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Math Help - Continuous & Periodic

  1. #1
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    Continuous & Periodic

    Let f: \mathbb{R} \longrightarrow \mathbb{R} be continuous and f(x)=f(x+1)=f(x+\sqrt{2}), for all x \in \mathbb{R}. Prove that f is constant.

    Source: Berkely Preliminary Exam
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    Senior Member pankaj's Avatar
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    f(x)=f(x+1)

    Also, f(x)=f(x+\sqrt{2})

    If T is periodic with period 1 and also periodic with period \sqrt{2} then \sqrt{2} must be an integral multiple of 1 which is impossible.

    This means that f(x) is periodic but its period is undefined.

    The situation is similar to
    f(x)=c
    where c is constant.Here f(x) is periodic but its period is not defined.

    Thus f(x) may be a constant function.

    But my solution lacks rigour.
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Let f: \mathbb{R} \longrightarrow \mathbb{R} be continuous and f(x)=f(x+1)=f(x+\sqrt{2}), for all x \in \mathbb{R}. Prove that f is constant. Is there a real analogue of Liouville's Theorem? Because  \mathbb{R} \subseteq \mathbb{C} .

    Source: Berkely Preliminary Exam
    We can say that  f is bounded and uniformly continuous. Maybe we can say that  f is a doubly periodic function (e.g.  f(z) = f(z+u) = f(z+v) ). And somehow use Liouville's theorem to show that it is constant? The imaginary parts would be  0 .

    Or going from the first idea...if  f is bounded, then  |f(x)| \leq M for some  M>0 and for all  x \in \mathbb{R} . So suppose for contradiction that  f was not constant. Then it is monotonic increasing or decreasing. But this contradicts the fact that  f is bounded. Hence  f is constant.
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  4. #4
    Senior Member pankaj's Avatar
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    I think I have it.

    f(x)=f(x+1)

    x+1=x+nT,where T is the fundamental period or the smallest period and n is any positive integer.

    T=\frac{1}{n}

    Similarly, f(x)=f(x+\sqrt 2) results in

    x+\sqrt 2=x+mT,where m is a positive integer and T is the fundamental period.

    T=\frac{\sqrt 2}{m}

    \frac{1}{n}=\frac{\sqrt 2}{m}

    m=\sqrt 2 n

    which is impossible since m and n are integers.Therefore T is undefined.But since the function repeats itself it can only be constant
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    Quote Originally Posted by pankaj View Post
    I think I have it.

    f(x)=f(x+1)

    x+1=x+nT,where T is the fundamental period or the smallest period and n is any positive integer.

    T=\frac{1}{n}
    that's not always true. for example consider the function f(x)=\begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}. here we have f(x+r)=f(x), \ \forall x \in \mathbb{R}, \ \forall r \in \mathbb{Q}. of course, f is not continuous!

    my point is that i don't see how "continuity" of f was used in your solution!
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    MHF Contributor Bruno J.'s Avatar
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    Consider the set S=\{a+b\sqrt{2}\: : \: a,b \in \mathbb{Z}\}. This set is dense in \mathbb{R}, and f must be equal to f(0) everywhere in S. Since f is continuous and S is dense, f(x)=f(0) for all real x and f is constant.

    To see that S is dense in \mathbb{R}, notice that it is a subgroup of \mathbb{R}, hence it suffices to prove that it is dense on some interval; we choose the interval [0,1]. Consider J=\{s-\lfloor s\rfloor\: : \: s\in S\}; this is a subset of S \cap [0,1]. Moreover if s_1-\lfloor s_1\rfloor = s_2-\lfloor s_2\rfloor then s_1=s_2 because \sqrt{2} is irrational. Hence there are infinitely many points of S in [0,1], hence two of them are arbitrarily close; but then since S is closed under addition, S is dense in \mathbb{R}, Q.E.D.
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    Quote Originally Posted by Bruno J. View Post

    Moreover if s_1-\lfloor s_1\rfloor = s_2-\lfloor s_2\rfloor then s_1=s_2
    this is not true! counter-examples: s_1=a_1+b\sqrt{2}, \ s_2=a_2+b\sqrt{2}, \ \ a_1 \neq a_2.


