Results 1 to 12 of 12

Math Help - continues prove question..

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    continues prove question..

    there is a function f(x) which is continues
    and every rational number "r" is its period of f(x)

    prove that f(x)=const

    ??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by transgalactic View Post
    there is a function f(x) which is continues
    and every rational number "r" is its period of f(x)

    prove that f(x)=const

    ??
    First let us establish a lemma. If f is continuous at c then there exists a neighborhood (c-\delta,c+\delta)~~\delta>0 such that for any x_1,x_2\in (c-\delta,c+\delta) the following must be true, |c-x_1|<|c-x_2|\implies|f(c)-f(x)|\leqslant|f(c)-f(x_1)|.

    Proof: Suppose the above is false, so that for any neighborhood of c there exists points x_1,x_2 in that neighborhood such that |x_1-c|<|x-c_2| and |f(c)-f(x_1)|>|f(c)-f(x_2)|{\color{red}(*)}. It is clear then from \color{red}(*) that |f(c)-f(x_1)| may be written as |f(c)-f(x_2)|+e~~e>0. So now in the definition of continuity choose \varepsilon=|f(c)-f(x_1)|+d~~0<d<e so there must exist a \delta>0 such that |x-c|<\delta\implies |f(c)-f(x)|<\varepsilon but this is a conradiction since |c-x_1|<|c-x_2|<\delta but |f(c)-f(x_1)|=|f(c)-f(x_2)|+e>\varepsilon

    So now back to the problem. Let us talk about our function on an interval [a,b]. Let c\in[a,b] and f(c)=\xi. So by the above lemma we must have that there is an interval such that |c-x_1|<|c-x_2|\implies |\xi-f(x_1)|\leqslant|\xi-f(x_2)| with x_1,x_2 elements of that interval. Now let x_2=c\pm r~~r\in\mathbb{Q}\implies f(x_2)=\xi. So we must then have that |c-x_1|<|c-x_2|\implies |f(c)-f(x_1)|\leqslant |\xi-\xi|=0 or that f(x_1)=\xi. So for any point of [a,b] there is a neighborhood such that f=\xi, thus we must have that f=\xi~~\forall x\in[a,b]

    Note: I forgot to mention one important fact...this construction is possible that since no matter how small the neighborhood in question is it must contain at least one rational
    Last edited by Mathstud28; December 24th 2008 at 03:34 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Mathstud28 View Post
    First let us establish a lemma. If f is continuous at c then there exists a neighborhood (c-\delta,c+\delta)~~\delta>0 such that for any x_1,x_2\in (c-\delta,c+\delta) the following must be true, |c-x_1|<|c-x_2|\implies|f(c)-f(x)|\leqslant|f(c)-f(x_1)|.
    This lemma somehow tells that a continuous function is locally monotonous. This fails to be true... Think of x\mapsto x\sin\frac{1}{x} for instance, at 0. It is continuous and it oscillates so the conclusion of the lemma can't be satisfied. The problem in the proof comes from the fact that there is no reason why x_1,x_2 would be in [c-\delta,c+\delta]. ( \delta is defined given x_1,x_2)

    There's much simpler (and correct): notice that for any rational q, since f is q-periodic, f(q)=f(0). In other words, f is constant on \mathbb{Q}. Let's say f(0)=c. Now, let x be any real number, rational or not. It is well known that there exists a sequence (q_n)_n of rational numbers that converges toward x. Since f is continuous, we conclude f(x)=\lim_n f(q_n)=c (the sequence (f(q_n))_n is constant, equal to c). qed.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Exclamation

    Quote Originally Posted by transgalactic View Post
    there is a function f(x) which is continues
    and every rational number "r" is its period of f(x)

    prove that f(x)=const

    ??
    Quote Originally Posted by Laurent View Post
    This lemma somehow tells that a continuous function is locally monotonous. This fails to be true... Think of x\mapsto x\sin\frac{1}{x} for instance, at 0. It is continuous and it oscillates so the conclusion of the lemma can't be satisfied. The problem in the proof comes from the fact that there is no reason why x_1,x_2 would be in [c-\delta,c+\delta]. ( \delta is defined given x_1,x_2)

