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Math Help - f being 0 on [-1,1]

  1. #1
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    f being 0 on [-1,1]

    If f:[-1,1] \to [-1,1] be a continuous function and if f satisfies f(2x^{2}-1)=2xf(x) . The prove that f \equiv 0 on [-1,1]
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Maybe you can show that the set : U= \left\{x:f(x)=0\right\} has an accumulation point. Then f\equiv 0

    We can easily show that \pm 1,\pm \frac{1}{2} are in U. Maybe we can construct a sequence x_n that must be in U.

    (EDIT: this is not true... I misremembered a theorem from Complex analysis,sorry)

    We must show U is dense.

    Didn't solve it yet myself. Also curious for a solution.
    Last edited by Dinkydoe; July 14th 2010 at 05:45 AM.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    My attempt...

    I defined the following function:

    g(x)=f(2x^{2}-1)-2xf(x)

    Now:
    f(1)=-2f(-1)
    f(1)=2f(1)

    Therefore:
    f(1)=-f(-1)

    Now:
    There is c \in [-1,1] so that: f(2c^{2}-1)=2cf(c)<br />

    Hence:

    g(-1)g(1)<0

    or:

    f(-1)f(1)>0, therefor f have no zeros...

    But:
    f:[-1,1] \to [-1,1]<br />

    so f has a point e \in [-1,1] which f(e)=e

    And I stuck here...


    Edit: (maybe therefor f(-1)f(1)=0 ==> f(1)=0=f(-1), and because the function is continuous f(x)=0 for every x in [-1,1])
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    My attempt...

    I defined the following function:

    g(x)=f(2x^{2}-1)-2xf(x)...
    Surely this is just the zero function, by definition!

    Now, note that actually im(f) \subseteq [-\frac{1}{2}, \frac{1}{2}] (why?).

    Next, you should notice that f(x) = \sqrt{2(x+1)}f(\sqrt{\frac{x+1}{2}}) and \sqrt{\frac{x+1}{2}} \in [-\frac{1}{2}, \frac{1}{2}]. So, if you can prove the result for f restricted to this interval then you are done. So, define,

    g:[-\frac{1}{2}, \frac{1}{2}] \rightarrow [-\frac{1}{2}, \frac{1}{2}]

    to be the restriction of f to [-\frac{1}{2}, \frac{1}{2}] ( g(x) = f(x) for x \in [-\frac{1}{2}, \frac{1}{2}]).

    Now, this function is continuous and defined over a closed interval, thus it must reach a minimum value. Let x be such that g(x) \leq g(c) for all c \in [-\frac{1}{2}, \frac{1}{2}].

    This now means that \sqrt{2(x+1)} \in [1, \sqrt{3}], and so g(x) = rg(y) where y=\sqrt{\frac{x+1}{2}} and r \geq 1. As g(y) \geq g(x) and multiplication by r can only make it bigger we have that g(x) = g(y) where y \in [\frac{1}{2}, \sqrt{\frac{3}{4}}] and r=1. As \sqrt{2(x+1)} = r=1 this means that x=-\frac{1}{2}. Note that g(-\frac{1}{2})= f(-\frac{1}{2}) = 0.

    Thus, it is now sufficient to prove that the maximum value of g(x)=0...which I can't seem to do...(proving the maximum value of f(x)=0 would also work as this would imply that the maximum value of g(x)=0.)
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  5. #5
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    Solve 2x^2-1=x you get x=-\dfrac{1}{2},1

    if you put x=1 you get

    f(1)=2f(1)

    It follows that f\equiv 0

    Nice and simple.
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  6. #6
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    Quote Originally Posted by fobos3 View Post
    Solve 2x^2-1=x you get x=-\dfrac{1}{2},1

    if you put x=1 you get

    f(1)=2f(1)

    It follows that f\equiv 0

    Nice and simple.
    I believe it just follows that f(1)=0
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  7. #7
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    Quote Originally Posted by MattMan View Post
    I believe it just follows that f(1)=0
    \dfrac{d}{dx}\left(f(2x^2-1)\right)=2f(x)+2x\dfrac{df}{dx}

    4x\dfrac{df}{dx}=2x\dfrac{df}{dx}+2f

    2x\dfrac{df}{dx}=2f

    \dfrac{1}{f}\dfrac{df}{dx}=\dfrac{1}{x}

    \ln f=\ln x+C

    f=Cx

    0=C because f(1)=0

    I don't know why I concluded f=0 from my earlier post. I blame hunger. Of course this is assuming f is differentiable
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    f is NOT differentiable!
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  9. #9
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    Quote Originally Posted by Also sprach Zarathustra View Post
    f is NOT differentiable!
    We have f(2x^2-1)=2xf(x)

    u=2x^2-1

    x=\sqrt{\dfrac{u+1}{2}}

    f(u)=2\sqrt{\dfrac{u+1}{2}}f(\sqrt{\dfrac{u+1}{2}}  )

    I define the operator Df(u)=2\sqrt{\dfrac{u+1}{2}}f(\sqrt{\dfrac{u+1}{2}  })=f(u)

    D^\infty f(u)=2\sqrt[\infty]{\dfrac{u+1}{2}}f(\sqrt[\infty]{\dfrac{u+1}{2}})

    f(u)=2f(1)=0=2xf(x) for x\ne 0

    And if x=0 we put u=0 which yields x=\sqrt{\dfrac{1}{2}} and f(0)=0

    "I turn away with fright and horror from this lamentable evil of functions which do not have derivatives" Hermite

    P.S.

