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Suppose \begin{align} f(t)=\alpha g(t)+(1-\alpha) h(t) \end{align} for all $t \in \mathbb{R}$ and for some $\alpha \in (0,1)$.

Suppose that $f(t)$ and $g(t)$ are characteristic function. Does this imply that $h(t)$ is also a chararacteristic function?

Thank you.

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    $f$ is a convex combination of $g$ and $h$. So if $g$ and $h$ are characteristic functions, $f$ is also a characteristic functions (consider the mixture distribution). But $h$ is not a convex combination of $f$ and $g$ so $f$ and $g$ are characteristic functions do not imply $h$ is. – 2017-02-02
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    @BGM I am not sure that I follow your proof. Any counter examples you have in mind? – 2017-02-02

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Firstly recall the valid statement: If $F_1$ and $F_2$ are some distributions with characteristic functions $g(t)$ and $h(t)$, then $F=\alpha F_1+(1-\alpha)F_2$ is their mixture with characteristic function \begin{align} f(t)=\alpha g(t)+(1-\alpha) h(t). \end{align} Take $\alpha=0{,}5$ for simplicity. Then the equality $F=\alpha F_1+(1-\alpha)F_2$ turns to $2F=F_1+F_2$.

If conversely we take some distributions $F$ and $F_1$, the difference $F_2=2F-F_1$ is not obliged to be some distribution. Say, we can take distribution $F$ degenerate at $1$, and $F_1$ degenerate at $0$, and the difference $F_2=2F-F_1$ is not a distribution at all.

Return to characteristic functions. CF for $F$ equals $f(t)=e^{it}$, CF for $F_1$ is $g(t)=1$. The second summand $h(t)$ in the equality $$e^{it}=\frac12 \cdot 1 + \frac12 \cdot h(t)$$ is $h(t)=2e^{it}-1$. This is not CF since the absolute value of this function can be greater than $1$.

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    Thanks. Do you know what happens if we restrict the question to real and symmetric characteristic functions? – 2017-02-14
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    @Boby Nothing will change. You can find a suitable example. – 2017-02-14