Is $||x||_1$ topologically equivalent to the standard norm in $\ell^2$?

Question

For $x \in \ell^2$ we have the norm $||x||_1 = (\sum_{n=0}^{\infty} |x_n|^2)^{1/2}$ + $\sum_{n=0}^{\infty} (\frac{3}{4})^{n}|x_n|$ is it topologically equivalent to the standard norm $||x|| = (\sum_{n=0}^{\infty} |x_n|^2)^{1/2}$

Some thoughts: I see that the topology of $||x||_1$ is contained in the one of $||x||$ as $||x|| < ||x||_1$ for $x \in \ell^2$. I suspect that is does not happen the other way around but I cannot find any sequence that proves so, nor any positive constant so that $a||x||_1 < ||x||$ so that they would be equivalent.

Answer 1

Hint: let $a_n=\left(\frac{3}{4}\right)^n$. Then, from the Cauchy-Schwarz inequality, we have that $$(a|x) \leqslant \lVert a\rVert \lVert x\rVert$$

Answer 2

They are equivalent. Note that $\sum (\frac 3 4)^{n} |x_n| \leq \sum (\frac 3 4)^{n}=\frac {3/4} {1/4}=3$ whenever $\|x\| \leq 1$. Hence $\|x\|_1 \leq 4$ whenever $\|x\| \leq 4$. This implies that $\|x\|_1 \leq 4\|x\|$ for all $x$. [Just apply the previous inequality to $y=\frac x {\|x\|}$].