equivalence of Hilbert spaces the question might be stupid, but I am confused.
Let us consider the following Hilbert space $l_{2}^{W}$, the space of infinite sequences with a scalar product:
$$
<X,Y> = \sum_{i=1}^{\infty}x_{i}y_{i}w_{i},
$$
for some vector $W= (w_{1},w_{2}, \dots)$ with $0 < w_{i} < \infty$ and $\limsup w_{i} < \infty$. I attempt to show that all spaces $l_{2}^{W}$ are all equivalent, i.e. norms generated by the scalar product are equivalent:
for any $W_{1}$ and $W_{2}$ there exist $0 < C_{1} < C_{2}$ such that $C_{1} ||X||_{W_{2}} \leq ||X||_{W_{1}} \leq C_{2} ||X||_{W_{2}} $. 
Should one impose more constraints on the vector of weights then?
 A: It is sufficient to prove $C_1\|X\|_{\ell^2}\leq \|X\|_{W}\leq C_2\|X\|_{\ell^2}$ where $\|X\|_{\ell^2}$ is the common norm in $\ell^2$. Then $\|\cdot\|_{\ell^2}$ and $\|\cdot\|_W$ are equivalent. You can conclude that $\|\cdot\|_{W_1}$ and $\|\cdot\|_{W_2}$ are equivalent since both are equivalent to $\|\cdot\|_{\ell^2}$.
As Aweygan noticed it is crucial to assume $\inf_i w_i>0$. For $W=\left(\frac1i\right)_i$ is
$$
e_n=(0,\ldots,0,\underbrace{1}_{n\text{.th}},0,\ldots)
$$
convergent to $0$ in $\|\cdot\|_W$ while in $\|\cdot\|_{\ell^2}$ it hasn't even a convergent subsequence.
Now if $W=(w_i)_i$ a sequence where $0<\inf_i w_i$ and $\sup w_i<\infty$ you get
$$
\|X\|_W=\sum_{i=1}^\infty x_i^2w_i\geq\sum_{i=1}^\infty x_i^2(\inf_k w_k)=(\inf_k w_k)\sum_{i=1}^\infty x_i^2 =(\inf_k w_k)\|X\|_{\ell^2}
$$
and
$$
\|X\|_W=\sum_{i=1}^\infty x_i^2w_i\leq\sum_{i=1}^\infty x_i^2(\sup_k w_k)=(\sup_k w_k)\sum_{i=1}^\infty x_i^2 =(\sup_k w_k)\|X\|_{\ell^2}
$$
Defining $C_1=\inf_k w_k>0$ and $C_2=\sup_k w_k>0$ yields the claim. 