    There's much simpler (and correct): notice that for any rational q, since f is q-periodic, f(q)=f(0). In other words, f is constant on \mathbb{Q}. Let's say f(0)=c. Now, let x be any real number, rational or not. It is well known that there exists a sequence (q_n)_n of rational numbers that converges toward x. Since f is continuous, we conclude f(x)=\lim_n f(q_n)=c (the sequence (f(q_n))_n is constant, equal to c). qed.
    Once I was asked a question by my professor in real analysis lesson, now I feel that it is very similar to this problem.
    Let f be a continuous function, if f is of fixed value at irrational numbers, then f takes that fixed value on rational numbers too, and hence it is a constant function.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Laurent View Post
    This lemma somehow tells that a continuous function is locally monotonous. This fails to be true... Think of x\mapsto x\sin\frac{1}{x} for instance, at 0. It is continuous and it oscillates so the conclusion of the lemma can't be satisfied. The problem in the proof comes from the fact that there is no reason why x_1,x_2 would be in [c-\delta,c+\delta]. ( \delta is defined given x_1,x_2)

    There's much simpler (and correct): notice that for any rational q, since f is q-periodic, f(q)=f(0). In other words, f is constant on \mathbb{Q}. Let's say f(0)=c. Now, let x be any real number, rational or not. It is well known that there exists a sequence (q_n)_n of rational numbers that converges toward x. Since f is continuous, we conclude f(x)=\lim_n f(q_n)=c (the sequence (f(q_n))_n is constant, equal to c). qed.
    Uh, but f:x\mapsto x\sin\left(\frac{1}{x}\right) is not continuous at zero. And I think I stated my Lemma wrong. It was meant to be that there exists a neighborhood of c such that there is a subset of that neighborhood that such that the above is true. Supposing that this suffices it still shows the result of the question and I believe it works for your example.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Mathstud28 View Post
    Uh, but f:x\mapsto x\sin\left(\tfrac{1}{x}\right) is not continuous at zero.
    It is continuous at 0 (provided that you define f(0)=0).

    Quote Originally Posted by Mathstud28 View Post
    And I think I stated my Lemma wrong. It was meant to be that there exists a neighborhood of c such that there is a subset of that neighborhood that such that the above is true. Supposing that this suffices it still shows the result of the question and I believe it works for your example.
    The "subset" would have to be highly disconnected for such a result to be true. In any case, such an elaborate result is not needed here, as Laurent's simple proof shows.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Sorry Opalg and transgalactic for the fallacious proof. I always make things harder than they are. How about this alternate proof (this is more for my curiosity then for the OP)

    Suppose the claim is true and f is constant. We must merely show that f is differentiable everywhere and f'=0, in other words for all x we must show that \lim_{t\to x}\frac{f(t)-f(x)}{t-x}=0. But now consider letting t=x+ r~~r\in\mathbb{Q}. So the above is equivalent to showing \lim_{r\to{0}}\frac{f(x+r)-f(x)}{r}=0 but f(x+r)=f(x) so \frac{f(x+r)-f(x)}{r}=0 which finally in turn implies \lim_{r\to0}\frac{f(x+r)-f(x)}{r}=0
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Mathstud28 View Post
    Sorry Opalg and transgalactic for the fallacious proof. I always make things harder than they are. How about this alternate proof (this is more for my curiosity then for the OP)

    We must merely show that f is differentiable everywhere and f'=0, in other words for all x we must show that \lim_{t\to x}\frac{f(t)-f(x)}{t-x}=0. But now consider letting t=x+ r~~r\in\mathbb{Q}. So the above is equivalent to showing \lim_{r\to{0}}\frac{f(x+r)-f(x)}{r}=0 but f(x+r)=f(x) so \frac{f(x+r)-f(x)}{r}=0 which finally in turn implies \lim_{r\to0}\frac{f(x+r)-f(x)}{r}=0
    Just for your curiosity . There is one step that would require a justification: if \lim_{r\to{0},\ r\in\mathbb{Q}}\frac{f(x+r)-f(x)}{r} exists, why would \lim_{r\to{0}}\frac{f(x+r)-f(x)}{r} exist as well? This is quickly proved from the definition of the limit (cf. right below), but I think it should be underlined since this is where the continuity is needed, and thus this is the core of the proof.

