In my functional analysis textbook, there is a beginning of a proof that the inner product space $\ell ^ 2$ of square summable complex sequences is complete. Here is the beginning of the proof:
Let $\{x^j\}_{j=1}^{\infty}$ be a Cauchy sequence of sequences in $\ell ^ 2$, i.e. for every $ j \in \mathbb{N} $ we have a sequence $\{x^j_n\}_{n=1}^{\infty} \in \ell^2$ (which just means $\sum_{n=1}^{\infty} |x^j_n|^2 <\infty$). Since the sequence $\{x^j\}_{j=1}^{\infty}$ is Cauchy, for any $ \epsilon > 0 $ there is a $K \in \mathbb{N}$ such that for all natural $j,k > K$ one has: $$\big\|x^j – x^k \big\|^2 = \sum_{n=1}^{\infty} |x^j_n – x^k_n |^2 < \epsilon $$ and hence $ |x^j_n – x^k_n |^2 < \epsilon $ for any $ n \in \mathbb{N} $ so that for fixed $ n \in \mathbb{N} $ the sequence $\{x^j_n\}_{j=1}^{\infty} $ is Cauchy and thus converges, denote its limit by $$y_n = \lim_{j \to \infty} x_n ^ j $$
All we have left to do to show completeness is to show that $ y = \{y_n\}_{n=1}^{\infty} \in \ell^2 $ and that $\lim_{j \to \infty} \left\| y-x^j \right\| = 0 $. Hint: we are instructed to show that the sequence $\{ \lVert x^j \rVert \} _{j=1}^{\infty}$ is Cauchy, hence convergent and bounded.
Here is what I have thus far: we can use the reverse triangle inequality $$| \lVert x^j \rVert – \lVert x^k \rVert | \leq \lVert x^j-x^k \rVert $$ which indeed shows that $\{ \lVert x^j \rVert \} _{j=1}^{\infty} $ is a Cauchy, and hence convergent and thus bounded sequence. But I am stuck. I cannot show the two remaining tasks using the given hint.I realize there are references to this but none follow this approach which I am interested in completing using the given instructions. Can somebody please help me formally complete the proof? I thank all helpers.
Best Answer
You have $\sum\limits_{n=1}^{\infty} |x_n^{j}-x_n^{k}|^{2} <\epsilon$. This implies that each term in the sum is less than $\epsilon$. This means $(x_n^{j})_j$ is Cauchy sequence of real/complex numbers for each fixed $n$. Let $x_n$ be the limit of this sequence. Now, if $N$ is any positive integer then $\sum\limits_{n=1}^{N} |x_n^{j}-x_n^{k}|^{2} <\epsilon$. Letting $k \to \infty$ in this we get $\sum\limits_{n=1}^{N} |x_n^{j}-x_n|^{2} \leq \epsilon$ for $j >K$. Apply this for a particular $j$, say $j=K+1$ and conclude from this that $x=(x_n) \in \ell^{2}$. Finally we get $\|x^{j}-x\|^{2} \leq \epsilon$ (by letting $N \to \infty$). This finishes the proof.