## 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 we need to set up a new function that takes a point in the Euclidean space and a displacement vector in as inputs, and which gives a displacement vector in as its output. It must be linear in the displacement, meaning that we can view as a linear transformation from to . And it must satisfy a similar approximation condition, replacing the absolute value with the notion of length in : for every there is a so that if we have

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 at a point (if it exists) is a linear transformation 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 for we can write the components of as separate functions. That is, we say

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

First, from components to the vector-valued function. Clearly this definition of gives us a linear map from displacements in to displacements in . But does it satisfy the approximation inequality? Indeed, for every we can find a so that *all* the inequalities

are satisfied when . Of course, there are different s that work for each component, but we can pick the smallest of them. Then it’s a simple matter to find

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

On the other hand, if the differential exists then for every there exists a so that if we have

But then it’s easy to see that

and so each of the component differentials exists.

Finally, I should mention that if we also pick an orthonormal basis for the input space we can expand each component differential in terms of the dual basis :

Then we can write the whole differential out as a matrix whose entry in the th row and th column is . If we write a displacement in the input as an -dimensional column vector we find our estimate of the displacement in the output as an -dimensional column vector:

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

Comment by Jonathan Vos Post | October 6, 2009 |

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