Proving the non-existence of a sequence satisfying a set of inequalities

Question

There exist some type of conditions called sequential optimality conditions that usually require some inequalities to be ensured along with it, see short introdution. Based on a conjecture on this type of sequences, It would be enlightening to this conjecture to find an analytic function $\boldsymbol{c} : \mathbb{R}^{n} \rightarrow \mathbb{R}^{p}$ and sequences $\{\boldsymbol{\mu}^{k}\} \subset \mathbb{R}^p_{+}$, which means $\mu^k_i \geq 0$ for all $i\leq p$ and $k\in\mathbb{N}$, and a convergent sequence $\{\boldsymbol{x}^{k}\} \subset \mathbb{R}^n$ such that, for all $k\in\mathbb{N}$, it's true that $\boldsymbol{c}(\boldsymbol{x}^{k}) \geq \boldsymbol{0}$, \begin{gather*} \left( \sum_{i=1}^{p} c_{i}(\boldsymbol{x}^{k}) \mu_i^{k}\right)^{2\theta} \left( \sum_{i=1}^{p}\mu_i^{k}\right)^{2(1-\theta)} \leq M + \left( \sum_{i=1}^{p} (c_{i}(\boldsymbol{x}^{k}))^2 \right) \left( \sum_{i=1}^{p}(\mu_i^{k})^2\right),\\ \left( \sum_{i=1}^{p} c_{i}(\boldsymbol{x}^{k}) \mu_i^{k}\right) \geq \delta \end{gather*} and $$ \lim_{k\rightarrow \infty} \sum_{i=1}^{p} c_{i}(\boldsymbol{x}^{k}) = 0 $$ for some $\delta>0$ and $M>0$ and all $0<\theta<1$. Is it possible?

Answer 1

$\def\paren#1{\left(#1\right)}$The answer is negative. Suppose it is possible, define\begin{gather*} y_{k, n} = c_k(x_{1, n}, \cdots, x_{p, n}),\quad Y_n = \max_{1 \leqslant k \leqslant p} y_{k, n},\quad M_n = \max_{1 \leqslant k \leqslant p} μ_{k, n},\quad \forall 1 \leqslant k \leqslant p,\ n \geqslant 1 \end{gather*} then the conditions are $μ_{k, n}, y_{k, n} \geqslant 0$ and\begin{gather*} \small\paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }^{2θ} \paren{ \sum_{k = 1}^p μ_{k, n} }^{2(1 - θ)} \leqslant M + \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 },\quad \forall 0 < θ < 1,\ n \geqslant 1 \tag{1}\\ \sum_{k = 1}^p y_{k, n} μ_{k, n} \geqslant δ,\quad \forall n\geqslant 1\tag{2}\\ \lim_{n → ∞} \sum_{k = 1}^p y_{k, n} = 0. \tag{3} \end{gather*}

First, because\begin{align*} &\mathrel{\phantom{=}} \sup_{θ \in (0, 1)} \ln\paren{ \paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }^{2θ} \paren{ \sum_{k = 1}^p μ_{k, n} }^{2(1 - θ)} }\\ &= 2\sup_{θ \in (0, 1)} \paren{ θ\ln\paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} } + (1 - θ)\ln\paren{ \sum_{k = 1}^p μ_{k, n} } }\\ &= 2\max\paren{ \ln\paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }, \ln\paren{ \sum_{k = 1}^p μ_{k, n} } }, \end{align*} and by the Cauchy-Schwarz inequality,$$ \paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }^2 \leqslant \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 }, $$ so (1) is equivalent to\begin{gather*} \paren{ \sum_{k = 1}^p μ_{k, n} }^2 \leqslant M + \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 }. \quad \forall n \geqslant 1 \tag{1$'$} \end{gather*} Next, since $\sum\limits_{k = 1}^p y_{k, n} μ_{k, n} \leqslant pY_n M_n$ and $0 \leqslant Y_n \leqslant \sum\limits_{k = 1}^p y_{k, n}$, then$$ M_n \geqslant \frac{δ}{pY_n},\quad \lim_{n → ∞} Y_n = 0, $$ from (2) and (3), respectively. However by (1'),$$ M_n^2 \leqslant \paren{ \sum_{k = 1}^p μ_{k, n} }^2 \leqslant M + \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 } \leqslant M + p^2 Y_n^2 M_n^2, $$ thus for sufficiently large $n$ such that $Y_n < \dfrac{1}{p}$,$$ M \geqslant (1 - p^2 Y_n^2) M_n^2 \geqslant (1 - p^2 Y_n^2) \frac{δ^2}{p^2 Y_n^2} = δ^2 \paren{ \frac{1}{p^2 Y_n^2} - 1 }, $$ which is contradictory with $\lim\limits_{n → ∞} Y_n = 0$.