# Abstract Nonsense

## The Total Derivative of a Multilinear Function (Pt. I)

Point of Post: In this post we prove that a multilinear form is differentiable everywhere and compute its derivative.

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Motivation

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Now that we have the definition of the derivative for mappings $f:\mathbb{R}^n\to\mathbb{R}^m$ it’s time to get our hands a little dirty and compute something. In particular we aim at proving that the wide sweeping class of multilinear function on spaces of the form $\mathbb{R}^{n_1}\times\cdots\mathbb{R}^{n_p}$ are everywhere differentiable and compute their derivative. From this we will be able to recover as a corollary a lot of particulary (and important) functions are differentiable, in particular linear trnasofrmations, the functions of the form $\mathbb{R}\times\cdots\times\mathbb{R}\to\mathbb{R}$given by $(x,\cdots,z)\mapsto x\cdots z$ and $(x,\cdots,z)\mapsto x+\cdots+z$, the usual inner product on $\mathbb{R}^n$, and the determinant.

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Total Derivative of a Multilinear Form

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We begin with a small lemma that will allow us to bound a multilinear function on the boundary of a polydisc. Namely:

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Lemma: Let $K$ be a multilinear function from $\mathbb{R}^{n_1}\times\cdots\times\mathbb{R}^{n_p}$ to $\mathbb{R}^m$. Then, $\|K\|$ is bounded on $\mathbb{S}^{n_1}\times\cdots\times\mathbb{S}^{n_p}$

Proof: We’d be done if we knew $K$ was continuous (which we don’t, in fact it will be a corollary of the following theorem). So, we have to do this old school. Namely, let $s_1,\cdots,s_p\in\mathbb{S}^{n_1}\times\cdots\times\mathbb{S}^{n_p}$ then we know that we can write

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$\displaystyle s_k=\sum_{j_k=1}^{n_k}\alpha_{k,j_k}e_{k,j_k}$

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where $e_{k,j_k}$ is the $j_k^{\text{th}}$ element of the canonical ordered basis on $\mathbb{R}^{n_k}$. But, by definition of a multilinear function we then have that

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$K\displaystyle (s_1,\cdots,s_k)=\sum_{j_1=1}^{n_1}\cdots\sum_{j_p=1}^{n_p}\prod_{r=1}^{p}\alpha_{r,j_r}K\left(e_{1,j_1},\cdots,e_{p,j_p}\right)$

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and thus clearly (applying the triangle inequality) since $\left|\alpha_{k,j_k}\right|\leqslant 1$ for all $k\in[p],j_k\in[n_k]$

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$\displaystyle \left\|K(s_1,\cdots,s_p)\right\|\leqslant \sum_{j_1=1}^{n_1}\cdots\sum_{j_p=1}^{n_p}\left\|K\left(e_{1,j_1},\cdots,e_{p,j_p}\right)\right\|$

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and since $(s_1,\cdots,s_p)$ was arbitrary the conclusion follows. $\blacksquare$

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So, with this in mind:

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References:

1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

May 23, 2011 -

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