A question on sequence of functions.

  • #1
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i have the follow question:
Does there exist a continuous function g:R->R that is not periodic, and for every sequence of real numbers {a_k} the sequence of functions g_k(x)=g(x+a_k) has a subsequence converging uniformly on R.

i think i found a suitable function, i just need to be sure it's fine:
i defined g_k(x)=g(x)=sin(e^x)
this function isn't periodic, but if i use Arzela-Ascoli lemma in here, then first of all, this function is uniformly bounded by 1, and i can show that it's equicontinuous, i mean all i need to show is that:
for every e'>0, there exist d>0, such that for every x,y in R whihc satisfy
|x-y|<d and for every n then |f_k(x)-f_k(y)|<e'.
here i can use the fact that |sin(e^(a_k+x))-sin(e^(a_k+y))|<=|e^a_k||e^x-e^y|
e^a_k converges, and e^x is continuous, so we can prove the equality from this, am i correct?

thanks in advance.
 

Answers and Replies

  • #2
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now when i look at this again, i dont think my example is correct, cause im not guaranteed that e^(a_k) converges at all.
so can you find an example or the answer that there isnt one function such as this.
 
  • #3
matt grime
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Let's suppose any continuous function with extra properties will do.

If the a_k are bounded, then they have a convergent subsequence, so use that. If not, then either a subsequence tends to +infty or -infty, so use that.

They're the only obviously 'nice' looking subsequences. Now, can we use a nice function like g(x)=exp(-x^2) to do anything? I chose that because it is bounded, which seems like the least you need for g, and tends to zero as |x| tends to infinty.
 
  • #4
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well, i think i need to expand the function youve given to include also the point x=0, which is no problem cause we can define it as f(0)=0 and otherwise f(x)=exp(-x^2).
obviously this function is continuous, the only question is, is it equicontinuous, well let's look at the difference: |e^-(x+a_k)^2-e^-(y+a_k)^2|=|e^(-(a_k)^2)|*|e^-(x^2+2xa_k)-e^-(y^2+2ya_k)|
basically i dont think i have got matters simpler cause now i have x^2+2xa_k in the exponent in the absolute brackets.
 
  • #5
matt grime
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Why is exp(-x^2) not defined (and equal to 1) at zero?

I had to google equicontinuous. It seems you mistyped something above - a sequence of functions is equicontinuous. You made it sound like a function was equicontinuous, which confused me.
 
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  • #6
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yes i thought it was e^-1/x^2. my mistake, sorry.

anyway, back to topic.
how should i prove that e^-x^2 fits the criteria in my question?
obvoulsy it's uniformly bounded, but the second is what's bugging me.

thanks in advance.
 
  • #7
Dick
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Hey Matt, he's trying to prove that a sequence of FUNCTIONS (arbitrary translates of the original function) converge. Sounds like you are going to have a hard time finding such a function. You can't apply Arzela-Ascoli because R isn't compact. (If g is periodic then you can just work over one period and that's compact).
 
  • #8
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the question asks if there exists such a function which isnt periodic which converges uniformly on R, obviously it's hard to find such a function but is it impossible?
 
  • #9
Dick
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the question asks if there exists such a function which isnt periodic which converges uniformly on R, obviously it's hard to find such a function but is it impossible?
Don't know. Let's think about it...
 
  • #10
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btw, i have another question which is obviously an arzela-ascoli question:
prove that there exists a subsequence {f_k_M} that converges uniformly to on the real line R, where f_k(x)=sin(x+a_k), a_k is a sequence of real numbers.
i proved that for the interval [0,2pi] there exists such subsequence by using arzela ascoli, now i think i only need to expand it to fit also the real line, so perhaps as we expand the interval but keep it compact, we still have a converging subsequence of f_k, but is it enough in order to prove for the whole real line?
 
  • #11
Dick
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If the sequence converges uniformly on [0,2pi] then it converges uniformly on the whole line. The values on the rest of the line just repeat the values in the interval [0,2pi].
 
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  • #12
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any new insight on the initial question, dick or matt?
 
  • #13
Dick
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No. And it's still annoying me.
 
  • #14
Dick
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Ok. The class of 'almost periodic' functions also has this property. I haven't managed to cobble together a proof - maybe you can do that??
 
  • #15
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when you say almost periodic, what to do you mean?
something like my example that sin(e^x), or something like this: e^(ax)*(Asin(wx)+Bcos(wx))?
 
  • #16
Dick
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when you say almost periodic, what to do you mean?
something like my example that sin(e^x), or something like this: e^(ax)*(Asin(wx)+Bcos(wx))?
Nope, not like those. You could look it up. One definition is that for all e>0 there exists a T such that |f(x+T)-f(x)|<e for all x.
 
  • #17
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Dick, I think that g(x)=c where c is constant will do the job, what do you think?
 
  • #18
Dick
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Dick, I think that g(x)=c where c is constant will do the job, what do you think?
Sure. But constant functions are periodic according to my definition.
 
  • #19
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Im not sure that it's periodic, cause there isn't a small non zero T such that g(x+T)=g(x), cause all of the values of g are the same.
 
  • #20
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Im not sure that it's periodic, cause there isn't a small non zero T such that g(x+T)=g(x), cause all of the values of g are the same.
g(x) = c

g(x+T) = c

Therefore, g(x+T)=g(x) for all T and x. So given e>0, |g(x+T)-g(x)|=0<e for all T, x. So the constant function not only is periodic, but has infinitely small periods
 
  • #21
Dick
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g(x) = c

g(x+T) = c

Therefore, g(x+T)=g(x) for all T and x. So given e>0, |g(x+T)-g(x)|=0<e for all T, x. So the constant function not only is periodic, but has infinitely small periods
It's EXTREMELY periodic.
 
  • #22
NateTG
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Do you consider things like
[tex]f(x)=\sin(x)+\cos(\pi x)[/tex]
periodic?
 
  • #23
Dick
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Do you consider things like
[tex]f(x)=\sin(x)+\cos(\pi x)[/tex]
periodic?
No. Should I? What's the period?
 
  • #24
NateTG
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Well, things like that make the need for aperiodicity seem a bit silly since one could simply dialate by an irrational number, and then add that to the original, for anything that isn't constant.

That said, it seems like any continuous aperiodic function [itex]f[/itex] that tends towards a constant [itex]c[/itex] function as it approaches either infinity, and that has a bounded derivative would meet the qualifications:

Since the derivative is bounded, it has some maximum norm [itex]D[/itex]. [itex]\delta=\frac{\epsilon}{D}[/itex] shows that the sequence of functions is uniformly equicontinuous.

Case 1: [itex]a_n[/itex] is unbounded.
Then there is a subsequence that goes to [itex]\infty[/itex], or [itex]-\infty[/itex] and the sequence of functions that corresponds to that subsequence goes to a constant function.

Case 2: [itex]a_n[/itex] is bounded.
Then clearly, there must be some limit point [itex]a[/itex], and subsequences that converge to that limit point. The corresponding functions converge to [itex]f_a(x)=f(x+a)[/itex]

So, an easy example is:
[tex]\frac{1}{e^{-x}+e^{x}}[/tex]
It's trivially aperiodic, the norm of the derivative is less than [itex]\frac{1}{2}[/itex], and the function nicely tends to zero at both infinities.
 
  • #25
Dick
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That example does not work. x_n=n does not have a convergent subsequence. On the other hand sin(x)+cos(pi*x) works great. It's 'almost periodic'. And that's a great example. Take any sequence on the reals and it has a convergent subsequence mod 2. Then that sequence has a convergent subsequence mod 2*pi.
 

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