The Unapologetic Mathematician

Mathematics for the interested outsider

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



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