This is a side question from Infinite-dimensional "algebraic varieties".

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 right side 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 $\operatorname{Ran}(L)$ as a subspace of $P^d(X)$? In particular, does this subspace precisely consist of all completely continuous (i.e. mapping weakly-convergent sequences to norm-convergent sequences) 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 $X'_\infty=l^\infty$. However, since $l^\infty$ is inseperable and has a complicated dual under $ZFC$, I'd rather exclude 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.

Progress $1$: 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 $P(v) \ne 0$. Examine the values of $P$ on the tails of $v$ to get a contradiction.), 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 sum of an unbounded sequence with a bounded sequence, which is unbounded (For $d \gt 2$ the coefficients in front of $e_j$ might be other numbers, e.g. in the case $d=3$, $P(e_i-e_j)$ is also involved.). Either case contradicts the continuity of $P$. We can see that $L$ is actually an isomorphism between $P^1(X_1)$ and $P^d(X_1)$, which gives an affirmative answer for the case $p=1$ considering the Dunford-Pettis property of $X_1$. (If my proof is correct, it would be surprising that this non-trivial isomorphism doesn't occur anywhere else. So I'm wondering if there exist some related references.)

Progress $2$: It seems that there are still many completely continuous polynomials outside $\operatorname{Ran}(L)$ when $d \ge p$. For example, any $Q(x)=a_0 x_0^2+a_1 x_1^2+\dotsb$ with $a_n \rightarrow 0$ is a completely continuous polynomial on $X_2$, but $Q \in \operatorname{Ran}(L) \iff (a_i) \in X_2$ (just as the difference between compact operators and Hilbert-Schmidt operators). It is then more meaningful to investigate the closure of $\operatorname{Ran}(L)$ (under the norm topology).

  • $\begingroup$ What is $l^p/c_0$ for $p<\infty$? $\endgroup$ Apr 27, 2022 at 13:24
  • $\begingroup$ @M.González The backslash means "or", not quotient space. $\endgroup$
    – Zerox
    Apr 27, 2022 at 13:38
  • 1
    $\begingroup$ Although it's clear what you meant in your comment, I mention that $/$ is a slash, not a backslash. But it's not so clear in the body, so I would encourage you just to say "or"—there's no character limit! Also, you might want to consider denoting your polynomials by something other than $p$ when discussing $l^p$ spaces …. $\endgroup$
    – LSpice
    Apr 27, 2022 at 16:49
  • 1
    $\begingroup$ @LSpice Thanks for your advice. I've changed some formulation. $\endgroup$
    – Zerox
    Apr 27, 2022 at 16:56

1 Answer 1


This post only partially answers your question. I hope you find it useful.

$\ell^1$ and $c_0$ have the Dunford-Pettis property. Thus, every polynomial on these spaces are weakly sequentially continuous (w.s.c.), see Proposition 2.34 in the book by Dineen.

$X=\ell^p$ is reflexive for $1<p<\infty$. Thus, every w.s.c. bounded bilinear map is of the form $A(x,y) = \langle Tx,y\rangle$ for some compact operator $T:X\to X^*$. In particular, when $p>2$, every bounded linear $T:\ell^p\to\ell^q$ is compact by Pitt's theorem; thus, every bilinear map and every degree 2 polynomial is w.s.c.

  • $\begingroup$ I remember I leant from somewhere else that degree-$d$ polynomials in $l^p$ for $p>d$ is weakly sequentially continuous. Is this statement correct? $\endgroup$
    – Zerox
    Apr 29, 2022 at 17:45
  • 2
    $\begingroup$ Yes, see Proposition 4.1 in doi.org/10.1006/jmaa.1997.5253 $\endgroup$
    – Onur Oktay
    Apr 29, 2022 at 21:55

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