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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?

AbracaDabra
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1 Answers1

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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 ${\tilde f(t)}$.

G Cab
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  • Would you please explain what is meant by ${\overline f }(t)$ ? I understand it not the same as just $\overline f$ – AbracaDabra Apr 18 '16 at 05:41
  • @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. – G Cab Apr 18 '16 at 10:16
  • :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? – AbracaDabra Apr 18 '16 at 13:59
  • @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. – G Cab Apr 19 '16 at 13:09
  • @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. – G Cab Apr 19 '16 at 14:03
  • Liked your answer.Beautiful one. Please consider upvoting if you liked it. – AbracaDabra Apr 19 '16 at 19:41
  • One more doubt . Can you tell me anything related to @DavidC.Ullrich 's comment? – AbracaDabra Apr 19 '16 at 19:44
  • @AbracaDabra Glad that I could help you to understand this interesting limit. I upvoted your post since the beginning. Concerning DavidC.Ulrich hint it's fully valid, but would be longer and less easy to get through. Actually my suggestion start in the same way, but instead of exploding all the components of $f$ and $g$, we just divide the first and the remaining. – G Cab Apr 19 '16 at 20:52
  • Would you care to explain @DavidC.Ullrich 's point (in the long way) in your answer?. Actually, I am beginner in Fourier Series, so it might as well be of great help to me. – AbracaDabra Apr 19 '16 at 21:57
  • @AbracaDabra we cannot extend our discussion further. I can give you just a hint. Take out the first harmonic from $\tilde f(t)$, i.e. put: $ \tilde f(t) = a_{,1} \cos t + b_{,1} \sin t + f^ * (t)$, see what happens repeating the procedure I indicated. – G Cab Apr 19 '16 at 22:25