2

also we have to find which is the greatest constant a and the smallest constant b that satisfies the inequality

$a||x||_{1}\le||x||\le b||x||_{1}$

so far I have only proven that you exist $b=1>0$ such that $||x||\le b||x||_{1}$

Observations: For $x\in \mathbb{R}^p$

$||x||=\displaystyle(\sum_{i=1}^p(x_i^2))^\frac{1}{2}$

$||x||_{1} = \displaystyle\sum_{i=1}^p|x_i|$

any suggestions to continue with the test are welcome. Thanks in advance.

ferphi
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  • Hint: Take the minimum over the unit sphere. (Of $\vert\vert$). – miraunpajaro Jul 16 '21 at 16:10
  • Does this answer your question? Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces – miraunpajaro Jul 16 '21 at 16:12
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    By Cauchy-Scwartz $\sum^p_{j=1}|x_j| = \sum^p_{j=1}|x_j|\cdot 1\leq \Big(\sum^p_{j=1}|x_j|^2\Big)^{1/2}\sqrt{p}$. By the trianggle inequality $|x|=|\sum^p_{j=1}x_j\mathbf{e}j|\leq\sum^p{j=1}|x_j||\mathbf{e}j|=\sum^p{j=1}|x_j|$. Here $\mathbf{e}_j$ is the vector with $0$ components except at the $j$-th component, which has value $1$. – Mittens Jul 16 '21 at 16:17
  • @oliver diaz Thank you very much, to determine which is the smallest and largest constant, I thought to define a set and then prove that each one corresponds to the highest and lowest of the set. Is the idea good? – ferphi Jul 16 '21 at 16:23
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    That $1$ is the best constant for $|x|\leq|x|1$ is trivial, since $x=\mathbf{e}_j$ gives you equality. That $\sqrt{p}$ is the best constant for $|x|_1\leq|x|$ take $x=\sum^p{j=1}\mathbf{e}_j$. In this case, $|x|_1=p$ and $|x|_p=\sqrt{p}$. This give you equality in $|x|_1\leq\sqrt{p}|x|$. – Mittens Jul 16 '21 at 16:28
  • @oliver diaz I understand. Thank you very much friend for your help. – ferphi Jul 16 '21 at 16:33

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