@YemonChoi Sorry for unfamilar of this rule.

I mean for a PDF $g$ on $[0,1]^2$, we can calculate its integer $\int_A g$ on the measurable subset $A$--and this value shows the proportion of $g$ distributed on $A$ compared to on $[0,1]^2$. For $f$ with $\int f>0$, the distribution (with respect to $f$) 's integer on $[0,0.5]\cdot [0,0.5]$ is just $\int_[0,0.5]\cdot [0,0.5] f/\int f$. However, when $\int f=0$ (as your example), could we do a samilar thing to know the proportion of $f$ distributed on $[0,0.5]\cdot [0,0.5]$ compared to on $[0,1]^2$? Thanks very much! @NateRiver

May I ask a question: For a distribution which is supposed on a 0-measured subset of $[0,1]^2$, how could we calculate its integer on a measureable subset of $[0,1]^2$? i.e., let $f:[0,1]^2\to [0,1]$ be a measureable function with $\int f=0$. Could we consider the distribution as something like $f(x,y)/(\int f)$ and calculate its integer on $[0,0.5]\cdot [0,0.5]$? Thank you a very much for your time!

@FedorPetrov Thanks! By the way, may I ask that I want to prove when $k$ is very large, for any $d>0$ the number of $a_i$ having Hamming distance with $x_0$ leq $a-d$ is no less than the number of $a_i$ having Hamming distance with $x_0$ geq $a-d$, could I consider the distribution of $u$ as two Gaussian function's product? i.e., repersent $f(u)=\binom{k_0}{\frac{u - k' + k_0}{2}} \binom{k - k_0}{ \frac{u + k' - k_0}{2}}$ as $g(u)=c_1\exp\{c_2(u/2-k_0/2)^2+c_3(u/2-k/2+k_0/2)^2\}$, and by compare the inetger of $g$ in two intervals to compare the sum of $f$ in two intervals?

@FedorPetrov May I ask does these $k$ 0/1 random variable i.i.d.? And if $k$ is very very large, could we consider the distribution of $u$(the number of $a_i$ that equal to $u$) as two product of Gaussian(use Gaussian is easier to calculate than use Binomal)?

Thanks very much! I'm reading your answer.

Thanks! Now,I have an idea but not sure whether it's correct...for the weak version with $0.5k\leq k_0\leq k'\leq 0.5+o(1)k$ and $k$ very large(why consider this because when k large, #0-1 vector with (#0s>0.5+o(1)k)/(2^k) tend 0). By considering $f(u) = \binom{k_0}{\frac{u - k' + k_0}{2}} \binom{k - k_0}{\frac{u + k' - k_0}{2}}$ satisfies $\sum_{u\in a\pm d}f(u)/(\sum_{u\in a\pm d}1)\leq a$, thus for any $t\in\mathbb{R}^1$, $\sum_{a'_i \geq t} a'_i \geq \sum_{a_i \geq t} a_i.$ So we only need to solve this question: math.stackexchange.com/questions/4993913/…

Thanks very much! By the way, do you think this question can be finally proved (or its weak version, allow a contact times it)? Could convex property (something like Karamata's inequality) be applied here for its prove? thank you!

When $k' = k_0$, the distribution is symmetric, and the inequality holds with equality. Additionally, I feel that as $k'$ increases, the difference between the right side and the left side of the inequality grows, but I am unsure how to prove this. @IosifPinelis

well, because I guess the distribution function(f(x):=the number of a_i which equals x) of $\{a_i\}$ is symmetric(or have some "symmetric" property) but I'm not sure. If symmetric then its equal. @IosifPinelis

Can Karamata's inequality be applied here? Or are there other methods we could consider for this type of problem?

yes, thanks for your two kind answers and I voted both of your answers up!