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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 $\mathbb{R}$.

Pedro
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3 Answers3

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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 $$ \int_{-\infty}^{\infty} g(x) dx = \int_{-\infty}^{\infty} g^2(x) dx = \frac{A^2}{B}. $$

Will Jagy
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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 Mar 02 '12 at 03:53
  • @PeterT.off, that's all you need. It's just a fancy way of saying that the two integrals exist in the first place. – hmakholm left over Monica Mar 02 '12 at 03:56
  • @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 02 '12 at 03:57
  • @Will Indeed, it isn't. $$\int_{\mathbb{R}} \left| \frac{\sin t}{t} \right|dt$$ is not finite. – Pedro Mar 02 '12 at 03:58
  • @Peter: Yes, indeed; the set of functions to which you can do what Will and I show is the intersection $L^1\cap L^2$. – robjohn Mar 02 '12 at 08:03
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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 $\int_{-\infty}^\infty f(x)\:\mathrm{d}x\not=0$.

robjohn
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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 $\int f=\int f^2$.

hmakholm left over Monica
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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 Mar 02 '12 at 03:45
  • $0<g(x_0)<1$ 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. – hmakholm left over Monica Mar 02 '12 at 03:54
  • 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 Mar 02 '12 at 03:56
  • I like your alternative :-) (+1) – robjohn Mar 02 '12 at 16:17