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Let $p>1.$ Define the set

$$C=\left\{\mu=\{\mu_n\}_{n=0}^\infty: \{\mu_n\}\subseteq (0,1), \;\{\mu_n\} \textrm{ decreasing, } \sum_{n=0}^\infty\mu_n(n+1)^p =1\right\}.$$

Determine $$\inf_{\mu\in C}\; \sum_{n=0}^\infty\mu_n(n+1).$$

This problem arises when we try to find a sharp upper bound on the dual norm of the derivative of the penalty function in the Borwein-Preiss Variational Principle. I tried to use Chebyshev inequality(after a transformation), but unsuccessfully obtained the trivial bound $inf\geq 0$. If you find a lower bound, I will be happy too.

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The inf is $0$. Let $k \ge 1$ and $\epsilon_k = \frac{1}{1^p+2^p+\dots+k^p}$. Take $\mu_n = \epsilon_k$ for $0 \le n \le k-1$, and $\mu_n = 0$ for $n \ge k$. Then, $\mu \in C$ and $\sum_{n=0}^\infty \mu_n(n+1) = \frac{1+2+\dots+k}{1^p+2^p+\dots+k^p}$. This goes to $0$ as $k \to \infty$ (since $p > 1$).

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  • $\begingroup$ Note that $\mu \notin C$ because $\mu_n=0$ for $n\geq k.$ It is a good idea nevertheless. I think we could modify it. thanks $\endgroup$ – John D Aug 31 '17 at 20:53
  • $\begingroup$ It can easily be modified. Just modify $\mu$ a bit to be extremely close to $0$ after $k$. $\endgroup$ – mathworker21 Aug 31 '17 at 21:05

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