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I come up with the question when studying Fourier Analysis. I know that we have for sure in the Riemann sense,

Let $ f:\mathbb{R}\to\mathbb{R} $ be a real function. Let $ F $ be a primitive of $ f $ that is bounded on all of $ \mathbb{R}. $ Let $ f $ be periodic with period $ L. $ Then, $ F $ is also period with period $ L. $

So my thinking is that: if we have $f, g$ are two L-periodic Riemann integrable. Define their convolution as follows: $$(f*g)(x)=\frac{1}{L}\int_0^{L}f(y)g(x-y) dy,$$ then if their convolution is also periodic?
As mentioned above, somehow i can think that the primitive of a L-periodic function (this primitive is assumed to be bounded first) is also L-periodic function. Is there any connection between these.
Thanks in advance.

Minh
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  • $(f \star g)(t+L) = \int_{-\infty}^{\infty} f(\tau)g(t + L - \tau) d\tau = \int_{-\infty}^{\infty} f(\tau)g(t - \tau) d\tau = (f\star g)(t)$, if that is what you are asking? – BenCWBrown Nov 09 '18 at 02:18
  • The word convolution appears in the title but not in the question. – Kavi Rama Murthy Nov 09 '18 at 06:21
  • @BenCWBrown oh nice, i must be obsessed with the idea of the periodicity of the primitive, then i cant think in this direct way. Many thanks. – Minh Nov 10 '18 at 13:47
  • @KaviRamaMurthy i just edited the question in more detail. Thanks – Minh Nov 10 '18 at 13:47

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