    Suppose that a continuous function \psi on \mathbb{R}\setminus\{0\} is such that \lim_{r\to 0,\ r\in\mathbb{Q}} \psi(x)=\ell. Let \varepsilon>0. There is \delta>0 such that if |r|<\delta and r\in\mathbb{Q}\setminus\{0\} then |\psi(r)-\ell|\leq \varepsilon. Now, if |x|< \delta and x\neq 0, x is a limit of non-zero rational numbers r_n with |r_n|<\delta, so that |\psi(x)-\ell|=\lim_n |\psi(r_n)-\ell|\leq \varepsilon (since this inequality holds for every n). This proves that \lim_{x\to 0}\psi(x)=\ell.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Laurent View Post
    Just for your curiosity . There is one step that would require a justification: if \lim_{r\to{0},\ r\in\mathbb{Q}}\frac{f(x+r)-f(x)}{r} exists, why would \lim_{r\to{0}}\frac{f(x+r)-f(x)}{r} exist as well? This is quickly proved from the definition of the limit (cf. right below), but I think it should be underlined since this is where the continuity is needed, and thus this is the core of the proof.

    Suppose that a continuous function \psi on \mathbb{R}\setminus\{0\} is such that \lim_{r\to 0,\ r\in\mathbb{Q}} \psi(x)=\ell. Let \varepsilon>0. There is \delta>0 such that if |r|<\delta and r\in\mathbb{Q}\setminus\{0\} then |\psi(r)-\ell|\leq \varepsilon. Now, if |x|< \delta and x\neq 0, x is a limit of non-zero rational numbers r_n with |r_n|<\delta, so that |\psi(x)-\ell|=\lim_n |\psi(r_n)-\ell|\leq \varepsilon (since this inequality holds for every n). This proves that \lim_{x\to 0}\psi(x)=\ell.
    Thank you Laurent. Two of my books had proven this result and I did not know whether or not to include it. I should have been more prudent...thank you
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    Quote Originally Posted by Laurent View Post
    This lemma somehow tells that a continuous function is locally monotonous. This fails to be true... Think of x\mapsto x\sin\frac{1}{x} for instance, at 0. It is continuous and it oscillates so the conclusion of the lemma can't be satisfied. The problem in the proof comes from the fact that there is no reason why x_1,x_2 would be in [c-\delta,c+\delta]. ( \delta is defined given x_1,x_2)

    There's much simpler (and correct): notice that for any rational q, since f is q-periodic, f(q)=f(0). In other words, f is constant on \mathbb{Q}. Let's say f(0)=c. Now, let x be any real number, rational or not. It is well known that there exists a sequence (q_n)_n of rational numbers that converges toward x. Since f is continuous, we conclude f(x)=\lim_n f(q_n)=c (the sequence (f(q_n))_n is constant, equal to c). qed.
    you say f(0)=c

    why (q_n)_n converges to x
    x is a variable it is not a contstant to which the sequence could converge to
    ??
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Of course I could be wrong so wait for Laurent or another senior member's confirmation, but what I think Laurent's solution points to is that if f:X\longmapsto\mathbb{R} with X\subset\mathbb{R} then \lim_{x\to p}f(x)=q\Longleftrightarrow \lim_{n\to\infty} f(p_n)=q for all p_n\subset X and p_n\to p\qquad p_n\ne p. Now using this result every number rational or irational is the limit of a sequence of rationals.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by transgalactic View Post
    you say f(0)=c

    why (q_n)_n converges to x
    x is a variable it is not a contstant to which the sequence could converge to
    ??
    I'm not sure I get what you mean.
    In the proof, I wrote "Let x be (...)" so that, from this point on, x is fixed, you can think of it as a constant, and I am allowed to introduce a sequence converging to x. More specifically, I use the fact that any real number (hence for instance, x) is the limit of a sequence of rational numbers. Finally, since I chose x to be any number, what I prove for x holds for any number.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. evenly continues question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 12th 2011, 03:39 PM
  2. Continues functions, Topology
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 9th 2011, 04:12 AM
  3. Replies: 1
    Last Post: February 22nd 2011, 07:47 AM
  4. Continues functions with 2 variables
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 3rd 2010, 01:18 PM
  5. continues prove question..
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 7th 2009, 12:23 PM

Search Tags


/mathhelpforum @mathhelpforum