0

I am trying to follow the proof that the sequence space $\ell^2$ is complete, e.g., Show that $l^2$ is a Hilbert space.

For any Cauchy sequence $\{x_n\}$ in $\ell^2$ we obtain a candidate limit by taking the pointwise limit of each coordinate, which we call $\tilde{x}$.

The answer in the link above argues that $\tilde{x} \in \ell^2$ as follows:

Consider $$ \sum_{k=1}^{\infty}|\tilde{x}(k)|^2 = \sum_{k=1}^{\infty} |\lim_{n \to \infty} x_n(k)|^2 = \lim_{n \to \infty} \sum_{k=1}^{\infty} |x_n(k)|^2 = \lim_{n \to \infty} || x_n ||^2. $$ The exchange of limits is justified, since the convergence of $\lim_{n \to \infty} x_n(k)$ is uniform over $k$. Finally, since $\{x_n\}$ is Cauchy, the inequality,

How can we see that the convergence of $\lim_{n \to \infty} x_n(k)$ is uniform over $k$? The only part of the argument that I am not confident about is the second equality in the quoted text.

neophyte_autodidact
  • 101

1 Answers1

2

First, we have $$\begin{split} |x_n(k)-\bar x(k)|^2 \le |x_n(k)-x_m(k)|^2 + |x_m(k)-\bar x(k)|^2\\ \le \left( \sum_j |x_n(j)-x_m(j)|^2\right) + |x_m(k)-\bar x(k)|^2 \\ = \|x_n -x_m\|_{l^2}^2 + |x_m(k)-\bar x(k)|^2 . \end{split}$$ Now take $\epsilon>0$. Then there is $N=N(\epsilon)$ such that $ \|x_n -x_m\|_{l^2}^2 \le \epsilon/2$ if $m,n>N$. Note that $N$ is independent of $k$. In addition, there is $M=M(\epsilon,k)$ such that $|x_m(k)-\bar x(k)|^2 < \epsilon/2$ for $m>M$.

Choose $n>N$. Then for all $k$, we can choose $m>\max(M,N)$ (depending on $k$), such that the above inequality gives us $$ |x_n(k)-\bar x(k)|^2 < \epsilon. $$ This is valid for all $n>N$ and all $k$.

So the trick in the proof above is to (1) use triangle inequality to introduce another element $x_m$ of the sequence, (2) use the Cauchy property to estimate $\|x_n-x_m\|$, (3) use the pointwise convergence to estimate $|x_m(k)-\bar x(k)|$.

daw
  • 54,637
  • 2
  • 44
  • 85
  • Thanks! Since $|x_n(k) - \bar{x}(k)| < \sqrt{\varepsilon}$ for all $k$ and the choice of $N$ did not depend on $k$, then can we say $x_n$ converges to $\bar{x}$ uniformly? I am unsure because we also needed to use $M$, which did depend on $k$. – neophyte_autodidact Sep 14 '21 at 10:56
  • The $M$ is only used in an intermediate step. It disappears. The definition of uniform convergence is: for all $\epsilon>0$ there is $N$ such that for all $n>N$ and all $k$ $|x_n(k)-\bar x(k)|<\epsilon$. Which is exactly what I have proven (apart from $\sqrt\epsilon$). – daw Sep 14 '21 at 11:33