I have been working trying to solve a problem. It is about equivalence of norms. I am very close to finishing the problem. I have obtained a lower bound and I still need to obtain the upper bound. I have written the steps I am following in case someone could help me. First, here is the problem:
Check if for $x \in l^{2}$ norm
$ \|x\|_{1}=\left(\sum_{n=1}^{\infty}\left|x_{n}\right|^{2}\right)^{\frac{1}{2}}+\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n}\left|x_{n}\right| $
is equivalent to the standard norm in $l^{2}$.
Okey, and here it is my solution:
- For the lower bound, we have
$$ \begin{aligned} \|x\|_1 &= \left(\sum_{n=1}^{\infty}\left|x_{n}\right|^{2}\right)^{\frac{1}{2}}+\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n}\left|x_{n}\right| \\ &= \|x\| + \sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n}\left|x_{n}\right| \\ &\geq \|x\| \end{aligned} $$
Because $$ \sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n}\left|x_{n}\right| \geq 0 $$.
So for the lower bound we have $1 \|x\| \leq \|x\|_1 $.
- For the upper bound, we have
$$ \begin{aligned} \|x\|_1^2 &= \left( \left( \sum_{n=1}^{\infty} |x_n|^2 \right)^{1/2} + \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n |x_n|\right)^2\\ &= \sum_{n=1}^{\infty} |x_n|^2 + \left( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n |x_n|\right)^2 + 2 \left( \left( \sum_{n=1}^{\infty} |x_n|^2 \right)^{1/2} \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n |x_n|\right) \\ &=\|x\|^2 + \left( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n |x_n|\right)^2 + 2 \|x\|\sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n |x_n| \\ &\leq \|x\|^2 + \left( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n |x_n|\right)^2 + 2 \|x\| \\ &\leq \|x\|^2 + \left( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^{2n} \right) \left( \sum_{n=1}^{\infty} |x_n|^2 \right)+ 2 \|x\| \\ &\leq \|x\|^2 + \left( \sum_{n=1}^{\infty} \left( \frac{4}{9} \right)^{n} \right) \|x\|^2 + 2 \|x\| \\ &= \|x\|^2 + \left( \frac{\frac{4}{9}}{1-\frac{4}{9}} \right) \|x\|^2 + 2 \|x\| \\ &= \|x\|^2 + \frac{4}{5} \|x\|^2 + 2 \|x\| \\ &= \frac{9}{5} \|x\|^2 + 2 \|x\|\\ &\leq \frac{9}{5} \|x\|^2 + 2 \end{aligned} $$
I do not know how to continue. In fact, there is one step where I have said that $2 \|x\| \leq 2$ and I do not know if it is correct. I have also used Cauchy-Schwarz inequality.
What do you recommend me? I appreciate your help.
In the upper bound, your first inequality is incorrect; you estimated $$\sum_{n=1}^\infty \left(\frac{2}{3}\right)^n |x_n| \leq 1$$ and also, as you mentioned yourself, $2||x|| \leq 2$ does not hold.
What you can do instead is to simply use Cauchy-Schwarz on the second term of your new norm: $$\sum_{n=1}^\infty \left(\frac{2}{3}\right)^n |x_n| \leq \sqrt{\sum_{n=1}^\infty \left(\frac{2}{3}\right)^{2n} }\cdot \sqrt{\sum_{n=1}^\infty |x_n|^2} = C||x||,$$ where $C = \sqrt{\sum_{n=1}^\infty \left(\frac{2}{3}\right)^{2n} }$ is finite. Hence $$||x||\leq ||x||_1 \leq ||x|| + C||x|| = (C+1)||x||.$$