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I have a question at hand (which may be easy to some) ,but unfortunately I don't know how to even begin with. Could someone help me?

If $f$ and $g$ are continuous, $2\pi$ periodic functions then prove that

$$\lim_{n\to \infty} {1\over 2\pi} \int_{-\pi}^\pi f(t)g(nt)dt=\left({1\over2\pi}\int_{-\pi}^\pi f(t)dt\right)\left({1\over 2\pi}\int_{-\pi}^\pi g(t)dt \right)$$

Is this even remotely connected to Cauchy-Schwarz?

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  • $\begingroup$ No, nothing much to do with Cauchy-Schwarz. If you know a little about Fourier series, one approach would be to do it first assuming that $f$ and $g$ are trigonometric polynomials... $\endgroup$ – David C. Ullrich Apr 17 '16 at 15:40
  • $\begingroup$ @DavidC.Ullrich : Would you care to explain in detail? $\endgroup$ – AbracaDabra Apr 19 '16 at 21:58
  • $\begingroup$ Which part are you stuck on, doing it for trigonometric polynomials or deducing the general case? $\endgroup$ – David C. Ullrich Apr 19 '16 at 23:04
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Interesting identity.
To grasp it, consider that the integrals on the right are just the average values of the two functions, i.e: $$ {1 \over {2\pi }}\int_{ - \pi }^\pi {f(t)\,dt} = {\rm avg}\left( {f(t)} \right) = \overline f $$ so that we can write: $ f(t) = \overline f \; + \tilde f(t)\quad \to \quad \int_{ - \pi }^\pi {\tilde f(t)\,dt} = 0$

and : $\quad \quad \quad \quad \quad \quad g(t) = \overline g + \tilde g(t) = \overline g + \tilde g(nt)\quad \left| {\;1 \le {\rm integer }\ n} \right.$
Therefore: $$ \eqalign{ & {1 \over {2\pi }}\int_{ - \pi }^\pi {f(t)\;g(n\,t)\,dt} = {1 \over {2\pi }}\int_{ - \pi }^\pi {\left( {\overline f \; + \tilde f(t)} \right)\;\left( {\overline g \; + \tilde g(n\,t)} \right)\,dt} = \cr & = {1 \over {2\pi }}\int_{ - \pi }^\pi {\overline f \;\overline g \,dt} + {1 \over {2\pi }}\int_{ - \pi }^\pi {\overline f \;\tilde g(n\,t)\,dt} + {1 \over {2\pi }}\int_{ - \pi }^\pi {\tilde f(t)\;\overline g \,dt} + {1 \over {2\pi }}\int_{ - \pi }^\pi {\tilde f(t)\;\tilde g(n\,t)\,dt} = \cr & = \overline f \;\overline g \;\,{1 \over {2\pi }}\int_{ - \pi }^\pi {dt} + \overline f {1 \over {2\pi }}\int_{ - \pi }^\pi {\;\tilde g(n\,t)\,dt} + \;\overline g {1 \over {2\pi }}\int_{ - \pi }^\pi {\tilde f(t)\,dt} + {1 \over {2\pi }}\int_{ - \pi }^\pi {\tilde f(t)\;\tilde g(n\,t)\,dt} = \cr & = \overline f \;\overline g \;\,{{2\pi } \over {2\pi }} + \overline f \;0 + \;\overline g \;0 + {1 \over {2\pi }}\int_{ - \pi }^\pi {\tilde f(t)\;\tilde g(n\,t)\,dt} = \cr & = \overline f \;\overline g \;\, + {1 \over {2\pi }}\int_{ - \pi }^\pi {\tilde f(t)\;\tilde g(n\,t)\,dt} \cr} $$

and the last integral is the average value of the "high-frequency" carrier $ {\tilde g(nt)}$ , amplitude modulated by ${\tilde f(t)}$.

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  • $\begingroup$ Would you please explain what is meant by ${\overline f }(t)$ ? I understand it not the same as just $\overline f$ $\endgroup$ – AbracaDabra Apr 18 '16 at 5:41
  • $\begingroup$ @AbracaDabra I indicated by ${\overline f }$ the average value (mean value) of $f(t)$, which is a constant, and of which you may think also to write as $\overline f (t) = \overline f = {\rm const}{\rm .}$. Otherwise, pls. better detail your perplexity. $\endgroup$ – G Cab Apr 18 '16 at 10:16
  • $\begingroup$ :How come $f(t)=2{\overline f}(t)$ ? I more thing, are you sure there aren't any typos in the answer? I mean could it be better formatted? $\endgroup$ – AbracaDabra Apr 18 '16 at 13:59
  • $\begingroup$ @AbracaDabra : I did not write that, I just split $f(t)$ into the constant average component $\overline f $, and the remaining which is the "variable" part, with null mean, and that I indicated with $\tilde f(t)$ (it's a tilde over $f$ not a bar). I do not see any typo error, but of course my formulation could be made better. I' ll try and edit it more explicitly. $\endgroup$ – G Cab Apr 19 '16 at 13:09
  • $\begingroup$ @AbracaDabra. I made some passages explicit, so I hope it is clear now. Unfortunately did not succeed manage TeX to show more clearly the distinction between the upper bar and tilde signs, so please pay attention to them. $\endgroup$ – G Cab Apr 19 '16 at 14:03

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