Isometry between $\mathcal{B}^n(\mathbb{K},Y)$ and $Y$

Question

Let $\mathcal{B}^n(X,Y)$ be the space of all continuous multilinear mappings of $X\times\cdots X$ ($n$-times) into $Y$.

I know that, if $Y$ is a $\mathbb{K}$-normed vector space and $n=1$, so exist an isometry between $\mathcal{B}(\mathbb{K},Y)$ and $Y$. The natural identification is just to see it as a multiplication (on the right) by the vector. If $A \in \mathcal{B}(\mathbb{K},Y)$, then for any $x\in \mathbb{K}$, we have $$A(x)=A(x\cdot 1) = xA(1).$$

So a linear operator $A\in \mathcal{B}(\mathbb{K},Y)$ can be identified with $A(1)\in Y$.

This is natural, what I'm wondering if I can generalize to any $n$, where $Y$ is a $\mathbb{K}$-vector space. Something like $$A(x_1,x_2,\cdots,x_n) = x_1 x_2 \cdots x_n A(1) \tag{*}\label{eq}$$

this identification is correct between $\mathcal{B}^n(\mathbb{K},Y)$ and $Y$?

I will post a slightly more detailed answer, I don't know exactly, but I think that for this to work the dimension of $\mathbb{K}$ must be 1.

For $n\ge 2$ the mapping is not injective:$A(x_1,\ldots, x_n)=0$ if $x_1=0.$ — Ryszard Szwarc, Apr 05 '23 at 07:49
@RyszardSzwarc I think you are wrong. The dimension would be 1. See for example, https://math.stackexchange.com/q/2117784/680679 — Mrcrg, Apr 05 '23 at 14:34

score 1 · Answer 1 · answered Apr 05 '23 at 17:54

1

Let $A:\mathbb{C}^n\to Y$ be multilinear. Then $$A(x_1,x_2,\ldots,x_n)=x_1A(1,x_2,\ldots, x_n)= x_1x_2A(1,1,x_3,\ldots, x_n)\\ =\ldots=x_1x_2\ldots x_nA(1,1,\ldots, 1)$$ Therefore each such map is determined by a unique element $y:=A(1,1,\ldots,1).$

answered Apr 05 '23 at 17:54

Ryszard Szwarc

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Isometry between $\mathcal{B}^n(\mathbb{K},Y)$ and $Y$

1 Answers1