# How convergence relates to equivalence of norms

Let $X$ be a normed linear space with two norms $||\cdot||_1$ and $||\cdot||_2$. Prove or disprove that this statements are equivalent:

• $||\cdot||_1$ and $||\cdot||_2$ are equivalent,
• $\{x_n\}$ converges in $||\cdot||_1$ iff $\{x_n\}$ converges in $||\cdot||_2\;\;$ $\forall \{x_n\}\in X$

The first implies the second trivially because of equivalence of topologies. But it implies something more: sequence converges to the same element in both norms. But I don't have such condition.

Is this a necessary condition to converge to the same in both norms, or this is just a consequent of "if and only if" in the statement of convergence, or there is a example disproving the equivalence?

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Suppose $\|\cdot\|_1$ and $\|\cdot\|_2$ are inequivalent, so one is unbounded with respect to the other; WLOG $\|\cdot\|_2$ is unbounded with respect to $\|\cdot\|_1$. That means you can find a sequence $x_n$ with $\|x_n\|_1 \leq 1$ for all $n$ but $\|x_n\|_2 \geq n$. Let $y_n = \frac{x_n}{\sqrt{n}}$. Then $y_n \to 0$ in $\|\cdot\|_1$, but $\|y_n\|_2 \to \infty$. So, for inequivalent norms you can always find a sequence which converges to 0 in one norm, but goes to infinity (hence doesn't converge to anything) in the other.

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Norms are uniquely defined by their convex unit balls, and they are equivalent if you can fit the unit ball of one norm inside that of the other norm, and vice versa. Also since you can freely translate things in a vector space, there is no loss of generality in assuming that $x_n \rightarrow 0$.
The sequence $x_n$ converges to zero for norm A if every one of these A-balls contains a tail of the sequence (and likewise for norm B). Since the balls are nested A inside B inside A, .., if it converges in one norm it has converge in the other.