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$f(x)=f(x+1)\ \forall x\Rightarrow \int_0^1f(x+t)dt=\int_0^1f(t)dt$, when $f$ is continuous on $[0,1]$. The proof it is not hard. My question is, this property remains true if $f$ is only integrable?

bob
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  • Yes, it's still true. In fact I don't really see how one can invoke continuity in the proof, other than to conclude that it's integrable. – Greg Martin Jun 03 '14 at 07:55
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    @GregMartin can you post a complete solution? Consider the function $F(x)=\int_x^{x+1}f(t)dt-\int_0^1f(t)dt$, this function it is derivable if $f$ is continuous, so $F'(x)=f(x+1)-f(x)=0$, hence is constant, but $F(0)=0$. – bob Jun 03 '14 at 08:26
  • ah good point. ok – Greg Martin Jun 03 '14 at 17:16

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Given $x\in\Bbb R$, write $k=\lfloor x\rfloor$ and $y=x-k$, so that $0\le y<1$. Then $f(u+k)=f(u)$ for all $u$, so $$ \int_0^1 f(t+x)\,dt = \int_0^1 f(t+y+k)\,dt = \int_0^1 f(t+y)\,dt. $$ Furthermore, \begin{align*} \int_0^1 f(t+y)\,dt &= \int_0^{1-y} f(t+y)\,dt + \int_{1-y}^1 f(t+y)\,dt\\ &= \int_0^{1-y} f(t+y)\,dt + \int_{1-y}^1 f(t+y-1)dt \end{align*} using $f(u)=f(u-1)$. Now make the change of variables $u=t+y$ in the first integral and $u=t+y-1$ in the second integral: $$ \int_0^{1-y} f(t+y)\,dt + \int_{1-y}^1 f(t+y-1)\,dt = \int_y^1 f(u)\,du + \int_0^y f(u)\,du = \int_0^1 f(u)\,du. $$

Fardad Pouran
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Greg Martin
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