Let $(E, \|\cdot \|)$ be a real Banach space. Let $\tau_1$ be the weak topology induced by $\|\cdot\|$. Let $[\cdot]$ be another norm on $E$ that is weaker than $\|\cdot\|$, i.e., there is $C>0$ such that $[u] \le C \|u\|$ for all $u\in E$. Let $\tau_2$ be the weak topology induced by $[\cdot]$.
In solving exercise 6.13.1 in Brezis' Functional Analysis, I would like to verify that $\tau_2 \subset \tau_1$. Could you please have a check on my below attempt?
Let $E^*_{\|\cdot \|}$ be the dual of $E$ w.r.t. $\|\cdot \|$. Let $E^*_{[\cdot ]}$ be the dual of $E$ w.r.t. $[\cdot ]$. Then $\tau_1$ is the initial topology on $E$ induced $E^*_{\|\cdot \|}$. Similarly, $\tau_2$ is the initial topology on $E$ induced $E^*_{[\cdot]}$. Then it suffices to prove that $E^*_{[\cdot]} \subset E^*_{\|\cdot \|}$. Let $T\in E^*_{[\cdot]}$. We will prove that $T \in E^*_{\|\cdot \|}$. Clearly, $T$ is linear. It remains to prove that $T$ is continuous w.r.t. $\|\cdot\|$. Let $x, x_n \in E$ such that $\|x_n-x\| \to 0$. Because $[\cdot]$ is weaker than $\|\cdot\|$, we get $[x_n-x] \to 0$. Because $T \in E^*_{[\cdot]}$, we get $Tx_n \to Tx$. This completes the proof.