# Regarding “stronger” norms

Let $X$ be a normed linear space. Show that a norm $\|\cdot\|_{1}$ is stronger than a norm $\|\cdot\|_{2}$ if and only if for any sequence $\{x_{n}\} \subset X$, $\|x_{n}\|_{1} \to 0$ always implies $\|x_{n}\|_{2} \to 0$.

My work:

$\Longrightarrow$ Suppose $\|\cdot\|_{1}$ is stronger than $\|\cdot\|_{2}$. This means that there exists some $M > 0$ such that $\|x\|_{2} \leqslant M\|x\|_{1}$ for all $x \in {X}$. Let $\{x_{n}\} \subset X$ be any sequence such that $\|x_{n}\|_{1} \to 0$. It follows that $M\|x_{n}\|_{1} \to 0$ which necessarily implies $\|x_{n}\|_{2} \to 0$. (Can someone verify this?)

$\Longleftarrow$ Suppose that $\|x_{n}\|_{1} \to 0$ always implies $\|x_{n}\|_{2} \to 0$. This implies that $x_{n}$ is a Cauchy sequence under both norms. Thus, for some $\epsilon_{1}, \epsilon_{2} > 0$, there exists $N_{1}, N_{2} > \mathbb{N}$ such that $\|x_{n} - x_{m}\|_{1} < \epsilon_{1}$ and $\|x_{n} - x_{m}\|_{2} < \epsilon_{2}$ for all $n,m > N_{1},N_{2}$, respectively. ....

My question is for the reverse direction, how do I connect this idea of a "stronger" norm knowing only that both norms converge to $0$.

## 1 Answer

For $\Rightarrow$, your argument is correct.

For $\Leftarrow$, you do not need Cauchy sequences (and this may not be helpful if the considered space is not complete for the involved norms). Instead, you can argue by contradiction and assume that $\lVert \cdot\rVert_1$ is not stronger than $\lVert \cdot\rVert_2$. This means that for each positive integer $n$, there exists $y_n\in X$ such that $\lVert y_n\rVert_2\gt n\lVert y_n\rVert_1$. Now we define $$x_n:=\frac 1{n\lVert y_n\rVert_1}y_n$$ to get the wanted contradiction.

Indeed, on one hand, we have $$\lVert x_n\rVert_1=\left\lVert \frac 1{n\lVert y_n\rVert_1}y_n\right\rVert_1=\frac 1{n\lVert y_n\rVert_1}\left\lVert y_n\right\rVert_1=\frac 1n$$ hence $x_n\to 0$ for the $\left\lVert\cdot\right\rVert_1$ norm.

On the other hand, since $\lVert y_n\rVert_2\gt n\lVert y_n\rVert_1$, we have $$\lVert x_n\rVert_2=\left\lVert \frac 1{n\lVert y_n\rVert_1}y_n\right\rVert_2=\frac 1{n\lVert y_n\rVert_1}\left\lVert y_n\right\rVert_2\gt 1,$$ hence we do not have $x_n\to 0$ for the $\left\lVert\cdot\right\rVert_2$ norm.

We thus constructed a sequence $(x_n)_{n\geqslant 1}$ such that $x_n\to 0$ for the $\left\lVert\cdot\right\rVert_1$ norm but not the for the $\left\lVert\cdot\right\rVert_2$ norm, and we proved that this is always possible if the norms $\left\lVert\cdot\right\rVert_1$ and $\left\lVert\cdot\right\rVert_2$ are not equivalent.

• Are the index $n$ and the constant multiple $n$ in $\|y_{n}\|_{2} > n \|y_{n}\|_{1}$ related? – clocktower Mar 13 '16 at 15:03
• Yes, it is the same. – Davide Giraudo Mar 13 '16 at 16:36
• I'm having trouble computing the contradiction. I've offered 50 bounty points to see this. – clocktower Mar 23 '16 at 12:53
• @clocktower I have added more details. – Davide Giraudo Mar 23 '16 at 13:40
• @clocktower, for $\Leftarrow$, you can also note that the identity operator from $(X,||\cdot||_1)$ to $(X,||\cdot||_2)$ is continuous, therefore, bounded. Hence the claim. In essence, the argument above tells that continuity implies boundedness. – zhoraster Mar 23 '16 at 15:30