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Suppose we are given a linear operator $L$ on a Banach space $X$. Is there any way to extend $L$ to a multi-linear operator $\mathcal{L}$ in such a way that $$\mathcal{L}(x_1, x_2^*, \ldots, x_n^*) = L(x_1)$$ for some fixed values of $x^*_2, \ldots, x_n^*$?

If this seems too difficult, any insight on how to canonically associate $L$ to a multi-linear operator would also be interesting.

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    $\begingroup$ What are you suggesting for the new domain of this multilinear operator? $X^{\otimes n}$? (I presume it's an endomorphism) You need to provide more information as to what you want. $\endgroup$
    – David Roberts ♦
    Commented Feb 13, 2011 at 23:15
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    $\begingroup$ Kate's first suggestion, making $\mathcal L$ depend only on its first argument, isn't multi-linear. The simplest solution, quite close to Kate's second suggestion, is to use $L(x_1)f(x_2)\dots f(x_n)$ where $f$ is some non-zero linear functional on $X$. Then take `$x_2^*=\dots=x_n^*=y$, where $y$ is chosen so that $f(y)=1$. $\endgroup$
    – Andreas Blass
    Commented Feb 13, 2011 at 23:50
  • $\begingroup$ My previous comment was originally intended to be a comment on Kate's answer, but MO tells me I can't comment there. $\endgroup$
    – Andreas Blass
    Commented Feb 13, 2011 at 23:50
  • $\begingroup$ @Andreas: I just realized that there is a problem with the first suggestion and I have deleted the answer, please put your comment as an answer, it seems that it is what Dan is looking for. $\endgroup$
    – Kate Juschenko
    Commented Feb 13, 2011 at 23:56
  • $\begingroup$ Thanks Professor Blass, that seems to work! And many congrats to Tim on his recent offer! Tim and I share an advisor. $\endgroup$
    – user7807
    Commented Feb 13, 2011 at 23:57

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Following Kate's suggestion, I'm posting my previous comment as an answer. Choose any non-zero linear functional $f$ on $X$ and any vector $y\in X$ such that $f(y)=1$. Define $\mathcal L(x_1,x_2,\dots,x_n)$
to be the product $L(x_1)f(x_2)\dots f(x_n)$ and take all of $x_2^*,\dots x_n^*$ to be $y$.

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