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Suppose that $f:[0,1] \to \mathbb{R}$ is $1$-periodic, meaning that $f(x)=f(x+1) \ \forall x$, and $\int_0^1 f(t) dt < \infty$. Does this mean that $f$ is bounded ?

For general function, i know this is not true, for example $f(x) = x^{-1/2}$ which is not bounded on $(0,1]$ yet its integrable there and $\int_0^1 t^{-1/2} dt = 2$.

But i could not prove my conjecture but i searched for unbounded 1-periodic functions and found that they all are not integrable on $[0,1]$ such as $\tan(2\pi x)$.

Please help.

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    $\begingroup$ Take $\frac{1}{\sqrt{x}}$ and extend it $1$-periodically to $\mathbb{R}$ (and give some random value to $f(0)$). No need to have a simple "analytical formula". Simply saying what I did here uniquely defines a periodic function and it's easy to compute it's value for all $x$ from this definition. E.g. $f(3.14) = f(0.14) = \frac{1}{\sqrt{0.14}}$. $\endgroup$ – Winther May 24 '18 at 19:13
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    $\begingroup$ If you want a simple example, you could note that $\tan(2 \pi x)$ blows up like $1/(1/4-x)$ near $x=1/4$, which is just like $x^{-1}$ near $x=0$, and similarly near $x=3/4$. So the square root of this should work, i.e. $\sqrt{|\tan(2 \pi x)|}$ is integrable, unbounded and periodic. $\endgroup$ – Nate Eldredge May 24 '18 at 19:36
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You can truncate your own example $f(x)=\frac1{\sqrt x}$ to $(0,1]$ and make it periodic using $$g(x)=\cases{f(x-\lfloor x\rfloor)&$x \in\Bbb R\setminus\Bbb Z$\cr 0& otherwise}$$

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Short answer: No! You gave an example for a semi-open interval. This means in order to extend this example to the closed interval you just have to define $g\left(0\right)\,:=\, f\left(1\right)$ and $g\left(x\right)\,:=\, f\left(x\right)$ elsewhere.

In order to achieve boundedness you would need to impose some constraint on the function such as continuity.

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