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Given a family of complex numbers $(z_n)_n$ prove that there exists a subsequence $(z_{n_k})_k$ such that

$$\sum_{n} |z_{n}|\le \pi \left|\sum_{k} z_{n_k}\right|$$

I ailled to find a counter example of this inequality. what tricky me here is the constant $\pi$. How does that constant appears here? How can I prove this inequality?

Any hint or idea?

Guy Fsone
  • 23,903
  • Restricted to reals, you could get a constant of 2 by considering the negative sums and the positive sums (and then taking the bigger sum in absolute value as your subsequence); maybe something similar could work here? (never took complex analysis so I don't have any other ideas, and no clue where pi comes from) – E-A Nov 28 '17 at 09:26
  • The pointed-to question is about an estimate with factor $4 \sqrt 2$ (which is easier to obtain), but this answer proves it with factor $\pi$ (and argues that $\pi$ is the best possible factor). – Martin R Nov 28 '17 at 12:21
  • And here is the generalization to higher dimensions: Maximum subset sum of $d$-dimensional vectors – Martin R Nov 28 '17 at 12:37
  • Here is the same solution, but with an instructive comment: "Trying to find a subset with largest sum, it is evident that the best thing you can do is to take those vectors which lie in some half-plane." – Martin R Nov 28 '17 at 12:42

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