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Miklos Schweitzer 1971 Problem 5:

Let $\lambda_1\leq \lambda_2\leq\cdots$ be a positive sequence and let $K$ be a constant such that $$\sum\limits_{k=1}^{n-1} \lambda_k^2 < K\lambda_n^2\qquad(n\in\mathbb{N})$$ Prove that there exists a constant $K'$ such that $$\sum\limits_{k=1}^{n-1} \lambda_k < K'\lambda_n$$

$\textbf{My work}$: We will prove by induction. For $n=2$, we know $\exists K_1: \lambda_1^2< K_1\lambda_2^2\iff \lambda_1 < \sqrt{K_1}\lambda_2$ as the sequence consists of positive terms, and the second inequality holds if we let $K_1'=\sqrt{K_1}$. Now assume $\exists K'_{n-1}: \sum\limits_{k=1}^{n-1} \lambda_k<K_{n-1}'\lambda_n$. Then for the next index, we have $ \sum\limits_{k=1}^n \lambda_k<(K_{n-1}'+1)\lambda_n$. Additionally, we know that $\sum\limits_{k=1}^{n-1}\lambda_k^2<K_{n-1}\lambda_n^2$, so $\sum\limits_{k=1}^n \lambda_k^2<(K_{n-1}+1)\lambda_n^2\leq^* (K_n+K_{n-1})\lambda_{n+1}^2$ where $\leq^*$ follows because $\lambda_n<\lambda_{n+1}$, so $K_{n-1}\lambda_n^2<K_{n-1}\lambda_{n+1}^2$ and so we can add it to both sides without reversing the inequality. Hence we can reduce the above inequality to saying $\lambda_n\leq \sqrt{\frac{K_n+K_{n-1}}{K_{n-1}+1}}\lambda_{n+1}$. Thus $\sum\limits_{k=1}^n \lambda_k<(K_{n-1}'+1)\lambda_n<(K_{n-1}'+1)\sqrt{\frac{K_n+K_{n-1}}{K_{n-1}+1}}\lambda_{n+1}$ and so if we choose $K_n'=(K_{n-1}'+1)\sqrt{\frac{K_n+K_{n-1}}{K_{n-1}+1}}$ we have proven that we can find a $K'$ specific to the choice of $n$ such that $\sum\limits_{k=1}^{n-1}\lambda_k<K'\lambda_n$ if we assume we can find a $K_n: \sum\limits_{k=1}^{n-1} \lambda_k^2<K_n\lambda_n^2$.

My Question: As pointed out by Calvin Lin, this doesn't answer the original question as $K'$ varies based on the choice of $n$. I can't seem to use a similar argument for proving the original question and wondered if anyone could shed some light on a better way to approach the problem.

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  • $\begingroup$ Can you clarify what is the $K'$ value that you've found? It seems to me that $ K'_n$ might be unbounded. $\endgroup$
    – Calvin Lin
    Jan 5, 2021 at 6:22
  • $\begingroup$ I just indexed the constant $K'$ with respect to the $n$ chosen. I haven't thought about the boundedness bit yet. $\endgroup$ Jan 6, 2021 at 14:53
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    $\begingroup$ Ah. You said "we have proven the statement for all natural n", so I was confused about how you have proven the (original) statement which had a fixed $K'$. $\endgroup$
    – Calvin Lin
    Jan 6, 2021 at 14:55
  • $\begingroup$ Well I guess I just misunderstood the question as asking if for any $n$, could we find a $K'$ that works - thanks for pointing that out $\endgroup$ Jan 6, 2021 at 15:25

1 Answer 1

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Let us provide a solution without the assumption that the sequence is nondecreasing: we rewrite the condition as

$$ \lambda_n^2 \ge \frac{1}{K} \left( \sum_{j=1}^{n-1} \lambda_j^2\right).$$

Iterating this condition once (that is, using it for $\lambda_{n-1}^2$), we get

$$ \lambda_n^2 \ge \frac{1}{K} \left(1+\frac{1}{K}\right) \left(\sum_{j=1}^{n-1} \lambda_j^2 \right).$$

Reiterating this $r-$times, $r < n,$ we have

$$ \lambda_n^2 \ge \frac{1}{K} \left(1 + \frac{1}{K}\right)^r \left(\sum_{j=1}^{n-r} \lambda_j^2 \right) \ge \frac{1}{K} \left(1 + \frac{1}{K}\right)^r \lambda_{n-r}^2,$$

which is equivalent to

$$ \lambda_{n-r} \le \frac{\sqrt{K}}{(1+ (1/K))^{r/2}} \lambda_n, \forall 1 \le r < n.$$

Therefore,

$$\sum_{r=1}^{n-1} \lambda_{n-r} \le \sqrt{K} \lambda_n \left(\sum_{r=1}^{n-1} \frac{1}{(1+(1/K))^{r/2}}\right).$$

As the last sum is $\le \frac{1}{(1+(1/K))^{1/2}-1},$ we conclude the problem.

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