Let $(x^{n})$ be a sequence in $\ell^1$ and let $x\in \ell^1$ such that $x^{(n)}_j\to x_j$ for all $j\in \mathbb{N}$. Then it is not necessary that $x^{(n)}\to x$ in $\ell^1$. For example, $(e_n)$ does not converge to $0$ in $\ell^1$, but $e_n(j)\to 0$ for all $j\in \mathbb{N}$. I have to show that if $x^{(n)},x\in S_{\ell^1}$ (the unit sphere of $\ell^1$), then componentwise converge implies convergence in $\ell^1$.
It can be shown that componentwise convergence implies weak convergence $x^{(n)}\xrightarrow{weak}x$ and then by Shur's theorem $x^{(n)}\to x$. But I don't want to apply weak convergence. Can you please help me in solving the problem using elementary properties only.
Best Answer
Fix $\varepsilon>0$. Pick $n_0$ such that $$\displaystyle\sum_{n=n_0+1}^{\infty}|x^n|<\varepsilon.$$ Using the pointwise convergence at the first $n_0$ coordinates, fix $j_0$ such that, for any $j>j_0$, $$ |x^n-x_j^n|<\frac{\varepsilon}{n_0},\ \forall 1\leq n \leq n_0. $$ Fix $j>j_0$ and let us verify that $\|x-x_j\|<4\varepsilon$.
We have that $$ \sum_{n=1}^{n_0}|x^n-x^n_j|<\varepsilon, $$ but then, since $\|x\|=1$ and $\displaystyle\sum_{n=n_0+1}^{\infty}|x^n|<\varepsilon$, it follows that $$ \sum_{n=1}^{n_0}|x^n|>1-\varepsilon, $$ and consequently, using that $\|x_j\|=1$, $$ \sum_{n=1}^{n_0}|x^n_j| \geq \sum_{n=1}^{n_0}|x^n|-\sum_{n=1}^{n_0}|x^n-x^n_j|>1-2\varepsilon\ \Rightarrow\ \sum_{n=n_0+1}^{\infty}|x^n_j|<2\varepsilon. $$ Finally, joining all the facts above we get that $$ \|x-x_j\| = \sum_{n=1}^{n_0}|x^n-x^n_j| + \sum_{n=n_0+1}^{\infty}|x^n-x^n_j| < \varepsilon + \sum_{n=n_0+1}^{\infty}|x^n| + \sum_{n=n_0+1}^{\infty}|x^n_j| < 4\varepsilon. $$