The Unapologetic Mathematician

Mathematics for the interested outsider

An Existence Condition for the Differential

To this point we’ve seen what happens when a function f does have a differential at a given point x, but we haven’t yet seen any conditions that tell us that any such function df(x;t) 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

\displaystyle df(x;t)=\left[D_if\right](x)t^i

where D_if is the partial derivative of f in the ith coordinate direction. This is clearly linear in the displacement t, 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 x, 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 x. 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 D_if exist and are continuous at x. We have to show that for any \epsilon>0 there is some \delta>0 so that if \delta>\lVert t\rVert>0 we have the inequality

\displaystyle\left\lvert\left[f(x+t)-f(x)\right]-\left[D_if\right](x)t^i\right\rvert<\epsilon\lVert t\rVert

We’re going to take the difference f(x+t)-f(x) and break it into n terms, each of which will approximate one of the partial derivative terms.

First off, since each D_if is continuous at x, there is some \delta so that if \lVert t\rVert<\delta then \lvert\left[D_if\right](x+t)-\left[D_if\right](x)\rvert<\frac{\epsilon}{n}. In fact, there’s a \delta for each index i, 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 \lVert t\rVert is actually less than \frac{\delta}{2}. We’ll write t=\lambda u, where u is a unit vector and \lambda is a scalar so that \lvert\lambda\rvert=\lVert t\rVert<\frac{\delta}{2}. We’ll also write u in terms of our orthonormal basis u=u^ie_i.

Now we can build up our displacement direction u step-by-step as a sequence of vectors v_0=0, v_1=u^1e_1, and so on, stepping in the ith direction on the ith step: v_k=v_{k-1}+u^ke_k (not summing on k here). So we can break up the difference of function values as

\displaystyle f(x+\lambda u)-f(x)=\sum\limits_{k=1}^n\left[f(x+\lambda v_{k-1}+\lambda u^ke_k)-f(x+\lambda v_{k-1})\right]

So now each step only changes the kth coordinate, and the points at each end both lie within the ball of radius \frac{\delta}{2} around x, since each v_k is shorter than u, which has unit length. To look closer at the step from f(x+\lambda v_{k-1}) to f(x+\lambda v_{k-1}+\lambda u^ke_k), we introduce a new function of one real variable:

\displaystyle g(\alpha)=f(x+\lambda v_{k-1}+\alpha e_k)

for -\lvert\lambda u^k\rvert\leq\alpha\leq\lvert\lambda u^k\rvert. This lets us write our step as g(\lambda u^k)-g(0). It turns out that everywhere in this closed interval, the function g is differentiable! Indeed, we have

\displaystyle\frac{g(\alpha+h)-g(\alpha)}{h}=\frac{f(x+\lambda v_{k-1}+\alpha e_k+he_k)-f(x+\lambda v_{k-1}+\alpha e_k)}{h}

So as h goes to zero, we find g'(\alpha)=\left[D_kf\right](x+\lambda v_{k-1}+\alpha e_k), which exists because we’re in a small enough ball around x. Now the mean value theorem can be brought to bear, which says

\displaystyle g(\lambda u^k)-g(0)=\lambda u^kg'(\alpha_k)

for some -\lvert\lambda u^k\rvert\leq\alpha_k\leq\lvert\lambda u^k\rvert. And now the difference of function values can be written

\displaystyle\begin{aligned}f(x+t)-f(x)&=\lambda\sum\limits_{k=1}^nu^k\left[D_kf\right](x+\lambda v_{k-1}+\alpha_ke_k)\\&=\sum\limits_{k=1}^n\left[D_kf\right](x)t^k+\lambda\sum\limits_{k=1}^nu^k\left[\left[D_kf\right](x+\lambda v_{k-1}+\alpha_ke_k)-\left[D_kf\right](x)\right]\end{aligned}

since t^k=\lambda u^k.

Now \lvert\lambda v_{k-1}+\alpha_ke_k\rvert\leq\lvert\lambda\rvert+\lvert\lambda u^k\rvert<2\lvert\lambda\rvert<\delta, and so we find that the each of these differences of partial derivative evaluations is less than \frac{\epsilon}{n}. And thus

\displaystyle\left\lvert\left[f(x+t)-f(x)\right]-\sum\limits_{k=1}^n\left[D_kf\right](x)t^k\right\rvert<\lvert\lambda\rvert\epsilon=\epsilon\lVert t\rVert

which establishes the inequality we need.

October 1, 2009 - Posted by | Analysis, Calculus


  1. […] are both constant, they’re clearly continuous everywhere. Thus by the condition we worked out yesterday the differential of exists, and we […]

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  4. Hi, I was trying to follow your posts on multivariate differentiation (your blog is a great resource!) but I got stumped by the condition “what will be sufficient is to ask that not only do all partial derivatives exist at x, but that they themselves are continuous there.”

    Consider the function f(x,y) whose value is 0 at (0,0) and xy/(x^2+y^2) at other places. That function is not differentiable at (0,0). But the partial derivatives do seem to exist and be continuous there. When y=0, Dxf is 0 for all x. When x=0, Dyf if 0 for all y.

    Where I’m going wrong?

    Comment by Daniel Díaz Carrete | July 25, 2010 | Reply

  5. The function isn’t even continuous at the origin, so the partial derivatives can’t exist there. Further, you’re not calculating the partial derivatives, but rather the derivatives along two particular lines. The partial derivative is something that exists (or at least tries to exist) on the entire domain of f. For example, in this case we have

    \displaystyle\frac{\partial f}{\partial x}=\frac{y^3-x^2y}{(x^2+y^2)^2}

    You can easily verify that this is undefined at the origin, but you should also go back and see why the original function isn’t continuous there.

    Comment by John Armstrong | July 25, 2010 | Reply

  6. Suppose we “patch” the function at the origin by specifying f(0,0)=0 and then apply the limit definition of the derivative to find the partial derivatives there. Both of them seem to exist and have value 0 at (0,0).

    But if df/dx is 0 at (0,0) and 0 anywhere else along the line y=0, then it should be continuous at (0,0) shouldn’t it?

    Applying the same reasoning to df/dy, we have that both partial derivatives exist at (0,0) and are continuous there. Shouldn’t the function be differentiable at the origin, then? And yet it isn’t even continuous there.

    Comment by Daniel Díaz Carrete | July 25, 2010 | Reply

  7. It’s not continuous. There is no “patch” that will work. The derivatives simply do not exist there.

    Comment by John Armstrong | July 25, 2010 | Reply

  8. But is multivariate continuity at a point neccessary for the existence of partial derivatives at that point?

    In your post about partial derivatives you gave an example of a function which has partial derivatives at (0,0) but is not continuous there.

    Comment by Daniel Díaz Carrete | July 25, 2010 | Reply

  9. You’re right, I was wrong to phrase it like that.

    But in that example the partial derivatives exist but are not continuous there. And yes when I say “continuous” in the post above I mean “continuous”, not just “continuous along whatever special curves I carefully select.”

    Comment by John Armstrong | July 25, 2010 | Reply

  10. […] nos vamos a detener en demostrar este teorema. Dicha demostración aparece, por ejemplo, en este post del blog The Unapologetic Mathematician o en la página 168 del libro de Análisis Matemático de Carlos Ivorra (Teorema […]

    Pingback by Confusión entre necesario y suficiente: el caso de la diferenciabilidad - Gaussianos | Gaussianos | July 19, 2011 | Reply

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