Finding Weak* limit of a sequence in the space of sequences

Question

For $n \in \mathbb{N}$ we define the functional $\phi_n$ on $l^{\infty}$ by $\phi_n(x) = \frac{1}{n} \sum_{j=1}^{n} x_j$ where $x$ denotes the sequence $\{x_j \}^{\infty}_{j=1}.$

It is clear that $\phi_n$ is linear. Using the triangular inequality, and the definition of $\infty-$norm, I was also able to prove that $\phi_n$ is in fact in ($l^{ \infty}$).*

Now, I am stuck with the remaining part of the problem. The problem also says that the sequence $\{\phi_n\}^{\infty}_{n=1}$ has a weak* cluster point $\phi$ in ($l^{ \infty}$)*, and $\phi$ does not arise from an element of $l^1.$

This is an exercise (#19 section 6.2) from Folland's Real Analysis book. Any help/suggestions would be highly appreciated. Thank you so much. Stay safe.

Edit: Suppose $f: l^1 \rightarrow (l^{\infty})^*$ is given by $(f(y))(x) = \sum_{j=1}^{n} x_{j}y_{j}$ where $y$ denotes the sequence $\{y_j\}^{\infty}_{j=1}$ in $l^1.$ I want to prove that the functional $\phi$ is not in the image of $f$ i.e. $\phi \notin f(l^1).$ This what I mean by saying $\phi$ does not arise from an element of $l^1.$

Answer 1

Recall that in order to show that there is a weak$^*$ cluster point $\phi$ of the sequence, there has to be some subsequence weak$^*$ converging to $\phi$ i.e.

$$\forall x \in \ell^{\infty}: \phi_n(x) \rightarrow \phi(x)$$

In our case here, even the entire sequence converges:

Define $\phi(x) := \lim_{n \rightarrow \infty} \frac{1}{n} \sum^n_{i = 0} x_i$, then we have clearly the above requirement is fulfilled. It is left to show that $\phi \in \ell^{\infty}$. For this, let $x \in \ell^{\infty}$ be arbitrary. Then we have

$$ \vert \phi(x) \vert = \lim_{n \rightarrow \infty} \frac{1}{n} \vert \sum^n_{i = 0} x_i \vert \leq \lim_{n \rightarrow \infty} \frac{1}{n} n \Vert x \Vert_{\infty} \leq \lim_{n \rightarrow \infty} \Vert x \Vert_{\infty} = \Vert x \Vert_{\infty}$$

Hence $\phi$ has operator norm of less than $1$ and is thus continuous.

Finally, we show that $\phi$ does not arise from an element of $\ell^1$, we want to show that there does not exist an element $y \in \ell^1$ s.t. $\forall x \in \ell^{\infty}: \phi(x) = \sum_{i = 0}^{\infty} x_i y_i$.

For the sake of contradiction assume there was an element $y \in \ell^1$ as described above, then for $\delta^m_i$ which is defined by having a $1$ in the $m$-th place and $0$ otherwise, then every $m \in \mathbb{N}$ we have

$$\phi(\delta^m_i) = \lim_{n \rightarrow \infty} \frac{1}{n} = 0$$

and also for the supposed $y \in \ell^1$:

$$\sum_{i = 0}^{\infty} x_i y_i = y_m$$

which has to equal $0$ by assumption. Hence $y = 0$, and hence $\phi = 0$ which is a contradiction (which can be seen for example via the constant $1$-sequence).

Edit: The above is false. The limit $\phi(x) = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i = 0}^n x_i$ exists when $(x_i)_{i = 0}^{\infty}$ is convergent. Otherwise this cannot be guaranteed, as shown here.

The right answer is the following:

Note that for every $n \in \mathbb{N}$ the operator norm of $\phi_n$ is less than $1$. Hence $(\phi_n)_{n = 0}^{\infty} \subseteq B_1(0)$ which is compact by Banach-Alaoglu. Hence it has a convergent subsequence $(\phi_{n_k})_{k = 0}^{\infty}$ i.e. the sequence has a cluster point.

Note here that Banach-Alaoglu uses Tychonoff's Theorem of compactness, which in turn uses the axiom of choice. Hence we don't have an idea of what this cluster point looks like.

Now it is left to show that this cluster point is not induced by some $(y_i)_{i = 0}^{\infty} \in \ell^1$. To see this, assume there exists an $(y_i)_{i = 0}^{\infty} \in \ell^1$ s.t. $(\phi_{n_k})_{k = 0}^{\infty}$ converges weakly$^*$ to $(y_i)_{i = 0}^{\infty}$. Then we have that for every $(x_i)_{i = 0}^{\infty} \in \ell^{\infty}$ we have

$$ \vert \frac{1}{n_k} \sum_{i = 0}^{n_k} x_i - \sum_{i = 0}^{\infty} x_i y_i \vert \rightarrow 0$$

Then consider for every $j \in \mathbb{N}$ the sequence $\delta^{j}$ defined by $\delta^j_i = 1$ if $i = j$ and $\delta^j_i = 0$ otherwise for $i \in \mathbb{N}$. Then for $\delta^j \in \ell^{\infty}$ the above conditions gives

$$ \vert \frac{1}{n_k} - y_j \vert \rightarrow 0, ~~ n_k \rightarrow $$

when $n_k \geq j$. This implies that $y_j = 0$ and thus $y = 0$ which is a contradiction (as can be seen for example via the constant $1$-sequence).

My apologies for the incorrect answer before.

Answer 2

It is straightforward to show $\phi_n\in B_{X^*}$ which is weak* compact by Banach Alouglu. Any compact set with an infinite subset is such that the infinite subset has a cluster point [see Thm 28.1 in Munkre's topology].

Suppose the cluster point is $y\in l_1$. This is an open set in the weak star topology: $$V_1=\{x^* \: | \:\: |x^*(e_j)-y(e_j)|<1\}$$

Take $\phi_{n_1}\in V_1\setminus\{y\}$ (this is possible because we have a cluster boint). Because the weak star is Hausdorff, we may take $y\in \hat{V}_2$ without $\phi_{j}$ for all $j\in \{1,...,n_1\}$. We define:

$$V_2=\{x^* \: | \:\: |x^*(e_j)-y(e_j)|<1/2\}\cap \hat{V}_2$$

We take $\phi_{n_2}\in V_2\setminus\{y\}$. Because the weak star is Hausdorff, we may take $y\in \hat{V}_3$ without $\phi_{j}$ for all $j\in \{1,...,n_2\}$. Define:

$$V_3=\{x^* \: | \:\: |x^*(e_j)-y(e_j)|<1/3\}\cap \hat{V}_3$$

If we proceed inductively, it is clear we have built a sequence such that for every $k\in \mathbb{N}$:

$$\left|\frac{1}{n_k}-y_j\right|<\frac{1}{k}$$

Taking the limit in $k$ (notice the $n_k$ are incresaing natural numbers by construction) yields $y=0$. However, it is also clear there is $\phi_{n}\in \{x^* \:\:|\:\:|x^*(1,1,1...)-y(1,1,1,...)|<1/2\}$. Hence:

$$|\phi_n(1,1,1,...)-y(1,1,1,...)|=\left|1-0\right|<\frac{1}{2}$$

Which is a contradiction.