TITLE: $\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$ is a convex function on $[0,+\infty)$?
QUESTION [26 upvotes]: Let $f(x)=\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$, where $u\in\mathbb{Z}^+$ and $\binom{x}{l}=\frac{x(x-1)\dots(x-l+1)}{l!}$ for all $l\in\mathbb{Z}^+$. 
Then can we prove $f(x)$ is a convex function on $[0,+\infty)$?
Updates:
1) It was pointed out by @user44191 that, observing $\binom{x}{i}=\binom{x-1}{i}+\binom{x-1}{i-1}$, the question is equivalent to $\binom{x-1}{1}+\binom{x-1}{2}+\dots+\binom{x-1}{2u}$ is convex on $[0,+\infty)$.
2) Pointed out by @FedorPetrov @GeraldEdgar @H.H.Rugh:
For $x<0$ each summand $\binom{x}{2i}$ is obviously convex, thus the question is equivalent to $f(x)$ is convex on $\mathbb{R}$.
3) Pointed out by @WłodzimierzHolsztyń‌​ski:
It has $(\Delta^2 f_u)(x) =  1+ f_{u-1}(x-2)$, where $(\Delta f_u)(x)=f_u(x)-f_u(x-1)$. Then we can conclude that $f(x)$ is discrete convex.

REPLY [6 votes]: Here's an alternative proof based on probabilistic arguments (showing different aspects). Let 
$$f_n(x):=\sum_{j=0}^n   { x  \choose j}=[t^n]\,\frac{(1+t)^x}{1-t}\;\;,$$
and let $^\prime$ denote derivative with resp. to $x$.
We have to show that for even $n=2k$ the second derivative $f_{2k}^{\prime\prime}(x)=[t^{2k}]
\frac{(1+t)^x}{1-t}\,(\log(1+t))^2$ is nonnegative.
The basic observation used below is that $g(t):=\frac{\log(1+t)}{t}$ is the Laplace transform (LT) of a nonnegative random variable  possessing all moments. 
The relation $g(t)=\int_0^1 \frac{1}{1+st}\,ds$ shows that $g$ is the LT of $U\cdot X_1$,  where $U$ is uniform on $[0,1]$, $X_1$ is $\Gamma(1,1)=\exp(1)$ distributed
(i.e. has LT  $\frac{1}{1+t}$), and the factors 
are independent.
Case 1: 
$x<0$. In this case $(1+t)^x $ is the LT of  $\Gamma(1, -x)$ . Thus $\ell(t):=(1+t)^x g(t)^2$ is the Laplace transform of a nonnegative rv 
and (for $k\geq 1$)  $f_{2k}^{\prime\prime}(x)=[t^{2k-2}]\, \frac{1}{1-t} \ell(t)$ is an even degree MacLaurin sum of $\ell$, evaluated  at $1$, 
and therefore exceeds $\ell(1)>0$.
Case 2:
$x\geq 0$. Since $f_n(x+1)=f_n(x)+f_{n-1}(x)$ it will suffice to show that $f_n^{\prime\prime}(x)\geq 0$ for $x\in [0,1)$ and all $n$.
For $x=0$ all derivatives $f_n^{\prime\prime}(0)$ are nonnegative, since $a_r:=[t^r] (g(t))^2=2\,(-1)^r \frac{H_{r+1}}{r+2}$ , 
and $a_0=1>0$, $a_{2k}+a_{2k+1}\geq 0$.
Let $0<x<1$ and write $$(1+t)^x =1 +xt\,\frac{ (1+t)^x -1}{xt} =1 +xt\,h_x(t)$$ and accordingly
$$f_n^{\prime\prime}(x)=[t^{n-2}] \frac{1}{1-t}\left(1+ x t\,h_x(t)\right)g(t)^2=f_n^{\prime\prime}(0)+ x [t^{n-3}]\frac{h_x(t)g(t)^2}{1-t}\;\;.$$
Here $h_x$ is the LT of $U\cdot X_{1-x}$, where $U$ is uniform on $[0,1]$, $X_{1-x}$ is $\Gamma(1,1-x)$ (LT $(1+t)^{x-1}$) distributed, and the factors 
are independent, thus $h_x(t)g(t)^2$ is again the LT of a random variable $Z$ posessing all moments.
(1) If $n$ is odd, $n-3$ is even and $f_n^{\prime\prime}(x)\geq f^{\prime\prime}(0)$, since the second term is nonnnegative by the same argument as above.
(2) If $n$ is even, $n-3$ is odd and the second term is the $Z$-expection of a decreasing function, and will therefore not increase if $Z$ is replaced by a stochastically 
larger random variable. Replacing $U\cdot X_1$ for $U\cdot X_{1-x}$ in the first factor replaces $g$ for $h_x$, makes $Z$ stochastically larger and we get
 $$f_n^{\prime\prime}(x)\geq f_n^{\prime\prime}(0)+ x [t^{n-3}]\frac{g(t)^3}{1-t}=f_n^{\prime\prime}(0)+x\,f_n^{\prime\prime\prime}(0)\;\;.$$
If $f_n^{\prime\prime\prime}(0)\geq 0$ we're done. If  $f_n^{\prime\prime\prime}(0) <0$ it will suffice to show that 
$f_n^{\prime\prime}(0)+f_n^{\prime\prime\prime}(0)\geq 0$. 
This amounts to showing that the (even) partial sums of the coefficients 
of $c(t):=(\log(1+t))^2 + (\log(1+t))^3 $ are nonnegative, and this can be done
EDIT: some details:  the $n$-th coefficient $c_n=[t^n] c(t)$ of $c$ is $$c_n=\frac{(-1)^n}{n}\left(2 H_{n-1}-3(H_{n-1}^2-H_{n-1}^{(2)})\right)$$
Hence
$$c_n+c_{n+1}=\frac{(-1)^n}{n(n+1)}\left(-3 H_{n-1}^2 + 8 H_{n-1} + 3 H_{n-1}^{(2)} -2\right)$$
Thus (for $2k\geq 2$) the even partial sums fall until $2k=n+1=12$  and rise thereafter. Since $[t^{12}]\frac{c(t)}{1-t}=\frac{26647}{221760}>0.12$, all even partial sums are nonnegative.
Finally, note that this proof also shows that $f_n$ is convex on $[0,\infty)$ for odd $n$.