# If $\sum_{k=1}^{n-1}\lambda_k^2<K\lambda_n^2$ for some $K$, then $\sum_{k=1}^{n-1}\lambda_k<K'\lambda_n$ for some $K'$

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 $$\exists K'_{n-1}: \sum\limits_{k=1}^{n-1} \lambda_k. Then for the next index, we have $$\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, 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 $$K_{n-1}\lambda_n^2 and so we can add it to both sides without reversing the inequality. Hence we can reduce the above inequality to saying $$\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 if we assume we can find a $$K_n: \sum\limits_{k=1}^{n-1} \lambda_k^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.

• Can you clarify what is the $K'$ value that you've found? It seems to me that $K'_n$ might be unbounded. Jan 5, 2021 at 6:22
• I just indexed the constant $K'$ with respect to the $n$ chosen. I haven't thought about the boundedness bit yet. Jan 6, 2021 at 14:53
• 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 $K'$. Jan 6, 2021 at 14:55
• 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 Jan 6, 2021 at 15:25

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.