    If x=-\sqrt{\dfrac{u+1}{2}} then x=\sqrt{i\dfrac{u+1}{2}} which does not change the result.
    Last edited by fobos3; July 14th 2010 at 01:01 PM.
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  10. #10
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    I don' think that if u define Df(x) like that then that D^{\infty}f(x)=... will hold.
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  11. #11
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    Why not. By the way it should have been u instead of x I'll fix that in a mo.
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  12. #12
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    Quote Originally Posted by fobos3 View Post
    Quote Originally Posted by Also sprach Zarathustra View Post
    f is NOT differentiable!
    We have f(2x^2-1)=2xf(x)

    u=2x^2-1

    x=\sqrt{\dfrac{u+1}{2}}

    f(u)=2\sqrt{\dfrac{u+1}{2}}f(\sqrt{\dfrac{u+1}{2}}  )

    I define the operator Df(u)=2\sqrt{\dfrac{u+1}{2}}f(\sqrt{\dfrac{u+1}{2}  })=f(u)

    D^\infty f(u)=2\sqrt[\infty]{\dfrac{u+1}{2}}f(\sqrt[\infty]{\dfrac{u+1}{2}})

    f(u)=2f(1)=0=2xf(x) for x\ne 0

    And if x=0 we put u=0 which yields x=\sqrt{\dfrac{1}{2}} and f(0)=0

    "I turn away with fright and horror from this lamentable evil of functions which do not have derivatives" Hermite

    P.S.

    If x=-\sqrt{\dfrac{u+1}{2}} then x=\sqrt{i\dfrac{u+1}{2}} which does not change the result.
    All done now.
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  13. #13
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    Quote Originally Posted by fobos3 View Post
    We have f(x^2-1)=2xf(x)

    u=x^2-1

    x=\sqrt{1+u}

    f(u)=2\sqrt{1+u}f(\sqrt{1+u})

    I define the operator Df(u)=2\sqrt{1+u}f(\sqrt{1+u})=f(u)
    Fine until here..

    D^\infty f(u)=2\sqrt[\infty]{1+u}f(\sqrt[\infty]{1+u})

    f(x)=2f(1)=0
    If Df(u) = 2 \sqrt{1+u} f( \sqrt{1+u} ) then D^2 f(u) = D(D(f(u)) = D(2 \sqrt{1+u} \cdot f( \sqrt{1+u} ) ) =

    =4 \sqrt{1 + \sqrt{1 + u}} \sqrt{1+u} f( \sqrt{1 + \sqrt{1+u}} ) f( \sqrt{1+u} )

    So your conclusion is false.

    and for the general case if we let x_1 = a \in [-1,1] be arbitrary such that f(a) \ne 0, and then let x_{n+1} = \sqrt{ \frac{x_n + 1}{2} } we finally get:

    f(a) = \lim_{ n \to \infty} 2^n \cdot (x_2 x_3 x_4 \cdots x_n) \cdot f(x_n)

    Which gives us, well... not much (this is the same result you should've reached with your operator, you just miscalculated.)

    "I turn away with fright and horror from this lamentable evil of functions which do not have derivatives" Hermite

    P.S.

    If x=-\sqrt{1+u} then x=\sqrt{i(1+u)} which does not change the result.
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  14. #14
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    How about this then

    \lim_{h \to 0} f(2x^2-1+h)=\lim_{h \to 0} f(2(x+h)^2-1)=\lim_{h \to 0}(2x+2h)f(x+h)

    \lim_{h \to 0}f(2x^2-1+h)=\lim_{h \to 0}(2x+2h)f(x)

    then

    \lim_{h \to 0}\dfrac{f(2x^2-1+h)-f(2x^2-1)}{h}=\lim_{h \to 0}\dfrac{(2x+2h)f(x)-2xf(x)}{h}=\lim_{h \to 0}\dfrac{2hf(x)}{h}=2f(x)

    Therefore the derivative exists

    Edit

    (2x^2-1) \in [-1,1] which covers the entire interval
    Last edited by fobos3; July 14th 2010 at 02:24 PM.
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  15. #15
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    Quote Originally Posted by fobos3 View Post
    How about this then

    \lim_{h \to 0} f(2x^2-1+h)=\lim_{h \to 0} f(2(x+h)^2-1)=\lim_{h \to 0}(2x+2h)f(x+h)
    The first equality is not correct:

    2x^2 - 1 + h \neq 2(x+h)^2 - 1 = 2x^2 + 4hx + 2h^2 - 1

    \lim_{h \to 0}f(2x^2-1+h)=\lim_{h \to 0}(2x+2h)f(x)

    then

    \lim_{h \to 0}\dfrac{f(2x^2-1+h)-f(2x^2-1)}{h}=\lim_{h \to 0}\dfrac{(2x+2h)f(x)-2xf(x)}{h}=\lim_{h \to 0}\dfrac{2hf(x)}{h}=2f(x)

    Therefore the derivative exists

    Edit

    (2x^2-1) \in [-1,1] which covers the entire interval
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