So using the definition of periodic function which is there exist $p \neq 0$ such that $f(x+p) = f(x)$ for all $x\in \mathbb{R}$. I know that I only need to prove on the interval $[a,b]$ which other part of the function just repeats what it looks like on the interval $[a,b]$. Where do I go from here?
18016
A function $f:\mathbb{R}\to \mathbb{R}$ is periodic if there exits $p&gt;0$ such that $f(x+P)=f(x)$ for all $x\in \mathbb{R}$. Show that every continuous periodic function is bounded and uniformly continuous.   For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval $[x_0,x_0+P]$. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. However, I'm not allowed to state that theorem directly. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above?
42388
A UC Berkeley prelim exam problem asked whether an additive function $f\colon {\mathbb R} \to {\mathbb R}$, i.e. satisfying $f(x + y) = f(x) + f(y)$ must be continuous. The counterexample involved taking a positive-valued Hamel basis $X$ of $\mathbb{R}$ as a vector space over ${\mathbb Q}$, and then letting $f(x_1) =1$ and $f(x_2)=-1$ for two different $x_1,x_2 \in X$, and letting $f(x)$ be arbitrary for other $x \in X$, and then extending the function to all of ${\mathbb R}$ using the property of additivity. Then a sequence $a_n = {p_nx_1 + q_nx_2}$ could be found with rational $p_n,q_n$ such that $a_n \to 0$ but $\lim \limits_{n\to \infty} f(a_n) \neq 0$, showing discontinuity. But is Zorn's Lemma necessary to produce such an example? In other words, is Zorn's Lemma saying we can find a Hamel basis of ${\mathbb R}$ over ${\mathbb Q}$ equivalent to being able to construct a discontinuous additive function $f\colon {\mathbb R} \to {\mathbb R}$?
93195
I've seen lots of examples of periodic functions, but they all have one thing in common: They all involve at least one trigonometric term (e.g. $\sin\theta$, $\cos\theta$, etc.). My question is simple: are there any continuous, differentiable periodic functions that do not involve trigonometric terms? If not, why?
21472
Let $f:[0,1] \to \mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x \in [0,1]$.  Any hint/idea for proving that $f$ has to be constant?
2508
Find all (if that's possible) functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(f(x))=\sin x.$$  This is a problem a friend gave me. He said he saw it on a book a long time ago, but he couldn't remember what book. I literally have no idea how to solve this so I can't post my attempt here.
27870
Find all functions $f:\mathbb R \to \mathbb R$ that satisfy $$ f(x)  + 3 f\left( \frac {x-1}{x} \right) = 7x $$ for all nonzero $x$.  My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.
61633
I want to know if there exists a general method to find the period of the sum of two periodic trigonometric function. Example:     $$f(x)=\cos(x/3)+\cos(x/4).$$
65423
I have to prove that if a function f is convergent : $$ \lim\limits_{x\to +\infty} f(x) \in \mathbb{R}$$ and f is a periodic function : $$\exists T \in \mathbb{R}^{*+}, \forall x \in \mathbb{R}, f(x + T) = f(x)$$ Then f is a constant function.  Actually, this is obvious, but I can't figure out how to prove it formally ..  (I previously thought of this : , but it doen't work because I assume the periodicity hypothesis, and then I use it to prove a contradiction ..)  Thanks !
51915
Prove that there is no continuous functions $f:[0,1]\to R$, such that   $$ f(x)+f(x^2)=x. $$   My try. Assume that there is a continuous function with this property. Thus, for any $n\ge 1$ and all $x\in [0,1]$, \begin{align*} f(x)&amp;=x-f(x^2)=x-\big(x^2-f(x^4)\big)=x-x^2+\big(x^4-f(x^8)\big)=\cdots\\ &amp;=x-x^2+x^4-\cdots+(-1)^n\left(x^{2^n}-f\big(x^{2^{n+1}}\big)\right) \end{align*}  since $f(0)=0$ and $\displaystyle\lim_{n\to \infty}x^{2^{n+1}}=0$ for any $x\in(0,1)$,it follows by the continuity of $f$ that $\displaystyle\lim_{n\to \infty}f\big(x^{2^{n+1}}\big)=0$, hence $$f(x)=x-x^2+x^4-x^8+\cdots+(-1)^nx^{2^n}+\cdots$$ for any $x\in (0,1)$  Why do I have prove that there is exists such a continuous functions $f$? Maybe  my example is wrong? Why?  if my   method is wrong,then How prove this problem ? Thank you.
