a) By induction. Let $f$ be a polynomial that can be written as sum of $\deg f = n$ periodic functions $f_{1},f_{2},\ldots,f_{n}$ with periods $t_{1},t_{2},\ldots,t_{n}$. We have: $f(x) = f_{1}(x)+\ldots+f_{n}(x)$ Since $f_{1}(x)-f_{1}(x+t_{1}) = 0$, we get: $f(x)-f(x+t_{1}) = f_{2}(x)-f_{2}(x+t_{1})+\ldots+f_{n}(x)-f_{n}(x+t_{1})$ Note that $f(x)-f(x+t_{1})$ is a polynomial of degree $n-1$, and it's written as sum of $n-1$ periodic functions. So we can "go down" until a polynomial of degree 1 is written as sum of only one periodic function... but a polynomial of degree 1 is not periodic! b) Since $f(x)$ is periodic if and only if $af(x)+b$ is, with a,b nonzero constants, and the same for boundedness, we can suppose the polynomial to be x. Let $f_{1}(x)+f_{2}(x) = x$ for all x, and the period of $f_{1},f_{2}$ be $t_{1},t_{2}$. Without loss of generality we suppose $f_{1}(0) = 0$. Then $f_{1}(nt_{1}) = 0$ for all integers n, and $f_{2}(nt_{1}) = kt_{1}$ for all integers n, and $f_{2}(nt_{1}+mt_{2}) = nt_{1}$ for all integers n,m. If the ratio $\frac{t_{1}}{t_{2}}$ is rational, than $f_{1}(x)+f_{2}(x)$ is a periodic function, but this is impossible. Therefore $\frac{t_{1}}{t_{2}}$ is irrational, and by a well-known theorem $nt_{1}+mt_{2}$ can go as near as we want to every real. For each $\epsilon > 0$, there are integers n,m such that $0<nt_{1}+mt_{2}<\epsilon$, and we can also make $nt_{1}$ very large by going nearer to 0, and then multiplying n,m by a large factor (if $\epsilon < |t_{2}|$ we must have $n \ge 1$). Then we can also go near to each real a with a very large $nt_{1}$, to which corresponds a very large value of $f_{2}$ which can't be bounded in an interval. The same for $f_{1}$.

c) I think that here we need the Axiom of Choice. Take two nonzero reals $t_{1},t_{2}$ such that $\frac{t_{1}}{t_{2}}$ is irrational. Let $A = \{nt_{1}+mt_{2}| n,m \in \mathbb{Z}\}$. $a-b \in A$ is an equivalence relation among the reals. Let $[a]$ be the equivalence class of a, and let $c: \{[a]: a \in \mathbb{R}\}\rightarrow \mathbb{R}$ a choice function of the classes. Using the definition of our equivalence relation and the fact that $\frac{t_{1}}{t_{2}}$ is irrational, we show that for each $x \in \mathbb{R}$ there exist unique integers n,m such that $x = c([x])+nt_{1}+mt_{2}$. Then we define: $f_{1}(x) = f_{1}(c([x])+nt_{1}+mt_{2}) = c([x])+mt_{2}$ $f_{2}(x) = f_{2}(c([x])+nt_{1}+mt_{2}) = nt_{1}$ $f_{1},f_{2}$ are periodic, since obviously $f_{1}(x+t_{1})= f_{1}(c([x])+(n+1)t_{1}+mt_{2})= c([x])+mt_{2}= f_{1}(x)$ and the same for $f_{2}$. And also $f_{1}(x)+f_{2}(x) = c([x])+mt_{2}+nt_{1}= x$. d) This can be done by induction, using c). Suppose the statament is true for all polynomial with deg < n, and that you can choose freely the periods of the functions of which the polynomial is sum. (not completely freely, their ratio is irrational). Take the polynomial f with $\deg f = n$. Choose a number $t_{n+1}$ as you want. Consider $f(x)-f(x-t_{n+1})$. It is a polynomial of degree n-1. Then there exists periodic functions such that $f(x)-f(x-t_{n+1})= g_{1}(x)+\ldots+g_{n}(x)$. We first prove the lemma: if $g$ is a periodic function of period t, and $\frac{t}{t_{1}}$ is irrational, then there exists a function $g'$ such that $g(x) = g'(x)-g'(x-t_{1})$ and $g'$ is still periodic of period t. The proof is somehow the same of the point c)... with the same terminology, define: $g'(x) = g'(c([x])+nt+m t_{1}) = \sum_{i=0}^{m}g(c([x])+nt+it_{1})$ if $m \ge 0$, 0 is m = -1 and $\sum_{i=1}^{-m-1}g(c([x])+nt+-it)$ if $m<-1$. Then it's easily seen that g' satisfies $g'(x)-g'(x-t_{1}) = g(x)$ and $g'(x) = g'(x+t)$. Using the lemma, we get that there are periodic functions such that: $f(x)-f(x-t_{n+1}) = f_{1}(x)-f_{1}(x-t_{n+1})+\ldots+f_{n}(x)-f_{n}(x-t_{n+1})$ Now we have just to add the suitable function $f_{n+1}$. For each $0\le x < t_{n+1}$ define: $f_{n+1}(x) = f(x)-f(x-t_{n+1})-f_{1}(x)+f_{1}(x-t_{n+1})+\ldots-f_{n}(x)+f_{n}(x-t_{n+1})$ And then extend $f_{n+1}$ to the reals using periodicity, and check that $f(x) = f_{1}(x)+f_{2}(x)+\ldots+f_{n}(x)$ for all real x.

