Differential of a multilinear function

Question

This was a question that was posted and deleted.

Since it is a useful basic little lemma, let me post it such that it stays.

Let $T:\mathbb{R}^n\to \mathbb{R}$ be a multilinear function. Then $T$ is differentiable and its differential is equal to $$DT(x)(y)=\sum_{i=1}^{n}T([x,y,\{i\}])$$ where $[x,y,I]$ is the vector obtained from $x$ by replacing $x_i$ with $y_i$, for $i\in I$.

The question is how to prove this. I will add the proof below.

Answer 1

First we show that the function $DT(x)(y)$ as defined above is linear in $y$. In fact, if $y,z\in\mathbb{R}^n$ and $a,b\in\mathbb{R}$, then $$\begin{align}DT(x)(ay+bz)&=\sum_{i=1}^{n}T([x,ay+bz;i])\\&=\sum_{i=1}^{n}\left[aT([x,y;i])+bT([x,z;i])\right]\\&=aDT(x)(y)+bDT(x)(z)\end{align}$$

Now we can just compute

$$\begin{align}T(x+y)&=T(x)+DT(x)(x)\\&\phantom{{}=T(x)}+\sum_{1\leq i_1<i_2\leq n}T([x,y,\{i_1,i_2\}])\\&\phantom{{}=T(x)}+\sum_{1\leq i_1<i_2<i_3\leq n}T([x,y,\{i_1,i_2,i_3\}])\\&\phantom{{}=T(x)}+{}...\\&\phantom{{}=T(x)}+\sum_{1\leq i_1<i_2<...<i_n\leq n}T([x,y,\{i_1,i_2\}])\end{align}$$

To see this formula, try it in small dimension. $$\begin{align}T(x_1+y_1,x_2+y_2)&=T(x_1,y_1+y_2)+T(x_2,y_1+y_2)\\&=T(x_1,x_2)+T(x_1,y_2)+T(y_1,x_2)+T(y_1,y_2)\end{align}$$ It is similar to expanding $(x_1+y_2)(x_2+y_2)$.

Observe that, by multilinearity, each term in the summations is multiple of at least two $y_i$. This implies that $$T(x+y)-T(x)-DT(x)(y)=o(|y|)$$ This, together with being linear in the increment $y$, is the property that defines the differential of $T$ at $x$.