If $\displaystyle f:[-1,1] \to [-1,1]$ be a continuous function and if f satisfies $\displaystyle f(2x^{2}-1)=2xf(x)$ . The prove that $\displaystyle f \equiv 0$ on $\displaystyle [-1,1]$

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- Jul 14th 2010, 01:32 AM #1

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- Jul 14th 2010, 04:33 AM #2
Maybe you can show that the set : $\displaystyle U= \left\{x:f(x)=0\right\}$ has an accumulation point. Then $\displaystyle f\equiv 0$

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

- Jul 14th 2010, 05:18 AM #3
My attempt...

I defined the following function:

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

Now:

$\displaystyle f(1)=-2f(-1)$

$\displaystyle f(1)=2f(1)$

Therefore:

$\displaystyle f(1)=-f(-1)$

Now:

There is $\displaystyle c \in [-1,1]$ so that: $\displaystyle f(2c^{2}-1)=2cf(c)

$

Hence:

$\displaystyle g(-1)g(1)<0$

or:

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

But:

$\displaystyle f:[-1,1] \to [-1,1]

$

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

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])

- Jul 14th 2010, 06:17 AM #4
Surely this is just the zero function, by definition!

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

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

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

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

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

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

Thus, it is now sufficient to prove that the maximum value of $\displaystyle g(x)=0$...which I can't seem to do...(proving the maximum value of $\displaystyle f(x)=0$ would also work as this would imply that the maximum value of $\displaystyle g(x)=0$.)

- Jul 14th 2010, 09:56 AM #5

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- Jul 14th 2010, 10:05 AM #6

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- Jul 14th 2010, 10:48 AM #7

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$\displaystyle \dfrac{d}{dx}\left(f(2x^2-1)\right)=2f(x)+2x\dfrac{df}{dx}$

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

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

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

$\displaystyle \ln f=\ln x+C$

$\displaystyle f=Cx$

$\displaystyle 0=C$ because $\displaystyle f(1)=0$

I don't know why I concluded f=0 from my earlier post. I blame hunger. Of course this is assuming f is differentiable

- Jul 14th 2010, 11:01 AM #8

- Jul 14th 2010, 11:33 AM #9

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We have $\displaystyle f(2x^2-1)=2xf(x)$

$\displaystyle u=2x^2-1$

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

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

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

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

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

And if x=0 we put u=0 which yields $\displaystyle x=\sqrt{\dfrac{1}{2}}$ 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 $\displaystyle x=-\sqrt{\dfrac{u+1}{2}}$ then $\displaystyle x=\sqrt{i\dfrac{u+1}{2}}$ which does not change the result.

- Jul 14th 2010, 11:35 AM #10

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- Jul 14th 2010, 11:43 AM #11

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- Jul 14th 2010, 12:00 PM #12

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- Jul 14th 2010, 12:17 PM #13

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Fine until here..

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

$\displaystyle f(x)=2f(1)=0$

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

So your conclusion is false.

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

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

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 $\displaystyle x=-\sqrt{1+u}$ then $\displaystyle x=\sqrt{i(1+u)}$ which does not change the result.

- Jul 14th 2010, 01:08 PM #14

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How about this then

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

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

then

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

Therefore the derivative exists

Edit

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

- Jul 14th 2010, 01:30 PM #15

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The first equality is not correct:

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

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

then

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

Therefore the derivative exists

Edit

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