equivalent norms in $\ell^2$?

Question

For $x \in \ell^2$ we have the norm $||x||_2 = (\sum_{n=1}^{\infty}|x_{2n-1}|^2)^{1/2} + (\sum_{n=1}^{\infty}|x_{2n}|^2)^{1/2} $ is it equivalent to the standard norm which is $||x|| = (\sum_{n=0}^{\infty}|x_{n}|^2)^{1/2}$?

Answer 1

First, note that $||x||_2^2 = ||x||^2 + 2\left( \sum_{n,m=1}^{\infty}|x_{2n-1}|^2|x_{2m}|^2 \right)^{1/2} \geq ||x||^2$, and hence $||x||_2 \geq ||x||$.

Secondly, note $||x||_2 = \left(\sum_{n=1}^{\infty}|x|_{2n-1}^2 \right)^{1/2} + \left(\sum_{n=1}^{\infty}|x|_{2n}^2 \right)^{1/2} \leq 2\left(\sum_{n=1}^{\infty}|x|_{n}^2 \right)^{1/2} = 2||x||$.

Therefore, the norms are topologically equivalent.

Answer 2

$$u=\sqrt{\sum_{\text{odd }n}|x_n|^2}\quad v=\sqrt{\sum_{\text{even }n}|x_n|^2}$$

$$\sqrt{u^2+v^2}\leq\sqrt{u^2+2uv+v^2}=u+v$$

$$u+v=(1,1)\cdot(u,v)\leq\lVert(1,1)\rVert\lVert(u,v)\rVert=\sqrt2\sqrt{u^2+v^2}$$