Is this map continuous w.r.t. the weak topology?

Question

I am currently working with kernels and I've stumbled upon a map that I really want it to be continuous, but I'm not seeing how to prove it (or disprove it). Consider the following map $\Phi: l_2(\mathbb{N}) \to \mathbb{R}$ given by $$\Phi(w) = C w_{i_1}^{j_1} w_{i_2}^{j_2}...w_{i_n}^{j_n},$$ where $i_1,...,i_n, j_1,...,j_n \in \mathbb{N}$, $C$ is some constant, and $w_i$ denotes the $i$-th component of $w\in l_2(\mathbb{N})$. In other words, given a square summable sequence, we simply select a finite number of its components, exponentiate them, and multiply everything.

Now, assuming that $l_2(\mathbb{N})$ is given the weak topology, and that $B_R$ denotes the closed ball of radius $R$ in $l_2$, is there any hope that $\Phi \in C(B_R)$, i.e. $\Phi$ is a continuous function on $B_R$? Note that $B_R$ is compact, by the Banach-Alaoglu theorem. I'm really not seeing how can I prove that $\Phi$ is continuous or discontinuous on $B_R$. Any suggestions?

Answer 1

The weak topology is the smallest topology making all linear functionals continuous. In particular, the projections $π_i : l_2(\mathbb{N})\to \mathbb{R}$ given by $π_i(w) = w_i$ are continuous.

Moreover, any product of continuous maps is also continuous (multiplication on $\mathbb{R}$ is continuous). Hence, $f_{i,j}(w) = w_i^j$ is also a continuous map. Multiplying finitely many of these $f_{i,j}$ together gives you precisely your $\Phi$, which is thus continuous w.r.t. the weak topology on $B_R$.

Answer 2

I'll add an answer here just so I show the kernel I'm working with. In the comments someone pointed out that it doesn't make sense to work with kernels and weak topology. I don't see why is that. As a matter of fact, when I show that $k$ is a kernel I don't make any considerations about the topology in $l_2(\mathbb{N})$. So, let me know if you spot any fallacy.

I am considering a function $k: l_2(\mathbb{N}) \times l_2(\mathbb{N}) \to \mathbb{R}$ of the form $k(x,y) = K(\langle x,y\rangle_{l_2})$, where $K$ is an analytic function expressible globally by its Taylor series expanded at $0$, i.e. $$K(t) = \sum_{n=0}^\infty a_n t^n, \ \ \text{and so} \ \ k(x,y) = \sum_{n=0}^\infty a_n \langle x,y \rangle_{l_2}^n.$$

As to why this is a kernel, I am using a Lemma from this paper. Namely:

Lemma: Assume that $n\in \mathbb{N}$ is fixed. Then, for all $j \in \mathbb{N}_0^{\mathbb{N}}$ with $|j|:= \sum_{i =1}^\infty j_i=n$, there exists a constant $c_j \in (0,\infty)$ such that for all summable sequences $(b_i)_{i \in \mathbb{N}}\subset [0,\infty)$ we have $$\left( \sum_{i=1}^\infty b_i \right)^n = \sum_{j \in \in \mathbb{N}_0^{\mathbb{N}} : |j| = n} c_j \prod_{i=1}^\infty b_i^{j_i}.$$

Now, by simply noting that

\begin{align*} k(x,y) =& \sum_{n=0}^\infty a_n \langle x,y \rangle_{l_2}^n = \sum_{n=0}^\infty a_n \left( \sum_{i=1}^\infty x_i y_i \right)^n = \sum_{n=0}^\infty a_n \sum_{j \in \in \mathbb{N}_0^{\mathbb{N}} : |j| = n} \tilde{c}_j \prod_{i=1}^\infty (x_i y_i)^{j_i} = \\ =& \sum_{j \in \mathbb{N}_0^\mathbb{N}} a_{|j|} \tilde{c}_j \prod_{i=1}^\infty x_i^{j_i} \prod_{i=1}^\infty y_i^{j_i}, \end{align*}

and by setting $c_j = \sqrt{a_{|j|} \tilde{c}_j}$, we conclude that $\Phi: l_2 \to l_2(J)$ given by $$\Phi(w) = \left( c_j \prod_{i=1}^\infty w_i^{j_i} \right)_{j \in J},$$ is a feature map of $k$, where we use the convention $0^0 =1$. In other words, $k(x,y) = \langle \Phi(x), \Phi(y) \rangle_{l_2(J)}$, thus is a kernel.

Answer 3

Since $\ell ^2 (\mathbb N )$ is a reproducing kernel Hilbert space, pointwise evaluation is continuous. Therefore for each $n$, the following map from $\ell ^2 (\mathbb N) \to \mathbb R$ sending $f \mapsto f(n)$ is continuous. In other words, the map $(x_i)_{i\in \mathbb N} \mapsto x_n$ is continuous for each $n\in \mathbb N$. The fact that your function $\Phi$ is continuous follows from the fact that it is product of pointwise evaluation functionals.

Note that weak topology is the smallest topology which makes every functional continuous and this is not included the definition for rkHs because it is unnecessary. In weak topology, the sequence $(e_i)_{i\in \mathbb N}$ converges to $0$. This certainly is not the case in the $\ell ^ 2 (\mathbb N)$.