Prove every continuous periodic function in R has a fixed point It was recommended that I use induction along with setting g(x) = f(x)-x and applying archimedes to |f(0)|. I know how to use induciton but don't understand how to apply it to this proof.
 A: Let $p$ be the period, then $f(np)-np= f(p)-np$ for every integer $n$, it is clear that for some value of $n$ this is positive and for some value of $n$ this is negative.
We conclude that the continuous function $g(x)=f(x)-x$ has positive and negative values, and hence by the intermediate value theorem there is a value $x_0$ such that $g(x_0)=0$, so $f(x_0)=x_0$
A: if $f$ is bounded, i.e. for some $M \ge 0$
$$
|f(x)| \le M
$$
then for positive $x$ 
$$
f(x)-x \le M-x
$$
which can be made negative by choosing $x \gt M$
similarly for negative $x$
$$
f(x)-x \lt |x| - M
$$
which becomes positive for large enough $|x|$
if $f(x)$ is continuous, then so is $f(x)-x$ so there must be a fixed point by the IVT.
a periodic function continuous on $\mathbb{R}$ is bounded since its range is the continuous image of any compact set of the form $[t,t+p]$ where $p$ is the period and $t$ is arbitrary.
A: More generally,
this holds for every
eventually periodic function,
in which,
for any $x$,
there is a
$p(x) > x$ such that
$f(x) = f(p(x))$.
Instead of
$np$,
we use
$p(x)$ iterated $n$ times
and the same reasoning works.
A: A proof has already been given in the answers, but I think it's nice to see "why" this is true on an intuitive/graphical level.
First notice that if $f(0) = 0$, then $0$ is the fixed point you're looking for, and you're done.  Therefore, we can focus on the cases when either $f(0) > 0$ or $f(0) < 0$.  
The intuition I'll present is the same in either case, so let's pick $f(0) > 0$ for concreteness.
If $f$ has a fixed point, then by definition there exists some real number $x_*$ for which $f(x_*) = x_*$.  Now let $h$ be the function whose graph is the line through the origin with unit slope, namely $h(x) = x$.  Then the fixed point condition can be re-written as $f(x_*) = h(x_*)$.  In other words, $f$ has a fixed point if and only if it has a point of intersection with the line passing through the origin having unit slope.
Draw a graph of that line with unit slope passing through the origin.  Now put your pen down at a point on the positive $y$-axis which is the value of $f$ at $x= 0$.  Since the $f$ is periodic you should be able to draw the graph of $f$ all the way out to infinity to the right without picking up your pen and keep coming back to the value $f(0)$ an infinite number of times.  You'll now see that this requires you to intersect the line you drew at or between some finite number of periods.
In other words, you can find the number of periods it takes for the value of $h$ to be larger than $f(0)$ and then notice that if $x$ is larger than that, you'll never be able to get back to the value $f(0)$.
