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I don't know if this is a very simple question with a very difficult answer, but:

Is $y = \dfrac{\sin x}{x}$ the only function such that

$$\int_{-\infty}^{\infty} f(x) dx =\int_{-\infty}^{\infty} f^2(x) dx \text{ ?}$$

This is to say, if we start with the integral equation

$$\int_{-\infty}^{\infty} f(x) dx =\int_{-\infty}^{\infty} f^2(x) dx \text{ ?}$$

would we only find $f(x) = \dfrac{\sin x}{x}$, or we can expect other solutions?

Maybe I should have made this clear, but I'm talking about $f(x) \neq 0$ and $f(x)$ continuous in $\mathbb{R}$.

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Any function that only takes the values of 0 and 1 would satisfy the equation. –  Tpofofn Mar 2 '12 at 3:27
    
Perhaps Peter means to add continuous to his criterion.... –  user02138 Mar 2 '12 at 3:29
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@user2138 Indeed. Why don't you use a nickname? –  Pedro Tamaroff Mar 2 '12 at 3:31
    
Hmm.. when you write $f^2(x)$ do you mean $(f(x))^2$ or $f(f(x))$? –  Henning Makholm Mar 2 '12 at 3:50
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Of course $f^2$ means $f$ squared, but which of $x\mapsto (f(x))^2$ and $x\mapsto f(f(x))$ does "$f$ squared" mean to you? –  Henning Makholm Mar 2 '12 at 3:57
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3 Answers

up vote 10 down vote accepted

Take any $f$ where both integrals exist. Define $$ A = \int_{-\infty}^{\infty} f(x) dx, $$ then $$ B = \int_{-\infty}^{\infty} f^2(x) dx. $$ Now, define $$ g(x) = \frac{A}{B} \; f(x). $$ Then $$ \int_{-\infty}^{\infty} g(x) dx = \int_{-\infty}^{\infty} g^2(x) dx = \frac{A^2}{B}. $$

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So that's what robjohn is saying. Should I know in detail what $L^1$ and $L^2$ are? I've read they are spaces, such that $\int_{\mathbb{R}} |f^2(x)|$ is finite ($L^2$) and $\int_{\mathbb{R}} |f(x)|$ is finite ($L^1$), but that's about it. I know nothing about measure theory. –  Pedro Tamaroff Mar 2 '12 at 3:53
    
@PeterT.off, that's all you need. It's just a fancy way of saying that the two integrals exist in the first place. –  Henning Makholm Mar 2 '12 at 3:56
    
@Peter, that is enough for you to know. In fact, I took them out, I suspect $\mbox{sinc}$ is not in $L^1$ anyway. All that matters is that the two integrals exist. –  Will Jagy Mar 2 '12 at 3:57
    
@Will Indeed, it isn't. $$\int_{\mathbb{R}} \left| \frac{\sin t}{t} \right|dt$$ is not finite. –  Pedro Tamaroff Mar 2 '12 at 3:58
    
@Will Could you help me out here? –  Pedro Tamaroff Mar 2 '12 at 4:06
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Note that $$ \int_{-\infty}^\infty\frac{1}{1+x^2}\mathrm{d}x=\pi $$ and $$ \int_{-\infty}^\infty\frac{1}{(1+x^2)^2}\mathrm{d}x=\frac{\pi}{2} $$ However, if we scale this up, we get $$ \int_{-\infty}^\infty\frac{2}{1+x^2}\mathrm{d}x=2\pi $$ and $$ \int_{-\infty}^\infty\frac{4}{(1+x^2)^2}\mathrm{d}x=2\pi $$ This can be done for any function so that $f$ and $f^2$ are integrable and $\int_{-\infty}^\infty f(x)\:\mathrm{d}x\not=0$.

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I got it. Thanks! –  Pedro Tamaroff Mar 2 '12 at 4:04
    
Could you help me out here? –  Pedro Tamaroff Mar 2 '12 at 4:06
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Not by far.

In fact, take any nonnegative continuous $g$ such that $\int g^2 > \int g$. Then select some $x_0$ such that $0<g(x_0)<1$ and consider the functions $$ f_a(x) = \begin{cases}g(x) & x\in(-\infty,x_0] \\ g(x_0) & x\in(x_0,x_0+a] \\ g(x-a) & x \in (x_0+a,\infty) \end{cases}$$ for all $a>0$. As you increase $a$, both of $\int f_a^2$ and $\int f_a$ will increase in proportion to $a$, but $\int f_a$ will increase faster by a factor of $1/g(x_0)$. Since $f_0=g$, at some $a$ you will find $\int f_a^2=\int f_a$.

Alternatively, take the same $g$ and let $b=\frac{\int g}{\int g^2}$ Then $f(x)=bg(x)$ will also satisfy $\int f=\int f^2$.

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I'm not quite getting it. Why do we need $0<g(x_0)<1$? Are you letting $a$ tend to something, or are you using an "intermediate value"-like theorem? –  Pedro Tamaroff Mar 2 '12 at 3:45
    
$0<g(x_0)<1$ ensures $g(x_0)^2 < g(x_0)$, so when we splice $a$ units of horizontal constancy into the function at $x_0$, we will see $\int f_a$ increase by $a\cdot g(x_0)$ while $\int f_a^2$ increases by only $a\cdot g(x_0)^2$. Setting $a=\frac{\int g^2-\int g}{g(x_0)-g(x_0)^2}\frac{1}{g(x_0)}$ will then make the integrals equal. –  Henning Makholm Mar 2 '12 at 3:54
    
Ok. I'm not very lucid now, but I'll take my time to think about it and I'll get back to you. –  Pedro Tamaroff Mar 2 '12 at 3:56
    
I like your alternative :-) (+1) –  robjohn Mar 2 '12 at 16:17
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