# How can we characterize weak convergence in $(c, \Vert \, \Vert _{\infty})$?

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?

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

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 $$c$$.

• This is what I looked for. Good proof. However, what about the "converse" I wrote. Does It work as well? Jul 9, 2020 at 19:09
• 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. Jul 9, 2020 at 19:33