Show the relative product topology in $l^2$ is weaker than the metric topology formed by $\|\cdot\|_{l^2}$ Show the relative product topology in $\ell^2$ is weaker than the metric topology formed by  $\|\cdot\|_{\ell^2}$
Is the relative product topology $ \rho(x,y) = \sum_{i=0}^\infty x_iy_i $?  What is the other one?  They are not defined in my notes. I need to find a sequence that converges in the weak one, but not in the strong one

We have the set $X = \{(x_n) \in \mathbb{R}^\omega : \sum_{n \in \omega} |x_n|^2 < \infty\}$ of square summable sequences.
The standard analysis way of giving it a topology is to use a norm $\|x\|_2 = \sqrt{\sum_n |x|^2}$ and an induced metric $d_2(x,y) = \|x-y\|_2$.
So a sequence $x_n$ in $X$ then converges to $x$ exactly when $\|x_n -x\|_2 \to 0$ as $n \to \infty$ (as a sequence of reals).
On the other hand $X$ is a subset of $\mathbb{R}^\omega$ (all real-valued sequences) in the product topology and as such we can give it the subspace topology w.r.t. that. This means that it also inherits the convergence property of the product topology: $x_n \to x$ iff for each coordinate $k$ we have that 
$(x_n)_k \to x_k$ as $n \to \infty$.
Note that if $x_n \to x$ in norm it also converges in the product subspace topology : $|(x_n)_k - x_k| \le \|x_n - x\|_2$ for all $k,n$: 
$|(x_n)_k - x_k|^2 \le \sum_p |(x_n)_p - x_p|^2 = \|x_n - x\|^2_2$ and we take square roots on both sides.
The topology is strictly weaker because if we define $x_n$ to be the sequence with all $0$'s  and one $1$ at coordinate $n$, then in every coordinate the sequence of points becomes $0,0, \ldots,1,0, \ldots$ with only one $1$ so that converges coordinatewise to $0$ so $e_n \to 0$ in the product topology, but as $\|e_n - 0\|_2 = 1$ for all $n$ (we always have one term $1$ in the sum), $e_n \not \to 0$ in the norm sense.
In fact with some computations one can show:

$x_n \to x$ in the norm iff $x_n \to x$ in the product topology (i.e. coordinatewise) and $\|x_n\|_2 \to \|x\|_2$ in the reals.

