Let $\phi$ be the characteristic function of some probability distribution, $0<b$ and $0<c<1$. Suppose $|t|\geq b \implies |\phi(t)|\leq c$. Prove that $|t|<b \implies |\phi(t)|\leq 1-\frac{1-c^2}{8b^2}t^2$
This is from a qualifying exam, with no hints. I can't make progress on this one and I'd appreciate any hint or suggestion.
The first step is to prove the following auxiliary result:
Proof: Let $f$ be a real-valued characteristic function and denote by $\mu$ the associated probability measure. Clearly, $$1-f(t) = \int_{\mathbb{R}} (1-\cos(tx)) \, d\mu(x), \qquad t \in \mathbb{R}.$$ Using that
$$1-\cos(y) = 2 \sin^2 \left( \frac{y}{2} \right) \tag{2}$$
and
$$|\sin(2y)| = 2 |\sin y \cos y| \leq 2 |\sin y| \tag{3}$$
we find
$$\begin{align*} 1- \cos(tx) \stackrel{(2)}{=} 2 \sin^2 \left( \frac{tx}{2} \right) &\stackrel{(3)}{\geq} \frac{1}{2} \sin^2 (tx) \stackrel{(2)}{=} \frac{1}{4} (1-\cos(2tx)). \end{align*}$$ Integrating both sides with respect to $\mu$ yields $$1-f(2t) \leq 4 (1-f(t)), \qquad t \in \mathbb{R}.$$ Applying this inequality to $f(t) = |\phi(t)|^2$ proves the assertion. (I leave it to you to check that this is indeed a characteristic function.)
Iterating $(1)$, we find
$$1-|\phi(2^n t)|^2 \leq 4^n (1-|\phi(t)|^2) \tag{4}$$
for all $n \in \mathbb{N}$. For any $t \neq 0$, $|t|<b$ we can choose $n$ such that $2^{-n} b \leq |t| < 2^{-n+1} b$. Then, by assumption, $|\phi(2^n t)|^2 \leq c^2$ and therefore
$$1-c^2 \leq 1-|\phi(2^n t)|^2 \stackrel{(4)}{\leq} 4^n (1-|\phi(t)|^2).$$
By the very choice of $n$, we have $2^n \leq 2b/|t|$. Hence,
$$1-c^2 \leq 4 \frac{b^2}{t^2} (1-|\phi(t)|^2),$$
i.e. $$|\phi(t)|^2 \leq 1- \frac{(1-c^2)}{4b^2} t^2.$$
Finally, we use the elementary inequality
$$\sqrt{1-C t^2} \leq 1- \frac{Ct^2}{2}$$
which holds for any constant $C>0$ to conclude that
$$|\phi(t)| \leq \sqrt{1- \frac{1-c^2}{4b^2} t^2} \leq 1- \frac{1-c^2}{8b^2} t^2.$$