Proving that pointwise limit implies limit of a sequence in $\ell^p$

Question

Say $\{x^{(n)}\}$ is Cauchy in $\ell^p$ and $x$ is its pointwise limit. To argue that $x^{(n)} \to x$ would the following be correct:

We have to show that for any $\varepsilon > 0$ there exists $N \in \mathbb{N}$, s.t. $|x^{(n)} - x|_p < \varepsilon$, for all $n > N$.

Proof:

Choose $\varepsilon > 0$, then there exists $N_i \in \mathbb{N}$, s.t. $|x^{(n)}_i - x_i|^p < \varepsilon^{\ p} \cdot 2^{-i}$, for all $n > N_i$, since pointwise limit.

For any $J \in \mathbb{N}$ there exists $N_0 = \max(N_i)$, s.t. $\sum_{i=1}^{J}|x^{(n)}_i - x_i|^p < \varepsilon^{\ p} \cdot \sum_{i=1}^{J}2^{-i} < \varepsilon^{\ p}$, for all $n > N_0$. Note that for any $J$ we can always calculate $N_0$.

Taking the limit $J \to \infty$:

$$\lim_{J\to\infty} \sum_{i=1}^{J}|x^{(n)}_i - x_i|^p < \varepsilon^{\ p}$$

Hence $|x^{(n)} - x|_p < \varepsilon$, for all $n > N_0 \in \mathbb{N}$

Is my proof correct?

Answer 1

Sure. This is closely related to a famous theorem: given a sequence Cauchy in $L^p(X)$ for a general measure space $X$, you can find a "rapidly Cauchy" subsequence which converges pointwise a.e. The original sequence then converges in $L^p$ to this pointwise limit. Thus the proof of this theorem goes in two parts: propose a limit for a Cauchy sequence (already done for you in your problem) and then check that the given sequence converges to that limit.