Abstract Nonsense

Crushing one theorem at a time

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.

\text{ }


\text{ }

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.

\text{ }

Total Derivative of a Multilinear Form

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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)

\text{ }

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]

\text{ }

\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\|

\text{ }

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

\text{ }

So, with this in mind:

\text{ }

\text{ }


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 - Posted by | Analysis | , , , , ,


  1. […] Point of Post: This post is a continuation of this one. […]

    Pingback by The Total Derivative of a Multilinear Map (Pt. II) « Abstract Nonsense | May 23, 2011 | Reply

  2. […] function of is differentiable. Moreover, it’s clear that is differentiable since it’s linear. Noting then that we may conclude from the chain rule that is differentiable at and . That […]

    Pingback by Further Properties of the Total Derivative (Pt. I) « Abstract Nonsense | May 25, 2011 | Reply

  3. […] since each coordinate function is differentiable at and is differentiable at since it’s mutlilinear. Thus, by the chain rule we have that is differentiable at and . But, using the same logic as […]

    Pingback by Further Properties of the Total Derivative (Pt. II) « Abstract Nonsense | May 26, 2011 | Reply

  4. […] since and is open we actually have that for some . So, consider then the function . Now, since (recalling that constant maps and multillinear maps are differentiable) each of these functions are […]

    Pingback by The Mean Value Theorem for Multivariable Maps « Abstract Nonsense | June 11, 2011 | Reply

  5. […] Note that by definition , differentiating this (recalling the way derivatives work for multilinear forms) we have that , but since is a bilinear form this may be rewritten as . Thus, we either have that […]

    Pingback by Curvature of Plane Curves « Abstract Nonsense | September 24, 2011 | Reply

  6. […] basis is known as the Frenet frame for . Now, recalling that the cross product is bilinear and recalling how derivatives interact with bilinear functions we may easily deduce […]

    Pingback by Space Curves « Abstract Nonsense | October 5, 2011 | Reply

  7. […] Proof: The first and last one we have proven before (here and here respectively) and the second can be proved thinking of multiplication as a bilinear map and applying this result. […]

    Pingback by Complex Differentiable and Holmorphic Functions (Pt. II) « Abstract Nonsense | May 1, 2012 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: