# Weak convergence under stronger norm

Let $(X,|.|)$ be a Banach space, and $Y$ is a linear subspace of $X$ which is dense in $X$. Now if we have another norm $|.|_Y$ in $Y$ which is comparable to $|.|$ by $$|y|\leq C |y|_Y \ \ \ (y\in Y),$$ with $C$ constant. I know that if a sequence $y_n \to y$ in $|.|_Y$ then $y_n \to y$ in $|.|$.

If $y_n \to y$ in $|.|_Y$ weakly, can we say that $y_n \to y$ in $|.|$ weakly ?

So here's what I found: it suffices to show that $(X,|.|)'\subset (Y,|.|_Y)'$.
For $\phi\in (X,|.|)'$ we have $$|\phi(x)|\leq M|x|, \ \ \forall x\in X$$ then $$|\phi(x)|\leq M|x|, \ \ \forall x\in Y$$ by the inequality we have $$|\phi(x)|\leq MC|x|_Y, \ \ \forall x\in Y$$ which means that $\phi\in(Y,|.|_Y)'$. Now if the property $\phi(x_n)\to\phi(x)$ is true for $\phi \in (Y,|.|_Y)'$ then it is true for $\phi \in (X,|.|)'$ which is inside $(Y,|.|_Y)'$.
Note: The density of $Y$ in $X$ is not necessary.