When is this true? $$\lim_{r\to 0}\int_{-K}^K f(rx)dx=\int_{-K}^K \lim_{r\to 0} f(rx)dx$$ Is it true without the hypothesis of continuity of $f$?
Thank you.
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2What are your thoughts on this? Do you know about uniform convergence? (I assume you realize that at least, the limit inside the integral on the right must exist for this to be true.) – Harald Hanche-Olsen Aug 18 '13 at 14:48
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OP: What happened to this question? – Did Aug 18 '13 at 14:51
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Yes. You don't even need continuity, only that the limit exists (the value of $f(0)$ is irrelevant.) See my answer. – Harald Hanche-Olsen Aug 18 '13 at 20:54
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If $f$ is bounded, then it is a one line proof by the dominated convergence theorem. – abnry Aug 18 '13 at 21:35
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Hint: Consider the function $$ f(x)=\left\{\begin{array}{} 0&\text{if }x=\frac1n\text{ for }n\in\mathbb{Z}\\ 1&\text{otherwise} \end{array}\right. $$ For the Lebesgue integral, read about Lebesgue's Density Theorem.
robjohn
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I am unsure what your point is. There seems to be an implicit requirement that $f$ has a limit at $0$, or else the right hand side does not make sense. What did I miss? – Harald Hanche-Olsen Aug 18 '13 at 20:39
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I suppose the limit is assumed to exist? $\lim_{r \rightarrow 0} f(r)$ doesn't exist for the $f$ given. – abnry Aug 18 '13 at 21:37
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@HaraldHanche-Olsen: if the right-hand side exists, then the left-hand side exists and is equal. The only way to show that the equation is not true is for the left-hand side to exist and the right-hand side not to exist. That was what I was trying to hint. Stating that the equation holds would imply that if the left-hand side exists, then the right-hand also exists. – robjohn Aug 18 '13 at 21:46
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I see. See also my respons to your comment on my answer. (+1 btw.) – Harald Hanche-Olsen Aug 19 '13 at 12:35
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There must be an assumption that the limit inside the integral on the right exists for at least one $x$. But then a simple substitution shows that the limit $$ \lim_{r\to0}f(r)=:c $$ exists, and then $$ \lim_{r\to0}f(rx)=c\qquad\text{for all $x\ne0$}. $$ So the integral on the right has the value $2Kc$.
Now look at the left hand side. Let $\varepsilon>0$, and pick $\delta>0$ so that $$\lvert f(x)-c\rvert<\varepsilon\qquad\text{for $0<\lvert x\rvert<\delta$.}$$
Notice that then $$\lvert f(rx)-c\rvert<\varepsilon\qquad\text{for $0<\lvert x\rvert<K$, $0<r<\delta/K$.}$$ This shows that $f(rx)\to c$ uniformly on $[-K,K]\setminus\{0\}$ as $r\to0$, which is enough to get the required convergence.
Harald Hanche-Olsen
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The appearance of $\lim\limits_{r\to0}f(rx)$ on the right-hand side does not imply the existence of that limit. If it were stated that that limit exists, then the the equation would definitely be true. However, the left-hand side can exist without the right-hand side existing. – robjohn Aug 18 '13 at 23:04
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@robjohn: Absolutely true. The problem statement is a bit vague on the exact conditions. And to me, the notion that the existence of the limit on the left might imply the existence of the one on the right seemed so ludicrous I did not even entertain the possibility that this might be part of the problem. – Harald Hanche-Olsen Aug 19 '13 at 12:34
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One minor point, did you mean to say that "$f(rx)\to c$ uniformly on $[-K,K]\setminus{0}$"? Usually one talks about a family of functions converging uniformly. – robjohn Aug 19 '13 at 12:49
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