This is a side question from Infinite-dimensional "algebraic varieties", and I have also asked it on MSE without getting any response.

Denote by $X_p$ ($1 \le p \le \infty$) the Banach spaces of complex sequences with finite $p$-norm and limit $0$. Suppose $(x_i):=(x_0, x_1, \dotsc)$, then the "degree-$d$ Veronese map" can be defined as $$V_p^d: X_p \rightarrow X_p: (x_i) \mapsto \left( \left( \frac{d!}{l_{1}! \dotsb l_{d}!} \right)^{\frac{1}{p}} x_{k_1}^{l_1} \dotsb x_{k_d}^{l_d} \right)\quad\left(\frac{1}{\infty}:=0\right)$$ where $l_1, \dotsb, l_d$ runs over non-negative integer partitions of $d$, $0 \le k_1 \lt \dotsb \lt k_d$ and the coordinates of the image are arranged in increasing order of $l_1 k_1 + \dotsb + l_d k_d$. It can be shown that $V_p^d$ is well-defined and continuous since $\lVert V_p^d(x) \rVert_p = \lVert x \rVert_p^d$.

Suppose $A$ is a symmetric $d$-linear functional on some $X=X_p$. Define the degree-$d$ homogeneous polynomial w.r.t. $A$ as $P_A(x)=A(\underbrace{x, \dotsc ,x}_{d \text{ times}})$, and equip the vector space of degree-$d$ homogeneous polynomials on $X$, $P^d(X)$, with the norm $$\lVert P \rVert = \sup_{\lVert x \rVert = 1}\lvert P(x) \rvert$$ to make it a Banach space. It can be seen that $L:X^*=P^1(X) \rightarrow P^d(X):\psi \mapsto \psi \circ V_p^d$ is a norm-$1$ injective linear map. How to describe the image of this map as a subspace of $P^d(X)$? In particular, does this subspace precisely consist of all weakly sequentially continuous (i.e. the image of a weakly-convergent sequence is norm-convergent) polynomials? If not, I want to know the strongest continuity (e.g. weakly-continuous on every bounded set) that the polynomials in this subspace could reach.

Similar statements can also be made for $X=l^\infty$. However, since $l^\infty$ is inseperable and has a complicated dual under $ZFC$, I'd rather excluded it for a bonus question.

PS: I welcome answers of a full explicit description of all homogeneous polynomials on $X_p$, since the most "canonical" homogeneous polynomials $x_0^d+x_1^d+ \dotsb$ for $(x_i) \in X_p,d \ge p$ seem to have no weak-topology-based continuity and can not be reduced to some class of (continuous) linear functional.

Progress: Until now the only solved case is $p=1$. We only give a proof of the case $d=2$ (the other values are similar): Suppose $P \in P^2(X_1)$. Denote $P_{ij}$ by the Gateaux derivative of $P$ along $e_i,e_j$, where $e_i \in X_1$ is the $i$-th unit vector. It's not hard to show that $P_{ij} \equiv 0 \implies P \equiv 0$ by the continuity of $P$ (Suppose on the contrary that $P \ne 0$, then $\exists v \in X^1$ such that $P(v) \ne 0$. Look at the values of $P$ on the tails of $v$ to get a contradiction. This also works for other values of $p$ and $d$.), thus $P$ can be formally written as $$P((x_i))=\sum_{i=0}^{+\infty}a_ix_i^2 + 2\sum_{0 \le i \lt j}b_{ij}x_ix_j$$ Then it is sufficient to prove that $\{ a_i \} \cup \{ b_{ij} \}$ is bounded. If $\{ a_i \}$ is unbounded, then $P(e_i)=a_i$ is unbounded; if $\{ a_i \}$ is bounded but $\{ b_{ij} \}$ is unbounded, then $P(e_i+e_j)=2b_{ij}-(a_i+a_j)$ $(i \ne j)$ is the difference of an unbounded sequence with a bounded sequence, which is unbounded. Either case contradicts the continuity of $P$. We can see that $L$ is a linear homeomorphism between $P^1(X_1)$ and $P^d(X_1)$ which gives an affirmative answer of the case $p=1$ considering the Dunford-Pettis property of $X_1$.