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If $f \colon \mathbb R\to\mathbb R$ is a continuous function, which of the following statements implies that $f(0)=0$?

(A) $\int_0^1 f(x)^n \,dx\to 0$ as $n\to\infty$
(B) $\int_0^1 f\left(\frac xn\right) \,dx\to 0$ as $n\to\infty$
(C) $\int_0^1 f(nx) \,dx\to 0$ as $n\to\infty$
(D) $\int_0^1 f(x+n) \,dx\to 0$ as $n\to\infty$

I came across the above problem and do not know how to tackle it. Can someone point me in the right direction? Thanks in advance for your time.

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Have you tried calculating the limiting values for each of the integrals? Pick your favourite continuous functions (I suggest $f(x)=1/2$ as a starting point) and look at what the limits are. You should be able to rule out some of the possibilities. – Chris Taylor Jan 18 '13 at 11:26
Some hints: (A) - Take a function such that $f(0)=1$ and $f(x)\in [0,1)$ if $x\in (0,1]$. (B),(C),(D) - Change of Variables – Tomás Jan 18 '13 at 11:31
In fact for D there is no constraint on $f$ at any finite argument. Let $f(x)$ be arbitrary for $x \lt 10^9$ and zero thereafter. It meets D. – Ross Millikan Jan 19 '13 at 4:31
For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Jan 20 '13 at 9:07
up vote 4 down vote accepted

EDIT: upon your request, I'll add a further discussion.

I'll start with (B), which is both true and also harder than the rest of the cases. Note that after the substitution I mentioned we have $$n \int_0^{1/n} f(x)dx.$$ Now, since $f(x)$ is continuous, we can write $|f(x) - f(0)| < \epsilon$ for every $\epsilon > 0$ as soon as $x < \delta$ for some $\delta > 0$. In particular, pick $n$ big enough, so that $1/n < \delta$. Using this we can write $$\left|n \int_0^{1/n} \left(f(x) - f(0) \right) dx \right| \leq n \int_0^{1/n} |f(x) - f(0)| dx \leq n {1 \over n} \epsilon = \epsilon.$$ Therefore the limit of integrals in the OP equals $f(0)$ and in particular vanishes if and only if $f(0) = 0$.

For (C), my substitution leads to $${1 \over n} \int_0^n f(x)dx$$ and one can think of this as an average over the integral of length $n$. Therefore for this to vanish, it's sufficient that $f(x)$ decays sufficiently quickly (for simplicity take $f(x) = e^{-x}$, but anything subconstant works as well) and so it is certainly not necessary that $f(0) = 0$. Case (D) can be treated similarly. Again, it's enough that the function decays sufficiently quickly.

The case (A) is a little bit different but note that it's only important whether $f(x) <1$ or $f(x) > 1$ for $x \in (0,1)$ (why?).

For (A) take e.g. $f(x) = 1 - x$. Using this as inspiration, try to find all functions for which (A) fails.

For the rest of the cases, use substitutions and notice that the integrals can always be expressed in terms of the original function $f$.

(B) substitute $x = ny$;

(C) substitute $y = nx$;

(D) substitute $y = x + n$.

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It is very kind of you for the detailed explanation.I have got it.Thanks a lot sir. – user53386 Jan 19 '13 at 3:40

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