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$f:[0,1]\rightarrow \mathbb{R}$ is a continuous function which satisfies $$ f(x) + f(1-x) + f\left(\sqrt{x^2+(1-x)}\right) = 0 \text{ and } f\left(\tfrac12\right)=0. $$

Can someone give explicit examples of $f$, apart from the trivial solution, $f(x)=0$? And are there infinitely many solutions for $f$?

I can derive some properties like $f(\frac{\sqrt3}2)=0$ or $f(0)=-2f(1)$, but I can't generate particular examples. Continuity seems important here, but I can't see how to use it, for if we take $f(x)=0$, for $x\in (0,1)$, and $f(1)=1$, $f(0)=-2$ also works.

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  • $\begingroup$ Thanks @Goos. It must be late. $\endgroup$ – alex.jordan May 14 '14 at 7:23
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    $\begingroup$ I sense some trigonometry here. $\endgroup$ – evil999man May 14 '14 at 7:33
  • $\begingroup$ @Awesome Hmm. How? $\endgroup$ – Sawarnik May 14 '14 at 7:47
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    $\begingroup$ I just get a feeling, you know? I see $1/2,\sqrt 3/2$ and the I see domain of the function... $\endgroup$ – evil999man May 14 '14 at 7:48
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    $\begingroup$ Where does this problem come from? $\endgroup$ – user37238 May 14 '14 at 7:49
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First, notice that $\sqrt{x^2+1-x}\geq\sqrt{3}/2$ for all $x\in[0,1]$. Let $g:[1/2,\sqrt{3}/2]\to\mathbb R$ be an arbitrary continuous function such that $g(1/2)=g(\sqrt{3}/2)=0$. Then, define $f:[0,1]\to\mathbb R$ as follows: \begin{align*} f(x)=\begin{cases} 0&\text{if $x\in[0,1-\sqrt{3}/2]$,}\\ -g(1-x)&\text{if $x\in(1-\sqrt{3}/2,1/2]$,}\\ g(x)&\text{if $x\in(1/2,\sqrt{3}/{2}]$,}\\ 0&\text{if $x\in(\sqrt{3}/2,1]$.} \end{cases} \end{align*}

For example:

Example:

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    $\begingroup$ Fantastic answer! Are there any other solutions? $\endgroup$ – 6005 May 14 '14 at 7:58
  • $\begingroup$ Nice :) This also proves there are infinite solutions :) $\endgroup$ – Sawarnik May 14 '14 at 8:01
  • $\begingroup$ @Goos I don't know if there exist other solutions, but as Sawarnik notes, it does provide infinitely many solutions. ;-) $\endgroup$ – triple_sec May 14 '14 at 8:10
  • $\begingroup$ @Sawarnik There are infinitely many infinitely differentiable solutions. $\endgroup$ – 6005 May 14 '14 at 8:17
  • $\begingroup$ @Goos Indeed, you can take $g$ to be an arbitrary multiple of the translated and contracted version of the bump function. $\endgroup$ – triple_sec May 14 '14 at 8:24
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Here is the set of all possible $f$. The algebra gets a little gross, so bear with me.

First, let $f_1: \left[1 - \frac{\sqrt{3}}{2}, 1 \right] \to \mathbb{R}$ be an arbitrary continuous function such that $f_1(x) = 0$ on the closed interval $\left[ \frac{\sqrt{3}}{2}, \frac{7}{4} - \frac{\sqrt{3}}{2} \right]$, and $f_1(\tfrac12) = 0$, and finally, $f_1(\tfrac{\sqrt{3}}{2}) + f_1(\frac{7}{4} - \frac{\sqrt{3}}{2}) = 0$.

Second, define $f_2 : \left[1 - \frac{\sqrt{3}}{2}, 1 \right] \to \mathbb{R}$ by $$ f_2(x) = \begin{cases} f_1(x) & \text{if } 1 - \frac{\sqrt{3}}{2} \le x \le \frac{\sqrt{3}}{2} \\ - f_1\left( \frac{1 - \sqrt{4x^2 - 3}}{2} \right) - f_1\left( \frac{1 + \sqrt{4x^2 - 3}}{2} \right) & \text{if } \frac{\sqrt{3}}{2} \le x \le \frac{7}{4} - \frac{\sqrt{3}}{2} \\ f_1(x) & \text{if } \frac{7}{4} - \frac{\sqrt{3}}{2} \le x \le 1 \\ \end{cases} $$ Verify that $f_2$ is continuous, and satisfies $f_2(x) + f_2(1-x) + f_2(\sqrt{x^2 - x + 1}) = 0$ on its domain. This is a bit gross, so I'm skipping the details for now.

Finally, define $$ f_3(x) = \begin{cases} -f_2(1 - x) - f_2(\sqrt{x^2 - x + 1}) &\text{if } 0 \le x \le 1 - \frac{\sqrt{3}}{2} \\ f_2(x) &\text{if } 1 - \frac{\sqrt{3}}{2} \le x \le 1 \\ \end{cases} $$ Verify that $f_3$ is continuous, and satisfies $f_2(x) + f_2(1-x) + f_2(\sqrt{x^2 - x + 1}) = 0$ on its domain. This is not too hard given that we already showed it for $f_2$.

Now just set $f = f_3$.


The end result is a function $f$ such that $f_1 = f$ on $\left[1 - \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2} \right]$ and $\left[ \frac{7}{4} - \frac{\sqrt{3}}{2}, 1 \right]$, and $f$ satisfies the desired property. So we get to choose $f$ however we want on $\left[1 - \frac{\sqrt{3}}{2} , \frac{\sqrt{3}}{2} \right]$ and $\left[ \frac{7}{4} - \frac{\sqrt{3}}{2} , 1 \right]$, and then $f$ is decided for us everywhere else.

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