Characteristic function with modulus 1 implies degenerate distribution 
Let $X$ be a random variable with characteristic function $\phi(\ )$ satisfying $|\phi(t)|=1$ for all $|t|\leq 1/T$ with some $T>0$. Show that $X$ is degenerate, i.e., there is $c$ such that $P(X=c)=1$.

My try :
$|\phi(t)|^2=1 \implies (\mathbb{E}(\cos tX))^2+(\mathbb{E}(\sin tX))^2=1=\mathbb{E}(\cos^2 tX+\sin^2 tX)=\mathbb{E}(\cos^2 tX)+\mathbb{E}(\sin^2 tX)$ so we can say that $\sin tX=\mathbb{E}(\sin tX), \cos tX=\mathbb{E}(\cos tX)$, that is $\phi(t)=\rm{e}^{\rm{i}tX}$ for $|t|\leq 1/T$. But I cannot go anywhere from here, can someone help me? Thanks.
Edit : 
I found out this fact. Let $\psi(t)=|\phi(t)|^2$ which is a characteristic function and its real, and $\psi(t)=1, |t|\leq 1/T$. Now employ the inequality $\Re(1-\psi(t))\leq 4\Re(1-\psi(t/2))$ now apply this $n$ times we get $(1-\psi(t))\leq 4^n\left(1-\psi\left(\dfrac{t}{2^n}\right)\right)$ now for any $t$ we get rhs goes to $0$. But since $\psi$ is real it is also $\le 1$ so $\psi(t)=1$ for all $t$. Now we have $|\phi(t)|=1$ for all $t$. I know this is pretty pointless, but I don't understand any of the proofs given below, if someone would clearly explain why the sets mentioned have only one element in common, I would be grateful.
 A: Let $Y$ be an independent copy of $X$, i.e. $Y$ has the same distribution as $X$ and $Y$ is independent of $X.$ Define $Z=X-Y.$ The characteristic function of $Z$ is given by $$E(e^{itZ})=E(e^{itX})E(e^{-itY})=\phi(t)\phi(-t)=|\phi(t)|^2.$$
Therefore, for $t\in (-1/T,1/T),$ the characteristic function of $Z$ matches with the characteristic function of a random variable $Z'$ degenerate at $0.$ This implies that $Z$ has the same distribution as $Z'$, which means that distribution of $Z$ is degenerate at $0.$ In other words, $X = Y$ with probability 1.
If $F(x) = \Pr(X \leq x)$ be the CDF of $X,$ then using the independence of $X, Y$ and the fact that $Y=X$ with probability 1, we get $$F(x) = \Pr(Y = X \leq x) = \Pr(X \leq x)\Pr(Y \leq x) = F(x)^2$$
which implies that for every $x,$ $F(x)$ is either 0 or 1. This clearly says that $X$ must have a degenerate distribution.
A: I found an answer to this problem here:
http://www.les-mathematiques.net/phorum/read.php?12,488850,489042
As French language may be a problem, here's a translation of this solution:
Let $t\in\left[-\frac{1}{T},\frac{1}{T}\right]$. If $|\phi(t)|=1$, then there exists $\theta=\theta(t)$ such that $\phi(t)=e^{i\theta}$. So:
$$E[1-e^{i(tX-\theta)}]=0$$
This leads to:
$$E[\operatorname{Re}(1-e^{i(tX-\theta)})]=0$$
As $\operatorname{Re}(1-e^{i(tX-\theta)})\geq 0$, when then have:
$$e^{i(tX-\theta)}=1\text{ a.s}$$
Thus:
$$X\in\frac{\theta}{t}+\frac{2\pi\mathbb{Z}}{t}$$
Now we can do the same thing with $t'$ such that $|t'|\leq\frac{1}{T}$ and $t$ and $t'$ are rationally independent.
$$X\in\left(\frac{\theta}{t}+\frac{2\pi\mathbb{Z}}{t}\right)\cap\left(\frac{\theta}{t'}+\frac{2\pi\mathbb{Z}}{t'}\right)$$
which is a singleton since $t$ and $t'$ are rationally independent.
A: First fix $t$. From $$(\mathbb{E}(\cos tX))^2+(\mathbb{E}(\sin tX))^2 = \mathbb{E}(\cos^2 tX+\sin^2 tX)$$
you get
$$\textrm{var}{(\cos{tX})} + \textrm{var}{(\sin{tX})} = 0$$
Then, almost surely
$$\cos{tX} = c(t) \qquad\text{and}\qquad \sin{tX} = s(t)$$
for some constants $c(t)$ and $s(t)$ satisfying $c(t)^2+s(t)^2=1$.
Then you could say $X = \frac{1}{t}\arg(c(t)+is(t)) = 1$ almost surely.
I must admit though that the last step seems a bit weird to me. But I will post this anyway. I hope someone points a possible mistake I made.
