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Let F be a continuous periodic function on R^N. Let a,b be vectors in R^N. Also assume a is not parallel to b.

Does the limit of

$\varepsilon \int_0^{1/\varepsilon} F(as+b/\varepsilon) ds$

Exist as epsilon tends to 0? I think it is equal to the limit of

$\varepsilon \int_0^{1/\varepsilon} F(as) ds$

The latter limit does exist.

But I cannot prove it. ANY help would be appreciated.

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What do you mean by a periodic function, or a quasiperiodic function? What is the period? What does "quasi" mean? –  Will Sawin Jun 25 '12 at 17:27
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1 Answer 1

up vote 1 down vote accepted

Take $a=(1,0)$, $b=(0,1)$, $F(x,y)=sin(y)$. Then $\int_0^{1/\epsilon} F(as+b/\epsilon) ds= (1/\epsilon) \sin(1/\epsilon)$, so you are asking for the limit as $\epsilon$ goes to $0$ of $\sin(1/\epsilon)$. This limit does not exist.

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Thank you very much –  dcs24 Jun 25 '12 at 17:34
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