Suppose that $\sum_{i=1}^{n}\lambda_{i}=1$, where $\lambda_{i}>0$, and $\sum_{i=1}^{n}b_{i}^{2}=1$, where $b_{i}>0$. Does one have $\sqrt{n}\sum_{i=1}^{n}\lambda_{i}b_{i}\le B$ for some constant $B$ (independent of $n$)? Thanks.
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The question changed, here is an answer to the new question: No. Choose $\lambda_1 = 1, b_1 = 1$. Let $\lambda_k=b_k=0$ for all $k\neq 1$. Then you have $\sum_i \lambda_i b_i = 1$. Hence $\sqrt{n} \sum_i \lambda_i b_i = \sqrt{n}$. So you cannot have a bound independent of $n$. Old answer: Yes. The Cauchy-Schwartz-Bunyakovsky-$\cdots$ inequality gives $|\lambda^T b | \leq \|\lambda\| \|b\| = \|\lambda\| \leq 1$. More simply, since $\sum_i b_i^2 = 1$ we have $|b_i| \leq 1$, and then $\sum_i \lambda_i b_i \leq \sum_i \lambda_i |b_i| \leq \sum_i \lambda_i =1$. It follows that $\sqrt{n} \sum_i \lambda_i b_i \leq \sqrt{n}$, so take $B=\sqrt{n}$. |
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