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Let me recall $c = \{ (x_h)_{h \in \mathbb{N}} \subset \mathbb{R} \, | \, \lim_{h \to \infty} x_h = k < \infty \}$ the space of convergent sequences equipped with $\Vert \, \Vert_{\infty}$.

What are sufficient and necessary conditions on a sequence $(x^{(n)})_{n \in \mathbb{N}} \subset c$ to say $ x^{(n)} \rightharpoonup x \in c$?

I found something like this

$$ x^{(n)} \rightharpoonup x \iff \begin{cases} \sup_n \Vert x^{(n)} \Vert_{\infty} < \infty & (1)\\ \lim_{n \to \infty} x^{(n)}_h = x_h & (2)\\ \lim_{n \to \infty} \lim_{h \to \infty}x^{(n)}_h =\lim_{h \to \infty} x_h & (3) \end{cases} $$

However it looks to me not such an immediate proof

Partial Proof:

$\Rightarrow$: If $x^{(n)}$ weakly converges to $x$ then we know gratis $\sup_n \Vert x^{(n)} \Vert_{\infty} < \infty$ $(1)$. Moreover, $x^{(n)} \rightharpoonup x \iff \phi ( x^{(n)}) \to \phi (x)$ for every $\phi \in c^*$. The projections $\pi_h (x) = x_h$ lie in $c^*$ and this fact leads to us to $\lim_{n \to \infty} x^{(n)}_h = x_h$ $(2)$. About the condition $(3)$: it is a consequence of $(2)$ if we can exchange the limits. However, $x^{(n)}$ is dominated and it converges pointwise. Then, using Dominated convergence theorem adapted to sequences, we obtain $(3)$. Is the proof of $(3)$ correct?

$\Leftarrow$: If we call $k^{(n)} = \lim_{h \to \infty} x^{(n)}_h$, and $k =\lim_{h \to \infty} x_h$ then (3) tells us $k^{(n)} \to k$. Hence $(x^{(n)}_h - k^{(n)}) \to (x_h-k)$ for every $h$ because of (2). Moreover, $\sup_n \Vert x^{(n)} - k^{(n)} \Vert_{\infty} \leq \sup_n(\Vert x^{(n)} \Vert + \Vert k^{(n)} \Vert ) < \infty$ because of (1).

Now, using the characterization of weak convergence in $c_0$ (indeed $x^{(n)}_h - k^{(n)}$ and $x_h-k$ are in $c_0$), we discover $$ (x^{(n)} - k^{(n)}) \rightharpoonup (x-k) $$

Finally, $x^{(n)} \rightharpoonup x$.

Does this implication hold?

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  • $\begingroup$ And so how does weak convergence imply (3)? $\endgroup$ Jul 8, 2020 at 8:59
  • $\begingroup$ @ Kavi Rama Murthy are you sure? And where I use the hypothesis $x^{(n)} \subset c$? $\endgroup$ Jul 8, 2020 at 19:24
  • $\begingroup$ @Kavi Rama Murthy why did you delete the comments? $\endgroup$ Jul 9, 2020 at 7:29

1 Answer 1

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Crucial for us is the description of the dual space $c'$. Namely, we have that $c' \cong \ell^1$ via the isomorphism $\ell^1 \to c'$ given by $$(\alpha_k)_{k=0}^\infty \mapsto f, \quad f(x_k)_{k=1}^\infty := \alpha_0\left(\lim_{k\to\infty} x_k\right) + \sum_{k=1}^\infty \alpha_kx_k.$$

Now, assume that $x_n \rightharpoonup x$ in $c$.

  1. By the uniform boundedness principle we get that $\sup_{n\in\Bbb{N}} \|x_n\|_\infty < +\infty$.

  2. For every $k \in \Bbb{N}$ for the projection we have $\pi_k \in c'$ so $$x_n(k) = \pi_k(x_n) \xrightarrow{n\to\infty} \pi_k(x) = x(k).$$

  3. For the limit functional $L(x_n)_n := \lim_{n\to\infty} x_n$ we have $L \in c'$ so $$\lim_{k\to\infty} x_n(k) = L(x_n) \xrightarrow{n\to\infty} L(x) = \lim_{k\to\infty} x(k).$$

Conversely, suppose that $(x_n)_{n=1}^\infty$ is a bounded sequence in $c$ so that $(1)-(3)$ holds for some $x \in c$. Pick $f \in c'$ and we claim that $f(x_n) \to f(x)$. There exists some $(\alpha_k)_{k=0}^\infty \in \ell^1$ such that $f$ is of the above form.

The functions $g_n, g : \Bbb{N}_0 \to \Bbb{C}$ for $n \in \Bbb{N}$ given by $$g_n(k) = \begin{cases}\alpha_0 \left(\lim_{j\to\infty} x_n(j)\right), &\text{ if $k=0$},\\ \alpha_k x_n(k) &\text{ if $k>1$}. \end{cases}, \qquad g(k) = \begin{cases}\alpha_0 \left(\lim_{j\to\infty} x(j)\right), &\text{ if $k=0$},\\ \alpha_k x(k) &\text{ if $k>1$}. \end{cases}$$ are all dominated by the function $k \mapsto \alpha_k\left(\sup_{n\in\Bbb{N}} \|x_n\|_\infty\right)$ which is summable by $(1)$. Moreover, by $(2)$ and $(3)$, we have $g_n \to g$ pointwise so by the Lebesgue Dominated convergence theorem we get \begin{align*} \lim_{n\to\infty} f(x_n) &= \lim_{n\to\infty}\left(\alpha_0 \left(\lim_{j\to\infty} x_n(j)\right)+ \sum_{k=1}^\infty \alpha_kx_n(k)\right) \\ &= \lim_{n\to\infty}\sum_{k=0}^\infty g_n(k)\\ &= \sum_{k=0}^\infty \lim_{n\to\infty} g_n(k)\\ &= \sum_{k=0}^\infty g(k)\\ &= \lim_{n\to\infty}\left(\alpha_0 \left(\lim_{j\to\infty} x(j)\right)+ \sum_{k=1}^\infty \alpha_kx(k)\right)\\ &= f(x). \end{align*} Since $f\in c'$ was arbitrary, we conclude $x_n \rightharpoonup x$ in $c$.

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  • $\begingroup$ This is what I looked for. Good proof. However, what about the "converse" I wrote. Does It work as well? $\endgroup$ Jul 9, 2020 at 19:09
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    $\begingroup$ Your proof works, but expressions such as $x-k$ are not well-defined. You noticed that $$x_n(k) - L(x_n) \xrightarrow{n\to\infty} x(k)-L(x)$$ and hence if we denote $e = (1,1,1, \ldots)$ we have $$x_n-L(x_n)e \xrightarrow{c_0} x-L(x)c$$ weakly so in particular it also converges weakly in $c$. Of course also $L(x_n) e \xrightarrow{c} L(x)e$ weakly so by adding two limits we get $x_n \xrightarrow{c} x$ weakly. $\endgroup$ Jul 9, 2020 at 19:33

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