# 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.

September 29, 2009 Posted by | Analysis, Calculus | 10 Comments