Show that if $\phi_{X}(t)=1$ in a neighborhood of $0$, then $X=0$ a.s. 
Let $\phi(t),t\in\mathbb{R}$, be the characteristic function of a random variable $X$. Show that if $\phi(t)=1$ in a neighborhood of $0$, then $X=0$ a.s.

The problem comes with the following hint: Show that $1-Re(\phi(2t))\le4(1-Re(\phi(t)))$ for $t\in \mathbb{R}$. I am stumped by this one, I am not even sure where to begin or how to prove\use use the hint, any help here would be greatly appreciated.
 A: Here is another method: start by noticing that since $\cos \leq 1$,
$$\Re \phi(t) = \mathbb{E}[\cos(tX)] \leq 1, \quad \forall t \in \mathbb{R}. \tag{1}$$
Now let $\def\eps{\varepsilon} (-\eps, \eps)$ be an interval on which $\phi = 1$. Then, using the hint, we have for $t \in (-\eps, \eps)$
$$0 \leq 1- \Re \phi(2t) \leq 4(1-\Re \phi(t)) = 0.$$
The first inequality follows from $(1)$ and the last equality from the assumption. This proves that $\Re \phi(2t) = 1$ for every $t \in (-\eps,\eps)$. Reiterating this process, we see that
$$\Re \phi(t) = 1, \quad \forall t \in \mathbb{R}.$$
But recall that $|\phi(t)| \leq 1$. This forces $\phi(t) = 1$ for every $t \in \mathbb{R}$. Since the characteristic function determines the distribution, it follows that $X = 0$ a.s.
A: Let's assume that $\varphi(t) = 1$ for any $t \in [0,\delta]$. Then in particular $\varphi(\delta)=1$. We'll show that it is the case that $\mathbb P(X \in \{\frac{2k\pi}{\delta} : k \in \mathbb Z \}) = 1$ . Let $\mu_X$ be distribution of $X$.
Note that $\varphi(\delta)=1$ means:
$$ 0 = 1 -\varphi(\delta) = 1 - \int_{\mathbb R} \cos(\delta x) d\mu_X(x) = \int_{\mathbb R} (1 - \cos(\delta x)) d\mu_X(x)$$
Since $1-\cos(\delta x) \ge 0$, we must have $\cos(\delta x) = 1$ , $x - d\mu_X$ almost surely, so that $x = \frac{2k\pi}{\delta}$ , $d\mu_X$ almost surely, which means $\mu_X( \{\frac{2k\pi}{\delta} : k \in \mathbb Z \})=1$.
Now note that we have only countable many points in set $\{\frac{2k\pi}{\delta} : k \in \mathbb Z \}$. For every $k \in \mathbb Z \setminus \{0\}$ we can find such $t_k \in (0,\delta)$ that $\frac{2k\pi}{\delta}$ is not equal to $\frac{2 m \pi}{t_k}$ for any $m \in \mathbb Z$ (because for every $m \in \mathbb Z$ there is at most one $s \in (0,\delta)$ such that $\frac{2m \pi}{s} = \frac{2k\pi}{\delta}$, but we have only countable many $m \in \mathbb Z$, but continuum-many $s \in (0,\delta)$, so there exists such $t_k$) which means that $\mu_X(\frac{2k\pi}{\delta}) = 0$ (for that given $k \in \mathbb Z \setminus \{0\}$, because $\varphi(t_k)=1$, so $\mu_X( \{ \frac{2m\pi}{t_k} : m \in \mathbb Z \}) = 1$, too). Since $k \in \mathbb Z \setminus \{0\}$ was arbitrary, and there are only countable many of them, we have $\mu_X( \{ \frac{2k \pi}{\delta} : k \in \mathbb Z \setminus \{0\} \} ) = 0$, so that $\mu_X(\{0\}) = 1$ what was to be proven.
EDIT: If you're interested, here's an approach with your hint. Let's prove it beforehand. $$ 1 - Re(\varphi(2t)) = \int_{\mathbb R} (1-\cos(2tx))d\mu_X(x) = 2\int_{\mathbb R} (1 - \cos^2(tx))d\mu_X(x) $$
It would be sufficient to show $1-\cos^2(s) \le 2(1- \cos(s))$ which is equivalent to $0 \le \cos^2(s) - 2\cos(s) + 1 = (\cos(s)-1)^2$, so true. Hence $$ 1- Re(\varphi(2t)) \le 4\int_{\mathbb R}(1 - \cos(tx))d\mu_X(x) = 4(1-Re \varphi(t))$$
Having lemma, it is pretty easy. Note that you have such $\delta$, that $\varphi(t) = 1$ for any $[-\delta,\delta]$. Now let's prove it is also the case for any $t \in [-2\delta,2\delta]$ using hint: Take $s \in [-2\delta,2\delta]$. We have $$ 1 - Re(\varphi(2s)) \le 4(1 - Re(\varphi(s)) = 0$$ since $s \in [-\delta,\delta]$. Moreover, $|\varphi(s)| \le 1$, so $\varphi(s) = 1$. Use it again, to prove the fact for any $[-2^k\delta,2^k\delta]$ getting $\varphi(t)=1$ for any $t \in \mathbb R$.
