural number?

the way we defined it g(0) = f(0)
 
  • #7


lanedance said:
now consider the function g(x) = x/d + f(0)?

could you convince yourself that g(x)>=f(x) for all x=nd, where n is a natural number?

the way we defined it g(0) = f(0)

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:
[tex]g(x)=x/\delta +f(x)[/tex]
In this case I do agree g(x) must be higher than or equal to f(x) itself, at least for every x>0.
 
  • #8


as f is uniformly continuous with e=1, you know there exists d>0 such that |f(x+d) -f(x)| < 1 for all x

now consider the function g(x) = x/d + f(0), clearly
g(x+d)-g(x) = ((x+d)/d + f(0)) - (x/d + f(0)) = 1

so g is increasing at least as quickly or faster than f
 
  • #9


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 continuity tells that:
[tex]\forall \varepsilon >0, \, \exists \delta>0:\, \forall x,y \in R, |x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon[/tex]

So there exists a particular [tex]\varepsilon '[/tex] such that there is a [tex]\delta '[/tex] for which [tex]\forall x,y \in R,|x-y|<\delta'[/tex] and [tex]|f(x)-f(y)|<\varepsilon '[/tex].
This means that for this fixed quantities, we can always check that the function's graph is contained in small boxes of side [tex]\delta' \,and\,\varepsilon '[/tex]. The problem is that we have now fixed these values, while the objective is to make a statement which means there is a greater box containing the function as [tex]x \rightarrow \infty[/tex]. Then my idea is that we can consider all those smaller boxew from a generic x (as big as we want) back to 0, moving with steps of length [tex]\delta '[/tex]: this means we make a number of steps [tex]n=|x|/\delta '[/tex].
So taken x>0, we surely have:
[tex]|f(x)-f(x-\delta ')|<\varepsilon '[/tex]
[tex]|f(x-\delta ')-f(x-2\delta ')|<\varepsilon '[/tex]
[tex]|f(x-2\delta ')-f(x-3\delta ')|<\varepsilon '[/tex]

[tex]|f(x-n\delta ')-f(0)|<\varepsilon '[/tex]
With this last passage, we get to f(0), or even a little before it. Anyway things should work well.
Let's sum all this inequalities:
[tex]|f(x)-f(x-\delta ')|+f(x-\delta ')-f(x-2\delta ')|+|f(x-2\delta ')-f(x-3\delta ')|+\ldots+|f(x-n\delta ')-f(0)|<n\varepsilon '[/tex]
using the generalised triangular inequality, this brings to:
[tex]|f(x)-f(0)|=|f(x)-f(x-\delta ')+f(x-\delta ')-f(x-2\delta ')+f(x-2\delta ')-f(x-3\delta ')+\ldots+f(x-n\delta ')-f(0)|<n\varepsilon '[/tex]
So:
[tex]|f(x)-f(0)|<n\varepsilon '[/tex], but before we stated [tex]n=|x|/\delta '[/tex].
This leads to [tex]|f(x)-f(0)|<|x|\displaystyle\frac{\varepsilon '}{\delta '}[/tex]
Using the fact that:
[tex]|\,|f(x)|-|f(0)|\,|<|f(x)-f(0)|<|x|\displaystyle\frac{\varepsilon '}{\delta'}[/tex], and noting that the RHS is surely positive as we move to +infinity, we can write
[tex]|f(x)|<f(0)+|x|\displaystyle\frac{\varepsilon '}{\delta '}=|x|(\displaystyle \frac{f(0)}{|x|}+\frac{\varepsilon '}{\delta '})[/tex]

We chose our [tex]\varepsilon '[/tex], so we actually cannot control the
[tex]\delta '[/tex], but if it is greater than d'>|x|, there's no problem since the thesis is automatically fulfilled. If it is [tex]\delta ' <|x|[/tex] we can put:

[tex]|f(x)|<|x|(\displaystyle \frac{f(0)}{|x|}+\frac{\varepsilon '}{\delta '})<|x|(\displaystyle \frac{f(0)}{\delta '}+\frac{\varepsilon '}{\delta '})<[/tex] which is a constant value identified with C.
In the end, we showed that:

[tex]|f(x)|<|x|C[/tex], for a generic x, as asked by the exercise.

I hope this proof is correct...
 

What does it mean for a function to be uniformly continuous?

Uniform continuity is a property of a function that states that for any given value of epsilon, there exists a corresponding value of delta such that the difference between any two input values within delta will result in an output difference within epsilon. In simpler terms, this means that the function's rate of change is consistent throughout its domain.

How can you determine if a function has finite slope towards infinity?

A function has a finite slope towards infinity if its rate of change does not approach a constant value as the input values approach infinity. In other words, the function's rate of change must not become infinitely steep or flat as the input values increase.

What are the implications of a function having finite slope towards infinity?

A function with finite slope towards infinity will have a consistent rate of change as the input values increase. This means that the function will not experience sudden changes or fluctuations in its behavior as the input values get larger. It also means that the function's behavior can be more easily predicted and analyzed.

Can a function be uniformly continuous and have finite slope towards infinity at the same time?

Yes, a function can have both properties simultaneously. A function that is uniformly continuous will have a consistent rate of change throughout its domain, and if this rate of change does not approach a constant value as the input values approach infinity, then the function will also have finite slope towards infinity.

How does having finite slope towards infinity affect the graph of a function?

The graph of a function with finite slope towards infinity will typically have a smooth and consistent appearance. This means that there will be no sudden changes or discontinuities in the graph, and the slope will not become infinitely steep or flat as the input values increase. Additionally, the graph will not exhibit any sharp turns or corners, as these would indicate a sudden change in the function's rate of change.

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