How to prove $H(H^{-1}(v)*p)$ is convex in $v$ for every $p\in [0,1]$

Question

In the proof of Mrs Gerber's lemma, there is a key step to prove $H(H^{-1}(v)*p)$ is convex in $v$ for every $p\in [0,1]$.I don't konw how to prove it.

Answer 1

$$ H^{-1}(r\times\nu_1+(1-r)\times\nu_2) \leq r \times H^{-1}(\nu_1)+(1-r) \times H^{-1}(\nu_2) (1) \\ H^{-1}(r \times \nu_1+(1-r) \times \nu_2)*p \geq (r \times H^{-1}(\nu_1)+(1-r) \times H^{-1}(\nu_2)) * p (2) \\ H^{-1}(r \times \nu_1+(1-r) \times \nu_2)*p \geq r \times H^{-1}(\nu_1) * p+(1-r) \times H^{-1}(\nu_2)) * p (3) \\ H(H^{-1}(r \times \nu_1+(1-r) \times \nu_2)*p) \leq H(r \times H^{-1}(\nu_1) * p+(1-r) \times H^{-1}(\nu_2)) * p) (4) \\ H(H^{-1}(r \times \nu_1+(1-r) \times \nu_2)*p) \leq r \times H(H^{-1}(\nu_1) * p)+(1-r) \times H(H^{-1}(\nu_2)* p) (5) $$

assume that $p$ is smaller than $1/2$. Similar derivations can be obtained for $p>1/2$.
(1) is derived because $H^{-1}(\nu)$ is a convex function of $\nu$,
(2) is derived because $u*p$ is a decreasing function of $u$ if $p<1/2$,
(3) is derived because $(r \times H^{-1}(\nu_1)+(1-r) \times H^{-1}(\nu_2)) * p= r \times H^{-1}(\nu_1)*p+(1- r) \times H^{-1}(\nu_2))*p$
(4) (tricky) is derived because it can be obtained that both sides of (3) are greater than $1/2$, and $H(x)$ is a decreasing function of $x$ if $x$ is greater than $1/2$.
(5) is obtained because $H(x)$ is a concave function of $x$.