my solution to c)

edit : oops, i just said a biiiiiiiiiiiiiig stupid thing.

maky wrote: my solution to c) (without AC) let $B$ be a Hamel-basis of $\mathbb{R}$ over $\mathbb{Q}$ (as a vectorial space), ... But this needs AC !

maky wrote: e) give an example of a function that can`t be written as a finite sum of periodic functions. Take $f(x)=e^{x}$ Demo: If $e^{x}$ is sum of $n$ periodic functions $e^{x}=\sum_{i=1}^{n}f_{i}(x)$, with $f_{i}(x+t_{i})=f(x)$ and $t_{i}>0$, then : $e^{x+t_{n}}=\sum_{i=1}^{n-1}f_{i}(x+t_{n})+f_{n}(x+t_{n})=\sum_{i=1}^{n-1}f_{i}(x+t_{n})+f_{n}(x)$ $e^{x+t_{n}}-e^{x}= \sum_{i=1}^{n-1}(f_{i}(x+t_{n})-f_{i}(x))$ $e^{x}=\sum_{i=1}^{n-1}\frac{f_{i}(x+t_{n})-f_{i}(x)}{e^{t_{n}}-1}$ $e^{x}=\sum_{i=1}^{n-1}h_{i}(x)$, with $h_{i}(x)=\frac{f_{i}(x+t_{n})-f_{i}(x)}{e^{t_{n}}-1}$, $h_{i}(x+t_{i})=h_{i}(x)$ and $t_{i}>0$ So, if $e^{x}$ is sum of $n$ periodic functions, $e^{x}$ is sum of $n-1$ periodic functions, which implies $e^{x}$ is sum of $1$ periodic function, which is false. So $e^{x}$ can't be sum of a finite number of periodic functions.

Nice! Here is my proof: Suppose a function $f$ is sum of n periodic functions $f_{1},\ldots,f_{n}$ with periods $t_{1},\ldots,t_{n}$, $t_{i}\neq 0$. Let $T = \{t_{1},\ldots,t_{n}\}$. Then you can see by induction that, for all real x, $\sum_{X \subset T}(-1)^{|X|}f\left( x+\sum_{t \in X}t \right) = 0$. I think that, with some conditions on the $t_{i}$, this can become a sufficient condition. If we put $f(x) = e^{x}$, we can factor the condition as: $e^{x}(e^{t_{1}}-1)(e^{t_{2}}-1)\cdots (e^{t_{n}}-1) = 0$ But this is nonzero because all $t_{i}$ are different from 0