Convex Combination again a characteristic function I also checked some other posts about similar problems but still not able to find a solution  for this problem.
Let $\phi(t)$ be a characteristic function. Is
$$\gamma (t) = \frac{1}{3} \phi(2t) + \frac{2}{3} \phi(\frac{t}{4})$$ also a characteristic function?
It seems like a convex combination of characteristic functions. I assume it is a characteristic function but do not know how to show it. I also checked the properties which (hopefully) holds in my calculations.
 A: It all boils down to one, not very hard fact:
Fact: If $\{\mu_k\}_{k \in \mathbb N_+}$ is a family of probability measures on $(E,\mathcal B(E))$, and $(p_k)_{k \in \mathbb N_+}$ is a sequence of wages (that is $p_k \in [0,1]$ for every $k \in \mathbb N_+$ and $\sum_{k=1}^\infty p_k=1$), then $\nu = \sum_{k=1}^\infty p_k\mu_k$ is also a probability measure. (Note we don't need the convex combination to be finite)
The proof isn't that hard, firstly note that $\nu$ is well defined on $(E,\mathcal B(E))$, since for every $A \in \mathcal B(E)$ we have $\mu_k(A) \le 1$, so the series converges for any such $A$. Obviously $\nu$ takes only non-negative values.
$\nu(E) =\sum_{k=1}^\infty p_k\mu_k(E) = \sum_{k=1}^\infty p_k = 1$
And lastly taking $A_1,A_2,...$ disjoint, we have:
$$\nu(\bigcup A_j) = \sum_{k=1}^\infty p_k \mu_k(\bigcup A_j) = \sum_{k=1}^\infty \sum_{j=1}^\infty p_k\mu_k(A_j) = \sum_{j=1}^\infty \sum_{k=1}^\infty p_k\mu_k(A_j)=\sum_{j=1}^\infty \nu(A_j)$$
Where we could change order of summation due to non-negative values.
Having this fact, it can be shown that $\gamma$ is a characteristic function of measure $$\nu = \frac{1}{3}\mu_{2X} + \frac{2}{3} \mu_{\frac{X}{4}}$$, where $\mu_Y$ means the distribution of $Y$
Indeed:
$$ \gamma(t)= \int_{\mathbb R}e^{itx}d\nu(x) = \int_{\mathbb R}e^{itx}d(\frac{1}{3}\mu_{2X}(x) + \frac{2}{3}\mu_{\frac{X}{4}}(x)) = \frac{1}{3}\int_{\mathbb R}e^{itx}d\mu_{2X}(x) + \frac{2}{3}\int_{\mathbb R}e^{itx}d\mu_{\frac{X}{4}}(x)$$
All you need to know now, is to know that when $\varphi$ is a characteristic function of $X$ (or equivalently $\mu_X$), then function $t \to \varphi(at)$ is a characteristic function of $aX$ (or again, equivalently $\mu_{aX}$), getting:
$$ \gamma(t) = \frac{1}{3}\varphi(2t) + \frac{2}{3}\varphi(\frac{t}{4})$$
