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Does there exist a continuous function $f:\mathbb{R}\to(0,\infty)$ that satisfies

$$\int_{\mathbb{R}} f^n d L^1 = 1 $$

for every natural $n$? ($L^1$ is Lebesgue measure on $\mathbb{R}$)

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uhmmmm... continues. It's not very continues in 0 and 1, is it? –  user19502 Feb 8 '13 at 21:01
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Do you have any thoughts/attempts to share? –  Jonas Meyer Feb 8 '13 at 21:02
    
I tried with $e^{-x^2}$, and I can construct a series of functions, $f_n$, that satisfy $\int_{\mathbb{R}} f_n^n dx=1$:$$\int_{-\infty }^{\infty } \left(\sqrt[2n]{\frac{n}{\pi }}e^{-x{}^{\wedge}2}\right)^n \, dx,$$, but I cannot think of one specific function –  user19502 Feb 8 '13 at 21:03
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1 Answer

up vote 5 down vote accepted

No. Suppose such $f$ does exist to reach a contradiction. Then $\sup\limits_{x\in \mathbb R}f(x)=\lim\limits_{n\to\infty}\left(\int_{\mathbb R} f^n dL\right)^{1/n}=1$, so $f(x)\leq 1$ for all $x$. Hence $f-f^2$ is a nonnegative function, and by continuity $f-f^2$ is not always zero. This implies (again using continuity) that $\int_{\mathbb R} f-f^2 dL>0$.

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Thanks. For some reason I cannot upvote you. –  user19502 Feb 8 '13 at 21:09
    
@user19502: You're going through initiation where you can't do much on the website until you earn more points. More info is in the FAQ. –  Jonas Meyer Feb 8 '13 at 21:10
    
Yeah, but I thought that thanking someone is not a privilege. –  user19502 Feb 8 '13 at 21:10
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And with a bit more work, you can prove that any measurable function with $\int f^n\ dx = 1$ for all positive integers $n$ is (almost everywhere) the indicator function of a set of measure $1$. –  Robert Israel Feb 8 '13 at 21:15
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@user19502: Yes, I added in parentheses subsequently that the last part also uses continuity. If a continuous function is positive at a point, then it is bounded below by a positive number in a neighborhood of that point. –  Jonas Meyer Feb 8 '13 at 21:17
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