Convergence Problem in Normed Space Probably easy, but I'm stuck atm: A sequence converges in norm 1 if and only if it converges in norm 2, for all sequences. Are the two norms necessarily equivalent?
 A: Yes, they are equivalent, since they induce the same topology.
To see this, let $C$ be any subset. Then $C$ is closed if and only if for all converging sequences $\{ x_n\} \subset C$, if $x_n \to x$ then $x \in C$. In particular $C$ is closed in the first topology if and only if it is closed in the other.
A: Consider a sequence $(x_n)_{n\ge0}$ that converges in $\Vert\cdot\Vert_1$ to $x$. Then, the new sequence  $(y_n)_{n\ge0}$ defined by $ y_{2n}=x_n$ and $y_{2 n+ 1}= x$ also converges in $\Vert\cdot\Vert_1$ to $ x$. Thus,  $(y_n)_{n\ge0}$ converges to some element $y$ in norm $\Vert\cdot\Vert_2$, so
$$\Vert\cdot\Vert_2-\lim_{ n\to\infty}x_{ n}=\Vert\cdot\Vert_2-\lim_{ n\to\infty}y_{2 n}=\Vert\cdot\Vert_2-\lim_{ n\to\infty}y_{2 n+1}=\Vert\cdot\Vert_2-\lim_{ n\to\infty}x=x$$
Thus, we have proved that if a sequence $(x_n)_{n\ge0}$  converges in $\Vert\cdot\Vert_1$ to $x$, then it converges also in $\Vert\cdot\Vert_2$ to  the same $x$. 
Interchanging the roles of the two norms we see that, conversely,
if a sequence $(x_n)_{n\ge0}$  converges in $\Vert\cdot\Vert_2$ to $x$, then  it converges also in $\Vert\cdot\Vert_1$ to  the same $x$. We conclude that the linear mapping $ x\mapsto x$ and its inverse are  continuous from $(E,\Vert\cdot\Vert_1)$ to
$(E,\Vert\cdot\Vert_2)$ which is equivalent to saying that the two norms are equivalent.
