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I learnt how to show the below inequality by C-S inequality: k is from $0$ to $\infty$

If $\sum a_{k}^{2}9^{k}\le 5$ then $\sum |a_{k}|2^{k}\le 3$.

next,I tried to show that 3 is the best possible constant in the last inequality.

Could you please help me to show how 3 is the best possible constant.

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  • $\begingroup$ Can you see what happens when the inequality on the left is an equality? $\endgroup$ – Ian Coley Feb 14 '14 at 8:03
  • $\begingroup$ What is the range of $k$? $\endgroup$ – 5xum Feb 14 '14 at 8:05
  • $\begingroup$ Also, would you explain how you showed your claim? $\endgroup$ – Ian Coley Feb 14 '14 at 8:06
  • $\begingroup$ $k$ must range from $0$ to $\infty$, for this claim to be true. Presumably you used Cauchy-Schwarz for your proof. Do you know under what circumstance the Cauchy-Schwarz inequality is in fact an equality? $\endgroup$ – Greg Martin Feb 14 '14 at 8:49
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Lemma. If $\omega>1$, and $\{b_n\}_{n\in\mathbb N}\subset \mathbb R$, then $$ (\omega^2-1)\Big(\sum_{k=1}^\infty \lvert b_k\rvert\Big)^2\le \sum_{k=1}^\infty \omega^{2k}b^2_k, $$ and $\omega^2-1$ is the optimal constant.

Proof. Using Cauchy-Schwarz (which is a special case for Hölder's inequality for $p=q=2$) we obtain \begin{align} \Big(\sum_{k=1}^N \lvert b_k\rvert\Big)^2&= \Big(\sum_{k=1}^N \omega^{-k}\big(\omega^{k}\lvert b_k\rvert\big)\Big)^2 \le \Big(\sum_{k=1}^N\omega^{-2k}\Big)\Big(\sum_{k=1}^N\omega^{2k}b_k^2\Big) \\&=\frac{\omega^{-2}-\omega^{-2N}}{1-\omega^{-2}}\Big(\sum_{k=1}^N\omega^{2k}b_k^2\Big) \le \frac{\omega^{-2}}{1-\omega^{-2}}\Big(\sum_{k=1}^N\omega^{2k}b_k^2\Big) \\&=\frac{1}{\omega^{2}-1}\Big(\sum_{k=1}^N\omega^{2k}b_k^2\Big)\le \frac{1}{\omega^{2}-1}\Big(\sum_{k=1}^\infty\omega^{2k}b_k^2\Big). \end{align} Thus $$ \Big(\sum_{k=1}^N\omega^{-2k}\Big)\Big(\sum_{k=1}^\infty\omega^{2k}b_k^2\Big) \le (\omega^{2}-1)\Big(\sum_{k=1}^\infty\omega^{2k}b_k^2\Big). \tag*{$\Box$} $$

In our case, $\omega=3/2$ and $b_k=2^ka_k$: $$ 5\ge\sum_{k=1}^\infty a^2_k9^k=\sum_{k=1}^\infty (2^ka_k)^2(3^k/2^k)^2\ge \Big((3/2)^2-1\Big) \Big(\sum_{k=1}^\infty 2^k\lvert a_k\rvert\Big)^2=\frac{5}{4} \Big(\sum_{k=1}^\infty 2^k\lvert a_k\rvert\Big)^2. $$ and hence $$ \sum_{k=1}2^k\lvert a_k\rvert\ \le 2, $$ and $2$ is the best estimate.

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  • $\begingroup$ thank you. is there a book you can recommend that i can learn these kind of inequalities. $\endgroup$ – 104078 Feb 14 '14 at 12:39
  • $\begingroup$ This inequality is definitely not a standard one. I came up to it up, using Lagrange multipliers. The inverse direction is more common, as it is related to Cauchy-Schwarz. The only books I would recommend are those related to Math Contests. $\endgroup$ – Yiorgos S. Smyrlis Feb 14 '14 at 12:47
  • $\begingroup$ our prof wants us to use the condition when hölder inequality holds for equality. i ve tried to solve it. but the $a_{k}$ I derive does not work. could you please help with this approach. i have tried so hard to figure it out. frustated... but then would not it contradict with your result $\le 2$. i am a bit confused $\endgroup$ – 104078 Feb 18 '14 at 18:42
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Let $b_k=3^ka_k$, then Cauchy-Schwarz says $$ \begin{align} \sum_{k=1}^na_k2^k &=\sum_{k=1}^nb_k\left(\frac23\right)^k\\ &\le\left(\sum_{k=1}^nb_k^2\right)^{1/2}\left(\sum_{k=1}^n\left(\frac49\right)^k\right)^{1/2}\\ &=\left(\sum_{k=1}^na_k^29^k\right)^{1/2}\frac2{\sqrt5}\sqrt{1-\left(\frac49\right)^n}\\ &=2\sqrt{1-\left(\frac49\right)^n}\tag{5} \end{align} $$ Note that equality holds in $(5)$ when $b_k=\lambda\left(\frac23\right)^k$ for some lambda; that is, when $a_k=\lambda\left(\frac29\right)^k$. This gives us $$ a_k=\frac{\frac52}{\sqrt{1-\left(\frac49\right)^n}}\left(\frac29\right)^k\tag{6} $$ which shows that $(5)$ is sharp.

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