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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-Schwarz-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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copper.hat, thank you for your answer, but in the question there is actually a coeffcient $\sqrt{n}$ in the front. – lovm Nov 15 '12 at 19:18
It is actually $\sqrt{n}|\lambda^T b |$ needed to be estimated. Thanks. – lovm Nov 15 '12 at 19:22
Just multiply both sides by the same positive number ($\sqrt{n}$)!!! – copper.hat Nov 15 '12 at 19:24
copper.hat, $B$ should be a constant. Sorry that I didn't make it clear. – lovm Nov 15 '12 at 19:24
It is a constant. Did you mean independent of $n$? If you choose $\lambda_1 = 1, b_1=1$ and all the rest zero, you see that $\sum_i \lambda_i b_i = 1$, so you have $\sqrt{n} \sum_i \lambda_i b_i = \sqrt{n}$. You cannot improve on this bound! – copper.hat Nov 15 '12 at 19:26

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