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Does there exist a non-negative continuous function $f:[0,1]\to \Bbb R$ such that

$\int _0^1 f^n \text{dx}\to 2$ as $n\to \infty $?

By Cauchy-Schwarz Inequality $\int _0^1 f^n dx\le (\int _0^1 f(x)dx)^n\implies \int _0^1 f(x)dx\to 2^{\frac{1}{n}}$ .

I am not getting anything from here.

Please give some hints.

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    It can certainly not happen if $f(x) \leqslant 1$ for all $x$. So there would have to be a point with $f(x_0) > 1$, but then … – 2017-02-27
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    Your statement "By Cauchy-Schwarz Inequality $\int _0^1 f^n dx\le (\int _0^1 f(x)dx)^n$" is false. Let $f(x) =x$ then the LHS is $\dfrac{1}{n+1}$ but the RHS is $\dfrac{1}{2^n}$, a contradiction. – 2017-10-24

1 Answers 1

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The answer is no. Let us prove it by contradiction. $f(x)$ needs to be strictly larger than $1$ on at least one point $x_+\in(0,1)$ - otherwise $\int _0^1 (f(x))^n \text{dx}\leq 1$ for all $n\geq 0$. Given that $f$ is continuous, there exist $\delta,\epsilon>0$ such that $f(x)\geq (1+\epsilon)$ for all $x:|x-x_+|\leq \delta$ (and such that $x_+\in(\delta,1-\delta)$). Then, since $f(x)\geq 0$, we have that $\int _0^1 (f(x))^n \text{dx}\geq 2\delta (1+\epsilon)^n$, which diverges to $\infty$ for $n\to \infty$.

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    Excellent ;thank you very much – 2017-02-27