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Thread: Prove that function is identically zero?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove that function is identically zero?

    Let $\displaystyle f(x)$ be a continuous function in $\displaystyle [-1, 1]$ and satisfies $\displaystyle 2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$. Prove that $\displaystyle f(x)$ is identically zero $\displaystyle \forall\ x\in [-1, 1]$.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Let $\displaystyle f(x)$ be a continuous function in $\displaystyle [-1, 1]$ and satisfies $\displaystyle 2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$. Prove that $\displaystyle f(x)$ is identically zero $\displaystyle \forall\ x\in [-1, 1]$.
    $\displaystyle 2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$

    2 cancels out both sides, why it is there?

    I think it may be $\displaystyle f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]$.
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  3. #3
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    Well, I'm having a go at $\displaystyle f(2x^2-1)=2xf(x)$, which seems more sensible. To start with, putting $\displaystyle -x$ for $\displaystyle x$ shows that $\displaystyle f(-x)=-f(x)$, so we can concentrate on the interval $\displaystyle [0,1]$.

    So put $\displaystyle x=\cos\theta$ for $\displaystyle \theta\in[0,\pi/2]$ to derive $\displaystyle f(\cos 2\theta)=2\cos\theta f(\cos\theta)$ (1).

    Putting $\displaystyle x=\sin\theta$ gives $\displaystyle f(-\cos 2\theta)=2\sin\theta f(\sin\theta)$, or $\displaystyle f(\cos 2\theta)=-2\sin\theta f(\sin\theta)$ (2).

    It is clear from (2) that $\displaystyle f(1)=0$.

    Putting $\displaystyle \theta=\pi/4$ in (1) and (2) gives $\displaystyle f(0)=\surd 2f(1/\surd2)=-\surd2f(1/\surd2)$, so now $\displaystyle f(\sin 0)=f(\sin \frac\pi4)=f(\sin\frac\pi2)=0$.

    Put $\displaystyle \theta=\pi/8$ in (1) and (2) now shows that $\displaystyle f(\sin\frac\pi8)=f(\sin\frac{3\pi}8)=0$.

    Now direct substitutions, the identity $\displaystyle \sin(\pi/2-\theta)=\cos\theta$ and previous results show that $\displaystyle f(\sin\frac\pi{16})=f(\sin\frac{3\pi}{16})=f(\sin\ frac{5\pi}{16})=f(\sin\frac{7\pi}{16})=0$.

    So far we know that

    $\displaystyle f(\sin 0)=f(\sin\frac\pi{16})=f(\sin\frac{2\pi}{16})=f(\s in\frac{3\pi}{16})=f(\sin\frac{4\pi}{16})=f(\sin\f rac{5\pi}{16})=f(\sin\frac{6\pi}{16})$ $\displaystyle =f(\sin\frac{7\pi}{16})=f(\sin\frac{8\pi}{16})=0$.

    You can see how to proceed. We can prove by induction that for all positive integers $\displaystyle n$, $\displaystyle f(\sin \theta)=0$ whenever $\displaystyle \theta\in S_n=\{\frac{k\pi}{2^{n+1}}:0\leq k\leq 2^n\}$.

    The function $\displaystyle g(\theta)=f(\sin\theta)$ is continuous on $\displaystyle [0,\pi/2]$ and vanishes on $\displaystyle S=\bigcup_{n=1}^\infty S_n$. Moreover $\displaystyle S$ is dense in $\displaystyle [0,\pi/2]$ (anyone else want to prove this?)

    It follows that $\displaystyle g(\theta)=0$ on $\displaystyle [0,\pi/2]$ and therefore $\displaystyle f(x)=0$ on $\displaystyle [0,1]$ and therefore on $\displaystyle [-1,1]$.
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