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Define for all $x = (x_1,x_2, \dots) \in \ell^1$ $$ \| x \|_* = 2 \left| \sum_{k=1}^{\infty} x_k \right| + \sum_{k=2}^{\infty} \left( 1+\frac{1}{k} \right) |x_k| \ . $$

Prove that $(\ell^1, \| \cdot \|_*)$ is a Banach space.

I proved that $\| \cdot \|_*$ is actually a norm on $\ell^1$.

Now let $(x_n)$, where $x_n = (x_{n,1}, x_{n,2}, \dots)$, be any Cauchy sequence in $(\ell^1, \| \cdot \|_*)$. Then $\| x_n - x_m \|_* \to 0 \ $ as $ \ n,m \to \infty$ implies $\sum_{k=2}^{\infty}|x_{n,k} - x_{m,k}| \to 0$. In particular that means that $(x_{n,k})_{n=1}^{\infty}$, where $k \geq 2$, is a Cauchy sequence in $\mathbb{C}$ and hence $x_{n,k} \to x_{*,k} \ $ as $n \to \infty \ $ for any $k \geq 2$.

Is it true, that $\sum_{k=1}^{\infty}|x_{n,k} - x_{m,k}| \to 0$ ?

If it is true, then we can try to set $y = (x_{*,1}, x_{*,2}, \dots)$. Now we need to show, that $y \in \ell^1$ and that $\| x_n - y \|_* \to 0$. To do so, we need to swap the order of limits in the expression $$\lim_{m \to \infty} \sum_{k=1}^{\infty}|x_{n,k} - x_{m,k}| \ .$$

Can we swap the order of limits? Why?

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    $\begingroup$ It's probably easier to prove it by showing that $\lVert\cdot\rVert_{\ast}$ and $\lVert\cdot\rVert_{\ell^1}$ are equivalent. $\endgroup$ – Daniel Fischer Jan 22 '17 at 13:39
  • $\begingroup$ @DanielFischer This question comes after the one above :) Ok, thanks for the idea, I can try it. $\endgroup$ – nightked Jan 22 '17 at 13:42
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    $\begingroup$ Hint for the use of Latex: \ell looks better than l. Compare $\ell^1$ vs $l^1$. $\endgroup$ – Tim B. Jan 22 '17 at 15:49
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Is it true, that $\sum_{k=1}^{\infty}|x_{n,k} - x_{m,k}| \to 0$ ?

You already know that

$$\sum_{k = 2}^{\infty} \lvert x_{n,k} - x_{m,k}\rvert \xrightarrow{m,n\to\infty} 0,$$

so what remains is the question whether $\lvert x_{n,1} - x_{m,1}\rvert \to 0$. Now

\begin{align} \lvert x_{n,1} - x_{m,1}\rvert &=\Biggl\lvert\sum_{k = 1}^{\infty} (x_{n,k} - x_{m,k}) - \sum_{k = 2}^{\infty} (x_{n,k} - x_{m,k})\Biggr\rvert \\ &\leqslant \Biggl\lvert\sum_{k = 1}^{\infty} (x_{n,k} - x_{m,k})\Biggr\rvert + \sum_{k = 2}^{\infty} \lvert x_{n,k} - x_{m,k}\rvert \\ &\leqslant \lVert x_n - x_m\rVert_{\ast}. \end{align}

If it is true, then we can try to set $y = (x_{*,1}, x_{*,2}, \dots)$. Now we need to show, that $y \in l^1$ and that $\| x_n - y \|_* \to 0$. To do so, we need to swap the order of limits in the expression $$\lim_{m \to \infty} \sum_{k=1}^{\infty}|x_{n,k} - x_{m,k}| \ .$$

Can we swap the order of limits? Why?

We can. If you can use some integration theory, you can justify everything with Fatou's lemma and the dominated convergence theorem.

But the more usual approach (since typically beginners in functional analysis aren't yet experienced in integration theory) to proving the assertion is to restrict to finite sums first, and then take the limit as the number of terms considered tends to $\infty$.

Since $(x_n)$ is a Cauchy sequence, it is bounded, say $\lVert x_n\rVert_{\ast} \leqslant C$ for all $n$. Then for every $N\in \mathbb{N}$ with $N \geqslant 2$ and all $m$, we have

$$\sum_{k = 2}^N \lvert x_{m,k}\rvert \leqslant C.$$

Letting $m\to \infty$ in that yields

$$\sum_{k = 2}^N \lvert y_k\rvert \leqslant C.$$

Now letting $N\to \infty$ we obtain

$$\sum_{k = 2}^\infty \lvert y_k\rvert \leqslant C,$$

so $y \in l^1$. Having established $y\in l^1$, we can now show $x_n \to y$. Fix $\varepsilon > 0$ and choose an $N\in \mathbb{N}$ such that $\lVert x_n - x_m\rVert_{\ast} \leqslant \varepsilon/4$ for all $m,n \geqslant N$. Then letting $m \to \infty$ in

$$\sum_{k = 2}^K \biggl(1 + \frac{1}{k}\biggr)\lvert x_{n,k} - x_{m,k}\rvert \leqslant \frac{\varepsilon}{4}$$

and afterwards letting $K \to \infty$ gives

$$\sum_{k = 2}^{\infty} \biggl(1+\frac{1}{k}\biggr)\lvert x_{n,k} - y_k\rvert \leqslant \frac{\varepsilon}{4}$$

for all $n \geqslant N$.

Then

$$2\Biggl\lvert\sum_{k = 1}^{\infty} (x_{n,k} - y_k)\Biggr\rvert \leqslant 2 \lvert x_{n,1} - y_1\rvert + 2\sum_{k = 2}^{\infty} \lvert x_{n,k} - y_k\rvert \leqslant 2\lvert x_{n,1} - y_1\rvert + \frac{\varepsilon}{2}$$

together with $x_{n,1} \to y_1$ gives $\lVert x_n - y\rVert_{\ast} \leqslant \varepsilon$ for all $n \geqslant N_1$, where $N_1 \geqslant N$ is chosen so that $\lvert x_{n,1} - y_1\rvert \leqslant \varepsilon/8$ for all $n \geqslant N_1$.

The argument is a wee bit cumbersome ;-)

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  • $\begingroup$ Thank you! And you are right about the equivalence: it is easier to show that $\| \cdot \|_*$ and $\| \cdot \|_{l^1}$ are equivalent $\endgroup$ – nightked Jan 22 '17 at 14:41

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