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Here is an interesting inequality that is simulated to be correct but I can not prove it. Can someone helps me?

The question is that: For any binary vector $\mathbf{y}\in\{0,1\}^n$, prove that \begin{equation} \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{y})\leq d\} \leq \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{1})\leq d\} \end{equation} holds for all $0\leq k \leq n-1$ and $0\leq d<\frac{n}{2}$, where $\nabla(\cdot)$ denotes the first-order difference operator (i.e., $\nabla(\mathbf{x})=(\mathbf{x}_2-\mathbf{x}_1,\mathbf{x}_3-\mathbf{x}_2,\cdots,\mathbf{x}_n-\mathbf{x}_{n-1})$), $H(\cdot,\cdot)$ denotes the Hamming distance, and $\mathbf{1}=(1,1,\cdots,1)^\top$ is the all-ones vector.

Remark 1: the condition $d<\frac{n}{2}$ is necessary, otherwise, if $d\geq \frac{n}{2}$, the simulation experiment can find the counter-example and thus the inequality does not hold.

Remark 2: the above inequality can not be sepearted for each $k,d$ case, i.e., the following inequalities $$ \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0= k, H(\mathbf{x},\mathbf{y})= d\} \leq \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0= k, H(\mathbf{x},\mathbf{1})= d\}, $$ $$ \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0= k, H(\mathbf{x},\mathbf{y})\leq d\} \leq \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0= k, H(\mathbf{x},\mathbf{1})\leq d\}, $$ and $$ \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{y})= d\} \leq \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{1})= d\} $$ are false that are verified by simulation experiments.

Remark 3: it is easy to see that $\#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{1})\leq d\}=\#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{0})\leq d\}$, where $\mathbf{0}=(0,0,\cdots,0)^\top$ denotes the all-zeros vector. This does not affect the correctness of the above inequality.

Remark 4: an alternative inequality is that, for any $\mathbf{y}\in\{0,1\}^n$, prove that $$ \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{y})\leq d\} \leq \#\{\mathbf{x}\in\{0,1\}^n \mid \|\nabla(\mathbf{x})\|_0\leq k, H(\mathbf{x},\mathbf{z})\leq d\} $$ holds for all $0\leq k\leq n-1$ and $0\leq d<\frac{n}{2}$, where $\mathbf{z}=(1,1,\cdots,1,0,\cdots,0)$ satisfying $\|\mathbf{z}\|_0=\|\mathbf{y}\|_0$. This inequality is also verified to be true in simulation experiments. If one can prove thus inequality, I can then prove the first inequality to be ture.

Remark 5: I have tried a lot efforts to prove this but all failed. E.g., I tried to construct a monotonic mapping from the left set to the right set, but failed! I tried to prove this inequality by using the mathematical induction, but failed!

Contact me: If there is anyone who are interested to this problem, please feel free to contact me for any questions or discussions, here or email. Email address: [email protected]; [email protected].

Best regards!

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  • $\begingroup$ Cross-posted from Math.SE $\endgroup$ Commented Apr 23 at 8:16
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    $\begingroup$ $(\nabla x)_i = x_i - x_{i-1}$ except for $(\nabla x)_0 = x_0$? $\endgroup$ Commented Apr 23 at 8:53
  • $\begingroup$ @mathworker21, I feel that the operator may be cyclic (modulo $n$ where $n$ is the length of the vector) so$ (\Delta x)_i=x_i-x_{(i-1)~mod~ n}.$ $\endgroup$ Commented Apr 23 at 9:45
  • $\begingroup$ Thanks for attentions. The first-order difference, here, is defined by $\nabla(\mathbf{x})=(\mathbf{x}_2-\mathbf{x}_1,\mathbf{x}_3-\mathbf{x}_2,\cdots,\mathbf{x}_n-\mathbf{x}_{n-1})$. $\endgroup$ Commented Apr 24 at 1:56
  • $\begingroup$ Can you help me to prove this? Or you can share this question to anyone you know who might be interested to this or can solve this. $\endgroup$ Commented Apr 24 at 2:00

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