Let $H$ be a Hilbert space, let $\mathcal{L}(H)$ denote the bounded linear functions from $H$ to itself and consider the two topologies on $\mathcal{L}(H)$:
(1) $\tau_1$ is the topology where a base is given by sets obtained by translating finite intersections of sets of the form $$ \left\lbrace T \in \mathcal{L}(H): \left| \sum_{i=1}^\infty (Tx_i| y_i) \right|<1 \right\rbrace. $$ Here, $(x_i)_{i \in \mathbb{N}}$ and $(y_i)_{i \in \mathbb{N}}$ are sequences in $H$ with $\sum_{i=1}^\infty \|x_i\|^2,\ \sum_{i=1}^\infty \|y_i\|^2 < \infty$. One can check that this is indeed a basis for a topology on $\mathcal{L}(H)$.
(2) $\tau_2$ is the topology where a base is given by sets obtained by translating finite intersections of sets of the form
$$
\left\lbrace T \in \mathcal{L}(H): \sum_{i=1}^\infty |(Tx_i| y_i)| <1 \right\rbrace.
$$
Again, $(x_i)_{i \in \mathbb{N}}$ and $(y_i)_{i \in \mathbb{N}}$ are sequences in $H$ with $\sum_{i=1}^\infty \|x_i\|^2,\ \sum_{i=1}^\infty \|y_i\|^2 < \infty$. Again, one can check that this is indeed a base (that is, closed under finite intersections).
Question: Are $\tau_1$ and $\tau_2$ the same topology on $\mathcal{L}$? If not, can one give an example demonstrating so?
Attempt: It seems to be clear that $\tau_1$-open sets are $\tau_2$-open sets because of the inequality: $$ \left|\sum_{i} (Tx_i|y_i) \right| \leq \sum_{i} |(Tx_i|y_i)|. $$ However, I cannot see any reverse inequality which makes me thing the topologies are distinct.
Thanks for reading, I'm new to Hilbert spaces and very confused.