Thread: f being 0 on [-1,1]

1. 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 $[-1,1]$

2. 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.

3. 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)
$

Hence:

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

or:

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

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

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])

4. Originally Posted by Also sprach Zarathustra
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 $g(x)=0$.)

5. 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.

6. Originally Posted by fobos3
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 $f(1)=0$

7. Originally Posted by MattMan
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

8. f is NOT differentiable!

9. Originally Posted by Also sprach Zarathustra
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.

10. I don' think that if u define Df(x) like that then that D^{\infty}f(x)=... will hold.

11. Why not. By the way it should have been u instead of x I'll fix that in a mo.

12. Originally Posted by fobos3
Originally Posted by Also sprach Zarathustra
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.

13. Originally Posted by fobos3
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 $D^2 f(u) = D(D(f(u)) = D(2 \sqrt{1+u} \cdot f( \sqrt{1+u} ) ) =$

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

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.

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

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

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

15. Originally Posted by fobos3

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