To this point we’ve seen what happens when a function does have a differential at a given point , but we 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 the differential must be given by the formula
where is the partial derivative of in the th coordinate direction. This is clearly linear in the displacement , so all that remains is to see whether the inequality in the definition of the differential can be satisfied.
We must have all partial derivatives to write down this formula, but that can’t be sufficient for differentiability, because if it were then having all partial derivatives would imply continuity, and we know that it doesn’t. What will be sufficient is to ask that not only do all partial derivatives exist at , but that they themselves are continuous there. Note, though, that I’m not asserting that this condition is not necessary for a function to be differentiable. Indeed, it’s possible to construct differentiable functions whose partial derivatives all exist, but are not continuous at . This is an example of the way that analysis tends to be shot through with “counterexamples”, as Michael was talking about recently.
Okay, so let’s assume that all these partial derivatives exist and are continuous at . We have to show that for any there is some so that if we have the inequality
We’re going to take the difference and break it into terms, each of which will approximate one of the partial derivative terms.
First off, since each is continuous at , there is some so that if then . In fact, there’s a for each index , but we can just take the smallest of all these, and that one will work for each index. From this point on, we’ll assume that is actually less than . We’ll write , where is a unit vector and is a scalar so that . We’ll also write in terms of our orthonormal basis .
Now we can build up our displacement direction step-by-step as a sequence of vectors , , and so on, stepping in the th direction on the th step: (not summing on here). So we can break up the difference of function values as
So now each step only changes the th coordinate, and the points at each end both lie within the ball of radius around , since each is shorter than , which has unit length. To look closer at the step from to , we introduce a new function of one real variable:
for . This lets us write our step as . It turns out that everywhere in this closed interval, the function is differentiable! Indeed, we have
So as goes to zero, we find , which exists because we’re in a small enough ball around . Now the mean value theorem can be brought to bear, which says
for some . And now the difference of function values can be written
Now , and so we find that the each of these differences of partial derivative evaluations is less than . And thus
which establishes the inequality we need.