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Define $$ X := \left \{ (a_n)_{n=0}^\infty :\sum_{n =0}^\infty |a_n| < \infty \right \} $$ with metrics $$ d_{l^1}((a_n)_{n=0}^\infty,(b_n)_{n=0}^\infty) := \sum_{n=0}^\infty |a_n - b_n| $$ $$ d_\infty((a_n)_{n=0}^\infty,(b_n)_{n=0}^\infty) := \sup_{n\geq 0} |a_n-b_n| $$

I want to show that these metrics are not equivalent. And here is my "proof" of that. I am not sure if this is correct.

Let $(x^{(n)})_{n=0}^\infty$ be a sequence in $X$ such that $$ x^{(n)}_k := \begin{cases} \frac 1 n & \text{ if } k \leq n \\ 0 & \text{ if k > n } \end{cases} $$ Define further $\tilde{x} \in X$ with $\tilde{x}_k = 0 $ for all $k \in \mathbb N$. Then $$ \lim_{n \rightarrow \infty} d_\infty (x^{(n)},\tilde{x}) = \lim_{n \rightarrow \infty} |1/n| = 0 $$ such that $(x^{(n)})_{n=0}^\infty$ converges to $\tilde{x}$ with respect to $d_\infty$. Assume now that there exists some (another) sequence $\tilde{x} \in X$ such that $\lim_{n \rightarrow \infty} d_{l^1}(x^{(n)},\tilde{x}) = 0$. Then we have $$ 0=\lim_{n \rightarrow \infty} \sum_{k=0}^n |1/n - \tilde{x}_k| + \sum_{k=n+1}^\infty |\tilde{x}_k| $$

Now we can split the limit and the right term will be $0$ because $\tilde{x} \in X$ and thus by the zero-test the claim follows. But I am not sure how to show that the left term is not $0$.

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Note: Maybe is this approach completly incorrect :) – Epsilon Jan 4 '13 at 22:55
up vote 4 down vote accepted

Let me slightly redefine your $x^{(n)}$ by letting $x^{(n)}_k=\frac1n$ if $k<n$ and $0$ otherwise. Then

$$d_{\ell^1}\left(x^{(n)},\tilde x\right)=\sum_{k\ge 0}\left|x^{(n)}_k\right|=n\cdot\frac1n=1$$

for all $n\in\Bbb N$, so $\left\langle x^{(n)}:n\in\Bbb N\right\rangle\not\to\tilde x$ in the topology generated by $d_{\ell^1}$. Thus, $d_{\ell^1}$ and $d_\infty$ generate different topologies on $X$, since (as you showed) $\left\langle x^{(n)}:n\in\Bbb N\right\rangle\to\tilde x$ in the topology generated by $d_\infty$. It’s not necessary to show that $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ isn’t convergent at all with respect to $d_{\ell^1}$, which is what you seem to be trying to do.

In fact it’s true that $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ isn’t convergent with respect to $d_\infty$. Suppose that $\tilde x\ne\hat x\in X$, so that $\hat x_m\ne 0$ for some $m\in\Bbb N$. Then for all $n\ge m$ we have

$$d_{\ell^1}\left(x^{(n)},\hat x\right)\ge\left|\frac1n-\hat x_m\right|\;,$$

and $$\left|\frac1n-\hat x_m\right|~\underset{n\to\infty}\longrightarrow~\left|\hat x_m\right|>0\;,$$

so the sequence does not converge to any non-zero sequence, either.

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I dont get why $d_{l^1}(x^{(n)},\tilde{x})$ is what you say. Where are the terms $\tilde{x}_k$ left ? – Epsilon Jan 4 '13 at 23:20
@André: I’m using your definition of $\tilde x$: $\tilde x_k=0$ for all $k\in\Bbb N$. – Brian M. Scott Jan 4 '13 at 23:25
Yeah. Thanks. Now I get it. Thank you for your help :) – Epsilon Jan 4 '13 at 23:26
@André: You’re welcome. – Brian M. Scott Jan 4 '13 at 23:27

You are on the right track until near the end. Note that $|x^{(n)}_k-\tilde{x}_k|\leq d_{\ell^1}(x^{(n)},\tilde{x})$ for each $k$, so $x^{(n)}_k\to \tilde x_k$ for each $k$, which implies $\tilde x_k=0$ for each $k$. So what is $d_{\ell^1}(x^{(n)},\tilde x)$?

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(And as Brian points out, it is enough to show that $x^{(n)}$ does not converge to the same $\tilde x$. Your approach, which I picked up here, shows that in fact the sequence doesn't converge at all in the $d_{\ell^1}$ distance.) – Jonas Meyer Jan 4 '13 at 23:04

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