A short question about the convexity of a function Let $x$ and $y$ be two numbers; $0\leq x \leq 1$ and $0\leq y \leq 1$ satisfying
$$\mathcal{X}\times \mathcal{Y}=\left\{(x,y):\sum^{\lfloor k\rfloor}_{i=0}\binom{n}{i}(1-y)^{i} y^{n-i} +\sum^{\lfloor k\rfloor}_{i=0}\binom{n}{i}x^i (1-x)^{n-i}=1\right\}$$
Define a function $f^n_k:[0,1]\rightarrow [0,1]$ with the mapping $\mathcal{X} \overset{f^n_k}{\mapsto} \mathcal{Y}$

Prove that for every $k<n/2$, the function $f_k^n$ is concave (or similarly for every $k>n/2$ the function $f_k^n$ is convex).

For simplicity, $n$ can be assumed odd.

Summary of what I've found (for $k<n/2$):
$f_k^n$ is obviously continous and passes through $(0,0)$ and $(1,1)$.
$f_k^n$ passes through all pairs $(x,y)$ for which $y>x$ holds (except at $(0,0)$ and $(1,1)$, therefore $f_k^n$ lies above $y=x$ line on $[0,1]^2$.
For every $f_k^n(x_0)=y_0$ there is an $\epsilon>0$ close point to $x_0$, namely $x_0+\epsilon$ such that $f_k^n(x_0+\epsilon)=y_0+\delta$ for some positive $\delta$.
This proves that $f_k^n$ is increasing above $y=x$ line, but I dont think that it is enough for concavity.
Here is the figure for $n=5$:
Details of what I've done:
The set defined above is the same with 
$$\mathcal{X}\times \mathcal{Y}=\left\{(x,y): B(k,n,1-y)=1-B(k,n,x)\right\}$$
where $B$ is the binomial c.d.f.
Now let $h_0(x;n,k)=1-B(k,n,x)$ and $h_1(y;n,k)=B(k,n,1-y)$. I am able to prove that both $h_0$ and $h_1$ are increasing functions of $x$ and $y$ respectively.
I know that in case $k=n/2$ then  $$h_0(x;n,n/2)=h_1(x;n,n/2)$$
One can see that we also have the following inequalities:
$$h_0(x;n,n/2-1)>h_0(x;n,n/2)$$
and
$$h_1(x;n,n/2-1)<h_1(x;n,n/2)$$
From this point i conclude that 
$$h_0(x;n,n/2-1)>h_1(x;n,n/2-1)\quad\forall x\quad\quad (1)$$
BUT the condition of the set says that I need to find all $(x,y)$ such that $$h_0(x;n,n/2-1)=h_1(y;n,n/2-1)\quad\quad (2)$$
If $x=y$ or $x>y$ is assumed to satisfy $(2)$ then this is a contradiction for $(1)$. Accordingly I conclude that $y>x$ must be true for all pairs $(x,y)\in[0,1]^2$ except $(0,0)$ and $(1,1)$, which always satisfy the equality $(2)$.
In the next step I assume the point $(x_0,y_0)$ is valid and satisfies $(2)$. Then II choose another point which is very close to $x_0$, i.e., $x_0+\epsilon$ for some $\epsilon>0$. Then since $h_0$ is an increasing function of $x$ then I will have $$h_0(x_0+\epsilon;n,n/2-1)>h_0(x_0;n,n/2-1)$$
this says that left hand side of $(2)$ increased with adding $\epsilon$ to $x_0$ and for $(2)$ to hold, right hand side must also increase. We are lucky because $h_1$ is also an increasing function of $y$ therefore if $(x_0,y_0)$ is a valid point satisfying $(2)$, then $(x_0+\epsilon,y_0+\delta)$ should also be valid for some positive $\delta$.
I used $h_0(x;n,n/2-1)$ for the case $k<n/2$ but I could use $h_0(x;n,n/2-a)$ for some $a$. The conclusion would not change.
Thanks:
Could you please help me?
Thanks in advance.
 A: Let
$$ g(n,k,x)=\sum_{j=0}^{k}\binom{n}{j}x^j(1-x)^{n-j}. $$
Assuming that $n$ is odd and $k<\frac{n}{2}$, and taking, for short, $g(x)=g(n,k,x)$, we just have to prove that
$$ g^{-1}(1-g(1-x)) $$
is a convex function on $[0,1]$. This is equivalent to proving that
$$ h(x)=g^{-1}(1-g(x)) $$
is a convex function on the same interval. 
$h(x)$ has an interesting property: it is an involutive map, since $h(h(x))=x$. Moreover, since $g$ is decreasing, we have that $1-g$ is increasing while $g^{-1}$ is decreasing, so $h$ is decreasing. By differentiation we have:
$$h'(x)\, h'(h(x)) = 1, $$
$$ h''(x)\, h'(h(x)) + h'(x)^2 h''(h(x)) = 0, $$
$$ h''(x)+h'(x)^3 h''(h(x)) = 0, $$
$$ h''(h(x)) + h'(h(x))^3 h''(x) = 0,$$
$$ \frac{h''(h(x))}{h''(x)} = -\frac{1}{h'(x)^3} \geq 0. \tag{1}$$
so the convexity/concavity of $h(x)$ over $[0,1]$ just depends on the convexity/concavity of $h(x)$ in a small neighbourhood of $1$, i.e. on the behaviour of $g$ in a small neighbourhood of $x=0$. Since $g(n,k,x)$ behaves like $1-\binom{n}{k+1}x^{k+1}$ in a neighbourhood of zero, we have that $h(x)$ is convex in a neighbourhood of $1$, so it is convex on $[0,1]$ due to $(1)$.
