Show that there exists $x_0$ such that $f(x_0+2 \pi)=f(x_0)$ 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?
 A: 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 $$f(x_{0}+2\pi)=f(x_{0}).$$
A: 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 $g(b) < 0$.
A: 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!
