F uniformly continuous -> finite slope towards infinity

[Stardust*]

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

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?

 

[lanedance]

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)?

 

[Stardust*]

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?

 

[lanedance]

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)

 

[Stardust*]

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 \varepsilon, I find my d such that y=x+d=d (for x=0). But if \varepsilon 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 \varepsilon, maybe the \psi=\sup_{i \in I}\{\varepsilon_i\}, where I is an appropriate set of index. But nothing suggests \psi cannot be infinity...

Anyway I would like to do something like this:
\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|
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 )?

 

[lanedance]

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)

 

[Stardust*]

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

 

[lanedance]

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

 

[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 continuity tells that:
\forall \varepsilon &gt;0, \, \exists \delta&gt;0:\, \forall x,y \in R, |x-y|&lt;\delta \Rightarrow |f(x)-f(y)|&lt;\varepsilon

So there exists a particular \varepsilon &#039; such that there is a \delta &#039; for which \forall x,y \in R,|x-y|&lt;\delta&#039; and |f(x)-f(y)|&lt;\varepsilon &#039;.
This means that for this fixed quantities, we can always check that the function's graph is contained in small boxes of side \delta&#039; \,and\,\varepsilon &#039;. 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 x \rightarrow \infty. 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 \delta &#039;: this means we make a number of steps n=|x|/\delta &#039;.
So taken x>0, we surely have:
|f(x)-f(x-\delta &#039;)|&lt;\varepsilon &#039;
|f(x-\delta &#039;)-f(x-2\delta &#039;)|&lt;\varepsilon &#039;
|f(x-2\delta &#039;)-f(x-3\delta &#039;)|&lt;\varepsilon &#039;

|f(x-n\delta &#039;)-f(0)|&lt;\varepsilon &#039;
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:
|f(x)-f(x-\delta &#039;)|+f(x-\delta &#039;)-f(x-2\delta &#039;)|+|f(x-2\delta &#039;)-f(x-3\delta &#039;)|+\ldots+|f(x-n\delta &#039;)-f(0)|&lt;n\varepsilon &#039;
using the generalised triangular inequality, this brings to:
|f(x)-f(0)|=|f(x)-f(x-\delta &#039;)+f(x-\delta &#039;)-f(x-2\delta &#039;)+f(x-2\delta &#039;)-f(x-3\delta &#039;)+\ldots+f(x-n\delta &#039;)-f(0)|&lt;n\varepsilon &#039;
So:
|f(x)-f(0)|&lt;n\varepsilon &#039;, but before we stated n=|x|/\delta &#039;.
This leads to |f(x)-f(0)|&lt;|x|\displaystyle\frac{\varepsilon &#039;}{\delta &#039;}
Using the fact that:
|\,|f(x)|-|f(0)|\,|&lt;|f(x)-f(0)|&lt;|x|\displaystyle\frac{\varepsilon &#039;}{\delta&#039;}, and noting that the RHS is surely positive as we move to +infinity, we can write
|f(x)|&lt;f(0)+|x|\displaystyle\frac{\varepsilon &#039;}{\delta &#039;}=|x|(\displaystyle \frac{f(0)}{|x|}+\frac{\varepsilon &#039;}{\delta &#039;})

We chose our \varepsilon &#039;, so we actually cannot control the
\delta &#039;, but if it is greater than d'>|x|, there's no problem since the thesis is automatically fulfilled. If it is \delta &#039; &lt;|x| we can put:

|f(x)|&lt;|x|(\displaystyle \frac{f(0)}{|x|}+\frac{\varepsilon &#039;}{\delta &#039;})&lt;|x|(\displaystyle \frac{f(0)}{\delta &#039;}+\frac{\varepsilon &#039;}{\delta &#039;})&lt; which is a constant value identified with C.
In the end, we showed that:

|f(x)|&lt;|x|C, for a generic x, as asked by the exercise.

I hope this proof is correct...