    Hence there are infinitely many points of S in [0,1], hence two of them are arbitrarily close
    what do you mean by "them"? whatever it means, obviously S \cap [0,1]=S_1 having infinitely many points doesn't imply that S_1 is dense in [0,1].
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  8. #8
    Senior Member pankaj's Avatar
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    'S' is dense in R

    What does this statement mean
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    Quote Originally Posted by pankaj View Post
    'S' is dense in R

    What does this statement mean
    Dense set - Wikipedia, the free encyclopedia
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    this is not true! counter-examples: s_1=a_1+b\sqrt{2}, \ s_2=a_2+b\sqrt{2}, \ \ a_1 \neq a_2.
    Oops. What I was thinking is that if s_1=b_1\sqrt{2}, s_2=b_2\sqrt{2} and s_1-s_2 \in \mathbb{Z} then s_1=s_2; since there are infinitely many choices of b, there are infinitely many points in S_1.

    what do you mean by "them"? whatever it means, obviously S \cap [0,1]=S_1 having infinitely many points doesn't imply that S_1 is dense in [0,1].
    The fact that S_1 contains infinitely many points does indeed imply that S is dense in \mathbb{R} : it implies that S_1 (and hence S) has an accumulation point, and any additive subgroup of \mathbb{R} having an accumulation point is dense in \mathbb{R}.
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  11. #11
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    Quote Originally Posted by Bruno J. View Post
    Oops. What I was thinking is that if s_1=b_1\sqrt{2}, s_2=b_2\sqrt{2} and s_1-s_2 \in \mathbb{Z} then s_1=s_2; since there are infinitely many choices of b, there are infinitely many points in S_1.



    The fact that S_1 contains infinitely many points does indeed imply that S is dense in \mathbb{R} : it implies that S_1 (and hence S) has an accumulation point, and any additive subgroup of \mathbb{R} having an accumulation point is dense in \mathbb{R}.
    Why is S_1 compact in \mathbb{R}?, or why is it closed? That's what you're using right?, that S_1 is sequentially compact
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Jose27 View Post
    Why is S_1 compact in \mathbb{R}?, or why is it closed? That's what you're using right?, that S_1 is sequentially compact
    Where do you see that I said S_1 is compact or closed? It is neither. I said that S_1 is dense in [0,1].

    To see this rigorously consider the map

    S\rightarrow [0,1]
    s\mapsto s-\lfloor s \rfloor

    This map, restricted to the subset of S consisting of elements of the form b\sqrt{2},\ b \in \mathbb{Z}, is an injection; moreover when s \in S, we have s-\lfloor s \rfloor \in S; hence there are infinitely many points of S in the bounded set [0,1]. Immediately from the Bolzano-Weierstrass theorem we know that S has an accumulation point. It is then an easy matter to show that any additive subgroup of \mathbb{R} having an accumulation point is dense in \mathbb{R}. To see this, suppose you have an open interval I of length L in \mathbb{R}. Then by the above there is an element of S with 0<|s|<L; taking all its multiples (which are all elements of S) we see that one of them must fall in I.
    Last edited by Bruno J.; July 1st 2009 at 10:00 PM. Reason: small mistake
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  13. #13
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    Quote Originally Posted by Bruno J. View Post
    Where do you see that I said S_1 is compact or closed? It is neither. I said that S_1 is dense in [0,1].

    To see this rigorously consider the map

    S\rightarrow [0,1]
    s\mapsto \lfloor s \rfloor

    This map, restricted to the subset of S consisting of elements of the form b\sqrt{2},\ b \in \mathbb{Z}, is an injection; moreover when s \in S, we have \lfloor s \rfloor \in S; hence there are infinitely many points of S in the bounded set [0,1]. Immediately from the Bolzano-Weierstrass theorem we know that S has an accumulation point. It is then an easy matter to show that any additive subgroup of \mathbb{R} having an accumulation point is dense in \mathbb{R}. To see this, suppose you have an open interval I of length L in \mathbb{R}. Then by the above there is an element of S with 0<|s|<L; taking all its multiples (which are all elements of S) we see that one of them must fall in I.
    I still don't get it, for example in I=[0,1] there are only two integers 0,1 and no elements of the form b\sqrt{2} with b\neq 0, the element \sqrt{2} -1 \in [0,1] and is the only one that falls in I from the set A= \{ \sqrt{2} +a : a \in \mathbb{Z} \} and the set B= \{ b\sqrt{2} -1 : b \in \mathbb{Z} \} and S=A+B, but how do you conclude that there are infinitely many elements from S in [0,1]
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  14. #14
    MHF Contributor Bruno J.'s Avatar
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    You should read carefully what I wrote.

    \sqrt{2} = 1.41421...
    2\sqrt{2} = 2.82843...
    3\sqrt{2} = 4.24264...
    4\sqrt{2} = 5.65685...

    and the sequence

    \{0.41421...,\: 0.82843...,\: 0.24264...,\: 0.65685...,\: ...\}

    is a sequence of distinct elements of S, all in [0,1].
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  15. #15
    Senior Member pankaj's Avatar
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    I think it is time that solution be revealed
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