# The Unapologetic Mathematician

## Uniqueness of the Differential

Okay, for the moment let’s pick an orthonormal basis $\left\{e_i\right\}_{i=1}^n$ for our vector space $\mathbb{R}^n$. This gives us coordinates on the Euclidean space of points. It also gives us the dual basis $\left\{\eta^i\right\}_{i=1}^n$ of the dual space $\left(\mathbb{R}^n\right)^*$. This lets us write any linear functional $\lambda:\mathbb{R}^n\rightarrow\mathbb{R}$ as a unique linear combination $\lambda=\lambda_i\eta^i$. The component $\lambda_i$ measures how much weight we give to the distance a vector extends in the $e_i$ direction.

Now if we look at a particular point $x$ we can put it into our differential and leave the second (vector) slot blank: $df(x;\hphantom{\underline{x}})$. We will also write this simply as $df(x)$, and apply it to a vector by setting the vector just to its right: $df(x;t)=df(x)t$. Now $df(x)$ is a linear functional, and we can regard $df$ as a function from our space of points to the dual of the space of displacements. We can thus write it out uniquely in components $df(x)=\lambda_i(x)\eta^i$, where each $\lambda_i$ is a function of the point $x$, but not of the displacement $t$.

We want to analyze these components. I assert that these are just the partial derivatives in terms of the orthonormal basis we’ve chosen: $\lambda_i(x)=\left[D_{e_i}f\right](x)$. In particular, I’m asserting that if the differential exists, then the partial derivatives exist as well.

By the definition of the differential, for every $\epsilon>0$ there is a $\delta>0$ so that if $\delta>\lVert t\rVert>0$, then

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

Now we can write $df(x;t)$ out in components

$\displaystyle df(x;t)=\lambda_i(x)t^i$

Next for a specific index $k$ we can pick $t=\tau e_k$ for some value $\delta>\lvert\tau\rvert>0$. Then $\lVert t\rVert=\lvert\tau\rvert$, $t^k=\tau$, and $t^i=0$ for all the other indices $i\neq k$. Putting all these and the component representation of $df(x;t)$ into the definition of the differential we find

$\displaystyle\lvert\left[f(x+\tau e_k)-f(x)\right]-\lambda_k(x)\tau\rvert<\epsilon\lvert\tau\rvert$

Dividing through by $\lvert\tau\rvert$ we find

$\displaystyle\left\lvert\frac{f(x+\tau e_k)-f(x)}{\tau}-\lambda_k(x)\right\rvert<\epsilon$

And this is exactly what we need to find that $\left[D_{e_k}f\right](x)$ exists and equals $\lambda_k(x)$.

Therefore if the function $f$ has a differential $df(x)$ at the point $x$, then it has all partial derivatives there, and these uniquely determine the differential at that point.

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Posted by John Armstrong | Analysis, Calculus

## 10 Comments	»

1. [...] of showing that the differential of a function at a point — if it exists at all — is unique (and thus we can say “the” differential), we showed that given an orthonormal basis we [...]

Pingback by | September 30, 2009 | Reply

2. [...] haven’t yet seen any conditions that tell us that any such function exists. We know from the uniqueness proof that if it does exist, then given an orthonormal basis we have all partial derivatives, and [...]

Pingback by | October 1, 2009 | Reply

3. minor typo: epsilon-eta confusion at beginning. A akways wonder if this sort of thign is worth pointing out

Comment by Avery Andrews | October 1, 2009 | Reply

4. You may as well, since it’s easy enough to correct. Usually that sort of thing happens when I go back to review my earlier pieces, use old notation, then later in the process of writing decide to change it. The editor for WordPress isn’t the best, but I have to use it if I’m going to get any sort of preview.

Comment by | October 1, 2009 | Reply

5. [...] showed that these partial derivatives are the components of the differential (when it exists), and so there should be some connection between the two [...]

Pingback by | October 5, 2009 | Reply

6. [...] Chain Rule Since the components of the differential are given by partial derivatives, and partial derivatives (like all single-variable derivatives) [...]

Pingback by | October 7, 2009 | Reply

7. [...] and is given by the product of these matrices, and the entries of the resulting matrix must (by uniqueness) be the partial derivatives of the composite [...]

Pingback by | October 8, 2009 | Reply

8. [...] by uniqueness we can read off the partial derivatives of in terms of and [...]

Pingback by | October 12, 2009 | Reply

9. [...] First let’s look at the second-order differential of a real-valued function of variables . We’ll use the as a basis for the space of differentials, which allows us to write out the components of the differential: [...]

Pingback by | October 16, 2009 | Reply

10. [...] the partial derivative either does not exist or is equal to zero at . And because the differential subsumes the partial derivatives, if any of them fail to exist the differential must fail to exist as well. On the other hand, if [...]

Pingback by | November 23, 2009 | Reply

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