$x^{(n)}$ converges to $x$ in $\ell^1$ if and only if $\lim x^{(n)}_j=x_j$ for all $j\in\mathbb N$

Let $x^{(n)}$ and $x$ belong to the unit sphere of $\ell^1$. I want to prove that

$x^{(n)}\to x$ if and only if $\lim_{n\to \infty} x^{(n)}_j=x_j$ for all $j\in \mathbb N$.

It's clear that $x^{(n)}\to x$ implies $x^{(n)}_j\to x_j$ since $$|x^{(n)}_j-x_j|\leq \sum_{j=1}^\infty|x^{(n)}_j-x_j|\to 0.$$

However the other side is not easy for me. If $x^{(n)}_j\to x_j$ for all $j$ and $\epsilon>0$ is given then there exists $N_j$ such that if $n\geq N_j$, then $|x^{(n)}_j-x_j|< \epsilon$.

But this doesn't help. I think maybe we should find a $N$ such that for $n\geq N$ we have $|x^{(n)}_j-x_j| <\frac{\epsilon}{2^j}$ for all $j$.

And I don't know how $\sum_j |x^{(n)}_j|=\sum_j|x_j|=1$ could help.

• It is assumed in advance that $x^{(n)}$ converges? If not, you should be looking for a counterexample. – GEdgar Aug 9 '17 at 17:30
• @GEdgar No! the problem is $(x^{(n)}),x$ are in the unit sphere of $\ell^1$ and $\lim_{n}x^{(n)}_j=x_j$ for all $j$. Now how can we prove $(x^{(n)})\to x$? – mac Aug 9 '17 at 17:35
• "Unit sphere" means $\|x\| \le 1$ or $\|x\| = 1$ ? – GEdgar Aug 9 '17 at 17:38
• @GEdgar Means $\|x\|=1$ – mac Aug 9 '17 at 17:40

Hint: Choose $J$ such that $\sum_{j=1}^{J}|x_j|> 1-\epsilon.$ Because $J$ is now fixed, we'll have

$$\sum_{j=1}^{J}|x_{nj}-x_j| < \epsilon$$

for large $n.$ For such $n,$ we then have

$$\sum_{j=1}^{J}|x_{nj}| \ge \sum_{j=1}^{J}|x_j|-\sum_{j=1}^{J}|x_{nj}-x_j| > 1-\epsilon - \epsilon.$$

Thus we will have both $\sum_{j>J}|x_{j}|, \sum_{j>J}|x_{nj}|$ small for large $n.$

Here is a more general statement:

Proposition. (Scheffé’s lemma) Let $(X, \mathcal{F}, \mu)$ be a measure space. Suppose that $f_n, f$ are $L^1$-functions on $X$ such that $f_n \to f$ pointwise. Then the followings are equivalent:

1. $f_n \to f$ in $L^1$.
2. $\int|f_n| \, d\mu \to \int |f| \, d\mu$

In our case, we can set $(X, \mathcal{F}, \mu) = (\mathbb{N}, \mathcal{P}(\mathbb{N}), \text{counting measure})$ so that $\ell^1(\mathbb{N})$ is the $L^1$-space on $\mathbb{N}$. Since the second condition is satisfied for our sequence, the claim follows. So it remains to prove the above proposition.

• Proof. The direction $1 \Rightarrow 2$ is straightforward, so we prove the converse $2 \Rightarrow 1$. By the Fatou's lemma,

$$\liminf_{n\to\infty} \int_X (|f_n| + |f| - |f - f_n|) \, d\mu \geq \int_X \liminf_{n\to\infty} (|f_n| + |f| - |f - f_n|) \, d\mu.$$

By the assumption that $\int_X |f_n| \, d\mu \to \int_X |f| \, d\mu$, the LHS is

$$\liminf_{n\to\infty} \int_X (|f_n| + |f| - |f - f_n|) \, d\mu = 2\int_X |f| \, d\mu - \limsup_{n\to\infty} \int_X |f - f_n| \, d\mu.$$

On the other hand, the RHS is

$$\int_X \liminf_{n\to\infty} (|f_n| + |f| - |f - f_n|) \, d\mu = \int_X 2|f| \, d\mu.$$

Plugging back and simplifying, we get

$$\limsup_{n\to\infty} \int_X |f - f_n| \, d\mu \leq 0$$

and therefore $f_n \to f$ in $L^1$. ////

Remark. Notice that this proof is almost identical to the proof of the dominated convergence theorem, where the role of dominating function being replaced by the convergence assumption $\int_X |f_n| \, d\mu \to \int_X |f| \, d\mu$.

• Thanks. Where can I read about Scheffé’s lemma? could you please introduce a book? – mac Aug 9 '17 at 18:18
• @mac, As far as I know, it is not widely spread outside probability/statistician community despite its long history. A couple of sources I know is David Williams' Probability with Martingales, and Berry Simon's Real Analysis: A Comprehensive Course in Analysis, Part 1. I should comment, however, that this lemma is by no means covered seriously in any of these books. – Sangchul Lee Aug 9 '17 at 18:49