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Let $f:\mathbb R \to \mathbb R$ be continuous. Then which of the following statements implies that $f(0)=0$?

(A)$\lim_{n \to \infty}\int_{0}^{1}f(x)^n dx=0.$

(B)$\lim_{n \to \infty}\int_{0}^{1}f(x/n) dx=0.$

(C)$\lim_{n \to \infty}\int_{0}^{1}f(nx) dx=0.$

(D) None of the above.

(A)Suppose $f(x)=1-x, \int_{0}^{1}(1-x)^n dx=-\frac{1}{n+1},$ which converges to zero. I tried to find the counter example for the (B) and (C). I couldn't find. please help me.

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2 Answers 2

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(A) Consider the constant function $f(x) = 1/2$.

(B) $\int_0^1 f(x/n)\, dx = n \int_0^{1/n} f(y)\, dy \to f(0)$ (by the mean value theorem), hence condition (B) implies $f(0) = 0$.

(C) Consider $f(x) = \cos(2\pi x)$.

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    $\begingroup$ You are right. I thought about non-constant functions :'(. Thank you very much. $\endgroup$
    – user464147
    Dec 10, 2017 at 16:12
  • $\begingroup$ Added details for points (B) and (C). $\endgroup$
    – Rigel
    Dec 14, 2017 at 15:56
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    $\begingroup$ @Rigel For all $n\in N$, by the MVT for integrals, there exists a $c\in(0,1/n)$, st $\int_0^{1/n} f(y)dy=f(c)(1/n-0)$. So $\lim_{n \rightarrow \infty}n\int_0^{1/n} f(y)dy=\lim_{n \rightarrow \infty} f(c)=f(\lim_{n \rightarrow \infty}c)=f(0)$, since $f$ is continuous and $c\to0$ as $n\to \infty$. Is it right? $\endgroup$
    – ZSMJ
    Nov 6, 2018 at 14:42
  • $\begingroup$ For every $n$ there exists $c_n \in (0, 1/n)$ such that... $\endgroup$
    – Rigel
    Nov 7, 2018 at 7:10
  • $\begingroup$ Mr. Rigel can you tell me how to tackle this kind of problems? I mean where do you come up with the non-examples, is it through experience or something else? $\endgroup$
    – Itachi
    Dec 5, 2020 at 13:59
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B ) $0=\lim_{n\to \infty} \int_0^1f(\frac{x}{n})\, dx\\=\lim_{n\to \infty} f(\frac{c}{n})(1-0) for \, some \, c\in (0,1)\\= \lim_{n\to \infty}f(\frac{c}{n})\\= f(0) \, since \, f \, is \, continuous . $

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