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Suppose, I have a continuous periodic function $f$ such that $f(x)=f(x+1)$ for all real $x$. Then show that there exists a $x_0 \in \mathbb{R}$ such that $f(x_0+2 \pi)=f(x_0)$.

From the first glance, this seems like a problem of IVT. So, I take $g(x)=f(x+ 2 \pi)- f(x)$.

Now, I can't find two $a,b$'s such that $g(a).g(b)<0$. Can anyone help?

Arctic Char
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Maths Freak
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3 Answers3

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Define the function $g : \mathbb R \rightarrow \mathbb R$ by $g(x)=f(x+2\pi)-f(x).$ Since $f$ is continuous on $\mathbb R$, then so is $g$. Moreover by the extreme value theorem, it follows that $f$ attains its minimum and maximum for some $a,b\in\mathbb R.$ Then

$$f(a+2\pi)-f(a)\geq0\ \space\space\space\space\space\space\space\space\space\space\text{and}\space\space\space\space\space\space\space\space f(b+2\pi)-f(b)\leq 0 $$

So $g(a)\geq 0$ and $g(b)\leq 0,$ which means that there exists $x_{0}$ between $a$ and $b$ such that $g(x_{0})=0.$ Thus $$f(x_{0}+2\pi)=f(x_{0}).$$

Alessio K
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Professor Vector has a great answer. Let's see why.

Assume $f : \mathbb{R} \rightarrow \mathbb{R}$. We want to evaluate $$\int_0^1 g(x) dx.$$

Let's use Integral of periodic function over the length of the period is the same everywhere. So

$$ \int_0^1 g(x) dx = \int_0^1 f(x+2\pi) dx - \int_0^1 f(x)dx.$$ Let $z = x + 2\pi$, $dz = dx$. Then we can rewrite this as $$ \int_0^1 g(x) dx = \int_{2\pi}^{1+2\pi} f(z) dz - \int_0^1 f(x)dx.$$ Now use the stackexchange to rewrite this as

$$ \int_0^1 g(x)dx = \int_0^1 f(z)dz - \int_0^1 f(x)dx = 0.$$

The integral being $0$ forces it to either be zero or to have points $a,b \in [0,1]$ with $g(a) > 0$ and $g(b) < 0$.

User203940
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    One could make that a bit more self-contained: with $G(y)=\int^{y+1}_yf(x),dx,$ we have $G'(y)=0,$ so it is a constant, and $\int^1_0g(x),dx=G(2\pi)-G(0).$ –  Sep 18 '20 at 15:36
  • Fair. I just remembered this fact from Fourier analysis. – User203940 Sep 18 '20 at 15:40
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If $g(x) = 0$ for some $x\in\mathbb{R}$ we are done.

So suppose $g(x) > 0$ for all $x$ or $g(x) < 0$ for all $x\in\mathbb{R}$.

Since $-f$ verify the some hypothesis as $f$, we can take $g(x) > 0$ for all $x$.

By continuity and 1-periodicity of $g$, we deduce that there exists $\varepsilon > 0$ such that $g(x) \geq \varepsilon$, e.g. $$f(x+2\pi) \geq f(x) + \varepsilon$$ By induction we deduce :

$$f(2\pi n) \geq n\varepsilon + f(0) \quad\forall n \in \mathbb{N}$$

contradiction because $f$ is bounded!

user388615
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