# The Unapologetic Mathematician

## Vector-Valued Functions

Now we know how to modify the notion of the derivative of a function to deal with vector inputs by defining the differential. But what about functions that have vectors as outputs?

Well, luckily when we defined the differential we didn’t really use anything about the space where our function took its values but that it was a topological vector space. Indeed, when defining the differential of a function $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ we need to set up a new function that takes a point $x$ in the Euclidean space $\mathbb{R}^m$ and a displacement vector $t$ in $\mathbb{R}^m$ as inputs, and which gives a displacement vector $df(x;t)$ in $\mathbb{R}^n$ as its output. It must be linear in the displacement, meaning that we can view $df(x)$ as a linear transformation from $\mathbb{R}^m$ to $\mathbb{R}^n$. And it must satisfy a similar approximation condition, replacing the absolute value with the notion of length in $\mathbb{R}^n$: for every $\epsilon>0$ there is a $\delta>0$ so that if $\delta>\lVert t\rVert_{\mathbb{R}^m}>0$ we have

$\displaystyle\lVert\left[f(x+t)-f(x)\right]-df(x)t\rVert_{\mathbb{R}^n}<\epsilon\lVert t\rVert_{\mathbb{R}^m}$

From here on we’ll just determine which norm we mean by context, since we only have one norm on each vector space.

Okay, so we can talk about differentials of vector-valued functions: the differential of $f$ at a point $x$ (if it exists) is a linear transformation $df(x)$ that turns displacements in the input space into displacements in the output space, and does so in the way that most closely approximates the action of the function itself. But how do we define the function?

If we pick an orthonormal basis $e_i$ for $\mathbb{R}^n$ we can write the components of $f$ as separate functions. That is, we say

$\displaystyle f(x)=f^i(x)e_i$

Now I assert that the differential can be taken component-by-component, just as continuity works: $df(x)=df^i(x)e_i$. On the left is the differential of $f$ as a vector-valued function, while on the right we find the differentials of the several real-valued functions $f^i$. The differential exists if and only if the component differentials do.

First, from components to the vector-valued function. Clearly this definition of $df(x)$ gives us a linear map from displacements in $\mathbb{R}^m$ to displacements in $\mathbb{R}^n$. But does it satisfy the approximation inequality? Indeed, for every $\epsilon>0$ we can find a $\delta$ so that all the inequalities

$\displaystyle\left\lvert\left[f^i(x+t)-f^i(x)\right]-df^i(x)t\right\rvert<\frac{\epsilon}{n}\lVert t\rVert$

are satisfied when $\delta>\lVert t\rVert>0$. Of course, there are different $\delta$s that work for each component, but we can pick the smallest of them. Then it’s a simple matter to find

$\displaystyle\begin{aligned}\left\lVert\left[f(x+t)-f(x)\right]-df(x)t\right\rVert&=\left\lVert\left(\left[f^i(x+t)-f^i(x)\right]-df^i(x)t\right)e_i\right\rVert\\&\leq\left\lvert\left[f^i(x+t)-f^i(x)\right]-df^i(x)t\right\rvert\lVert e_i\rVert\\&=\sum\limits_{i=1}^n\left\lvert\left[f^i(x+t)-f^i(x)\right]-df^i(x)t\right\rvert\\&<\sum\limits_{i=1}^n\frac{\epsilon}{n}\lVert t\rVert\\&=\epsilon\lVert t\rVert\end{aligned}$

so if the component functions are differentiable, then so is the function as a whole.

On the other hand, if the differential $df(x)$ exists then for every $\epsilon>0$ there exists a $\delta>0$ so that if $\delta>\lVert t\rVert>0$ we have

$\displaystyle\lVert\left[f(x+t)-f(x)\right]-df(x)t\rVert<\epsilon\lVert t\rVert$

But then it’s easy to see that

$\displaystyle\begin{aligned}\left\lvert\left[f^k(x+t)-f^k(x)\right]-df^k(x)t\right\rvert&=\sqrt{\left\lvert\left[f^k(x+t)-f^k(x)\right]-df^k(x)t\right\rvert^2}\\&\leq\sqrt{\sum\limits_{i=1}^n\left\lvert\left[f^i(x+t)-f^i(x)\right]-df^i(x)t\right\rvert^2}\\&=\lVert\left[f(x+t)-f(x)\right]-df(x)t\rVert\\&<\epsilon\lVert t\rVert\end{aligned}$

and so each of the component differentials exists.

Finally, I should mention that if we also pick an orthonormal basis $\tilde{e}_j$ for the input space $\mathbb{R}^m$ we can expand each component differential $df^i(x)$ in terms of the dual basis $dx^j$:

$\displaystyle df^i(x)=\frac{\partial f^i}{\partial x^1}dx^1+\dots+\frac{\partial f^i}{\partial x^m}dx^m=\frac{\partial f^i}{\partial x^j}dx^j$

Then we can write the whole differential $df(x)$ out as a matrix whose entry in the $i$th row and $j$th column is $\frac{\partial f^i}{\partial x^j}$. If we write a displacement in the input as an $m$-dimensional column vector we find our estimate of the displacement in the output as an $n$-dimensional column vector:

$\displaystyle\begin{pmatrix}df^1(x;t)\\\vdots\\df^n(x;t)\end{pmatrix}=\begin{pmatrix}\frac{\partial f^1}{\partial x^1}&\dots&\frac{\partial f^1}{\partial x^m}\\\vdots&\ddots&\vdots\\\frac{\partial f^n}{\partial x^1}&\dots&\frac{\partial f^n}{\partial x^m}\end{pmatrix}\begin{pmatrix}t^1\\\vdots\\t^m\end{pmatrix}$

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Posted by John Armstrong | Analysis, Calculus

## 5 Comments	»

1. Nicely done! Once again, you’ve brought order out of chaos.

Comment by | October 6, 2009 | Reply

2. [...] since each of the several is differentiable we can pick our radius so that all of the [...]

Pingback by | October 7, 2009 | Reply

3. [...] derivatives of . This works out because when we consider differentials as linear transformations, the matrix entries are the partial derivatives. The composition of the linear transformations and is given by the product of these matrices, and [...]

Pingback by | October 8, 2009 | Reply

4. [...] just as we did for vector-valued functions, we’ll just take the differentials of each of these components separately, and then cobble [...]

Pingback by | October 16, 2009 | Reply

5. [...] some open region in . That is, if we pick a basis and coordinates of , then the function is a vector-valued function of real variables with components . The differential, then, is itself a vector-valued function [...]

Pingback by | November 11, 2009 | Reply

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