71236
Show that if $f(z)$ is an entire function such that it takes real values on the lines $Im(z)=0$ and $Im(z)=a$ (for some $a \in \mathbb{R}_{\neq 0})$, then $f$ is periodic with period $2 i a$.    So here is my thought process: First of all by reflection principle $f(\bar{z})=\overline{f(z)}$ so if our function $f$ is given in the strip $0 \leq Im(z) \leq a$, then its values on $-a \leq Im(z) \leq a$ follow by reflection principle. Then by imposing $f(z)=f(z+2 ia)$ we can intuitively see that this will define an entire function by looking at how the strips $na \leq Im(z) \leq 2na$ seamlessly stack on top of each other. Is this logical? And how would one prove this analytically?
3739
I want to prove:     For an integrable function $f(x)$ and periodic with period $T$, for every $a \in \mathbb{R}$, $$\int_{0}^{T}f(x)\;dx=\int_{a}^{a+T}f(x)\;dx.$$   I tried to change the values and define $y=a+x$ so that $dy=dx$ and the limits of the integrals are as we want, but I'm not sure how to use the fact that $f(x)$ is periodic.  Thanks a lot!
113180
For some function $f(x)$ there exist numbers $T \ne 0$ and $a$ such that for any $x \in D(f)$:       $$ x+T \in D(f) \\ x-T \in D(f) $$      and $f(x+T) = f(x) + a$.       $D(f)$ denotes the domain of $f$.      Prove that $f(x) = \varphi (x) + {a \over T}x$, where $\varphi (x)$ is a periodic function with period $T$.   Here is my try but I have strong doubts about its validity.  Assume that $f(x) = \varphi (x) + {a \over T}x$.  Lets see how $f(x)$ behaves in $x+T$:  $$ f(x+T) = \varphi (x+T) + {a \over T}(x+T) $$  Given $\varphi (x)$ is periodic, then $\varphi (x+T) = \varphi (x)$. Therefore:  $$ \begin{align} f(x+T) &amp;= \varphi (x) + {a \over T}(x+T) \\ &amp;=\varphi (x) + {ax \over T} +a \end{align} $$  But $f(x+T) = f(x) + a$ and then:  $$ f(x+T) = f(x) + a \\ f(x) + a =\varphi (x) + {ax \over T} + a \iff \\ \iff f(x) = \varphi (x) + {ax \over T} + a - a $$  So finally I arrived at where I started. Moreover the above shows that $f(x)$ is periodic which contradicts the initial statement that $f(x+T) = f(x) + a$  One may notice that $f(x + nT) = f(x) + na$ but I don't see how I could use that fact. What is the correct way of proving the above?   Please note that the restrictions in the question section is everything i'm given. So the function is just known to have some domain $D(f)$ and both $T$ and $a$ are not restricted to any particular set of numbers. Also that is from pre-calc section.
20656
I have a trigonometric function and I'm interested to know whether or not it has a period. At this stage I'm fairly certain that it is not periodic. However, I don't know how to prove it. Can anyone help?  This is the function:  $$g(x) = \sin(2 \pi (x-a) \times b \times \cos(2 \pi (x - a) \times c)) + d$$  Thanks
33713
$f(x)$ is positive and continuous function on $\mathbb{R}$ and, moreover, $\int_{-\infty}^{+\infty}f(x)dx=1$. $\alpha\in(0;1)$ and $[a;b]$ is the interval having a minimum length such that the $\int_{a}^{b}f(x)dx=\alpha$. Prove $f(a)=f(b)$.  With mean value theorem easy to show, that for every $\alpha\in(0;1)$, exist such $[a;b]$, so  $\int_{a}^{b}f(x)dx=\alpha$. The statement looks like Rolle's theorem, but I have no idea. I will be grateful for help.
20693
Let $f:[0,1] \longrightarrow \mathbb R$ be a continuous function, and let $g:\mathbb R \longrightarrow \mathbb R$ be a continuous and periodic function with period $1$. Prove that   $\displaystyle\lim_{n\to \infty}\int_0^1f(x)g(nx)dx=\left(\int_0^1f(x)dx\right)\left(\int_0^1g(x)dx\right)$.  any Ideas?
19806
How to show that the sequence $$x_n, n \geq 0, x_{n+2}=|x_{n+1}-x_{n-1}|, n \geq 1$$ with $x_0, x_1, x_2$ positive integers, not all null,  is periodic?  I tried to pick up the square but obtained equalities not helped.  Thanks so much for any suggestion.
51435
I'm having trouble with this problem from an old analysis qual:  Find a function $f$ such that for $p\in (1,\infty)$, $f$ is in $L^p(\mathbb{R})$ only when $p=4$.