Let $(E, \|\cdot \|)$ be a real Banach space. 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 $\tau$ be the norm topology induced by $[\cdot]$. Let $\sigma(E, E^*)$ be the weak topology induced by $\|\cdot\|$. Certainly, both $\tau$ and $\sigma(E, E^*)$ are weaker (coarser) than the norm topology induced by $\|\cdot\|$.
Is $\sigma(E, E^*)$ weaker than $\tau$, i.e., $\sigma(E, E^*) \subset \tau$?
Let $u_n, u\in E$ such that $[u_n-u] \to 0$. Let $T:E \to \mathbb R$ be linear and continuous w.r.t. the norm topology induced by $\|\cdot\|$. If the answer is affirmative, we will get $|Tu_n - Tu| \to 0$. I think there is no guarantee that $T$ is continuous w.r.t. $\tau$. So I guess the answer is negative. However, I could not come up with a counter example.
The answer is no.
Let $E=C[0,1]$ and let $\|.\|$ be the usual sup norm. Let $[f]=\int_0^{1}|f|$. The set $A=\{f: f(0)<1\}$ is weakly open in $(C[0,1],\|.\|)$ because $f \mapsto f(0)$ is a continuous linear functional on this space. In fact, $A$ is a neighborhood of $0$ (the constant function $0$). If this belongs to $\tau$ then there exists $\epsilon >0$ such that $\int_0^{1}|f| <\epsilon$ implies $f(0) <1$. Let $f(x)=\frac 4 {\epsilon} (\frac {\epsilon} 2-x)$ for $0 \leq x \leq \frac {\epsilon} 2$ and $0$ for $x >\frac {\epsilon} 2$. Then $\int_0^{1}|f| =\frac {\epsilon} 2$ and $f(0)=2>1$.