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.



$\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$ is a convex function on $[0,+\infty)$?
*This is not an answer to your question, is only an equivalent reformulation that seems promising. I write it as an answer only because of space constraints.*


For any nonnegative integer $n$ and any $\newcommand{\bR}{\mathbb{R}}$ $x\in\bR$ we define


$$ a\_n(x)=\sum\_{k=0}^n \binom{x}{2k}, \;\;b\_n(x)=\sum\_{k=0}^n \binom{x}{2k+1},$$


where for any nonnegative integer $m$ we set


$$
\binom{x}{m}:=\frac{x(x-1)\cdots (x-m+1)}{m!}.
$$


For $t\in (-1,1)$ and $x\in\bR$ the series


$$ F\_x(t):=\sum\_{k=0}^\infty \binom{x}{k}t^k $$


is the Taylor series at $t=0$ of the function $t\mapsto (1+t)^x$ and thus


$$ F\_x(t)=(1+t)^x,\;\;\forall |t|<1. $$


The generating series of the even binomial coefficients $\binom{x}{2m}$ is then


$$ F^0\_x(t)=\sum\_{m\geq 0}\binom{x}{2m}t^{2m}= \frac{1}{2}\Bigl(\, F\_x(t)+F\_x(-t)\,\Bigr)=\frac{(1+t)^x+(1-t)^x}{2}. $$


The generating series of the odd binomial coefficients $\binom{x}{2m+1}$ is then


$$ F^1\_x(t)=\sum\_{m\geq 0}\binom{x}{2m+1}t^{2m}= \frac{1}{2}\Bigl(\, F\_x(t)-F\_x(-t)\,\Bigr)=\frac{(1+t)^x-(1-t)^x}{2}. $$


The generating series of the sequence


$$a\_n(x)=\sum\_{k=0}^n \binom{x}{2k} $$


is


$$ A\_x(t)=\sum\_{n\geq 0} a\_n(x)t^{2n}=\frac{1}{1-t^2} F\_x^0(t)=\frac{(1+t)^x+(1-t)^{x}}{2(1-t^2)}. $$


The generating series of $b\_n(x)$ is
$$
B\_x(t)=\sum\_{n\geq 0} b\_n(x)t^{2n+1}=\frac{1}{1-t^2} F\_x^1(t)=\frac{(1+t)^x-(1-t)^{x}}{2(1-t^2)}. $$


We have $\newcommand{\pa}{\partial}$


$$\pa^2\_x A\_x(t)=\sum\_{n\geq 0} a\_n''(x) t^{2n}. $$


The problem is equivalent to showing that, for any $x\in \bR$, the Taylor coefficients at $t=0$ of the function


$$[0,1)\ni t\mapsto \pa^2\_xA\_x(t) $$


are nonnegative, i.e. for any $x$, the function $t\mapsto \pa^2\_x A\_x(t)$ is absolutely monotonic on the $t$-interval $[0,1)$; for definition and properties of absolutely monotonic functions see Chap. IV of Widder's classical monograph *The Laplace Transform*.


Now observe that


$$ \pa^2\_xA\_x(t)=\frac{(1+t)^x\bigl(\,\log(1+t)\,\bigr)^2+(1-t)^x\bigl(\,\log(1-t)\,\bigr)^2}{2(1-t^2)}. $$


Actually we only need to prove that the even degree Taylor coefficients at $t=0$ of the function


$$ t\mapsto G\_x(t)=(1+t)^x\frac{\log^2(1+t)}{1-t^2} $$


are positive for any $x\in\bR$.


**Remark.** As observed in comments to the question, we can instead study the convexity of the function


$$ c\_n(x)=\sum\_{j=0}^{2n}\binom{x}{j}.$$


The generating series
$$ C\_x(t) =\sum\_{n\geq 0} c\_n(x) t^{2n}, $$


is the even part of


$$ G\_x(t)=\frac{(1+t)^x}{1-t}, $$


i.e.,


$$ C\_x(t)=\frac{1}{2}\left( \frac{(1+t)^x}{1-t}+ \frac{(1-t)^x}{1+t}\right)=\frac{(1+t)^{x+1}+(1-t)^{x+1}}{2(1-t^2)} =A\_{x+1}(t). $$


