Sufficient conditions for differentiability in multi-variable calculus.

We will be focusing on sufficient conditions of differentiability of f:\Bbb{R}^2\to R. The theorem says that if f_x and f_y exist and are continuous at point (a,b), then f is differentiable at (a,b).

We have f(x,y), which we know is partially differentiable with respect to x and y, but may not be differentiable in general.

Differentiability at point a in the \Bbb{R}^n\to \Bbb{R} setting is described as \frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=\epsilon(h)

D is the matrix of partial derivatives with respect to the n independent variables in \Bbb{R}^n.

What does all this mean? This is something that confused me for some time, and is likely to be helpful for others with the same doubts.

This is the \epsilon-\delta definition of differentiation. We say \lim\limits_{h\to 0} \frac{f(a+h)-f(a)}{h}, if it exists, is the derivative of f at a. Let the limit be l. We’re effectively saying that  for any \epsilon>0, \left|\frac{f(a+h)-f(a)}{h}-l\right|<\epsilon if 0<h<\delta. Here h=|x-a|. However, this is not what we SEEM to say through this definition. What we seem to be saying is for any \delta>0 and 0<h<\delta, there exists \epsilon>0 such that \frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=\epsilon(h). Is there a difference? Yes. This will be illustrated below.

In our example of multi-variable differentiation, we’ve made \epsilon a function of h. Why is that? What we mean by that is not that \epsilon can be any function of h, like h+c, where c is a constant. What we mean is \lim\limits_{h\to 0} \epsilon=0, although this is not obvious from the fact that \epsilon is a function of h. Why should we explicitly mention the fact that \epsilon should tend to 0 as h\to 0? Because in the original definition \frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=\epsilon(h) we have made the argument that for any \delta>0 and 0<h<\delta, \exists \epsilon such that the difference between \frac{f(a+h)-f(a)}{\|h\|} and D \frac{h}{\|h\|} is \epsilon. Here, \epsilon may not converge to 0! We have just proven its existence, and none of its properties! For example, we could have said that for any \delta>0 and 0<h<\delta, \frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=2. Here \epsilon=2. We have proven the existence of \epsilon.

It is only when we specify that \lim\limits_{h\to 0} \epsilon=0 that we make the definition of the derivative clear- that it is \lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{\|h\|}. Such a limit is defined when for any \epsilon>0, there exists \delta>0 such that \left |\frac{f(a+h)-f(a)}{\|h\|}-l\right|<\epsilon when 0<h<\delta.

Now we come back to sufficient conditions of differentiability. Let a function f(x,y) be differentiable with respect to x and y at (a,b). This implies

f(a+\Delta x,b)-f(a,b)=\epsilon(\Delta x)\Delta x + f_x(a,b)\Delta x


f(a,b+\Delta y)-f(a,b)=\epsilon(\Delta y)\Delta y + f_y(a,b)\Delta y.

Adding these two, we get

f(a+\Delta x,b)+f(a,b+\Delta y)-2f(a,b)=\epsilon(\Delta x)\Delta x+\epsilon(\Delta y)\Delta y + f_x(a,b)\Delta x+f_y(a,b)\Delta y.

Can we say

f(a+\Delta x,b)-f(a,b)\approx f(a+\Delta x,b+\Delta y)-f(a,b+\Delta y),

assuming \Delta x and \Delta y are small enough? We have

f(a+\Delta x,b)+f(a,b+\Delta y)-2f(a,b)=\epsilon(\Delta x)\Delta x+\epsilon(\Delta y)\Delta y+f_x(a,b)\Delta x+f_y(a,b)\Delta y.

Now we use the property that the partial derivatives are continuous.

\frac{f(a+\Delta x,b)-f(a,b)}{h}=\frac{f(a+\Delta x,b+\Delta y)-f(a,b+\Delta y)}{h}\implies f(a+\Delta x,b)-f(a,b)=f(a+\Delta x,b+\Delta y)-f(a,b+\Delta y)+o(h).

o(h) is the correction factor which is utimately rectified. Remember that here h=\Delta x or \Delta y. It can’t be a combination of both, as only partial derivatives f_x,f_y are continuous.

We now have the formula

f(a+\Delta x,b+\Delta y)=f_x(a,b)\Delta x+f_y(a,b)\Delta y+\epsilon(\Delta x)\Delta x+\epsilon(\Delta y)\Delta y+o(h).

The rest of the proof is elementary, and can be found in any complex anaysis textbook (pg. 67 of Complex Variabes and Applications, Brown and Churchill). I have only explained the difficult step in the proof.

Reiterating the theorem, if f is is partially differentiable with respect to its independent variables at a particular point, and all those partial derivatives are continuous, then f is differentiable at that point.

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Graduate student

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