Reginald Metcalf

Answered

2022-01-17

Suppose $\frac{d}{dz}\left(f\left(z\right)\right)$ exists at a point z. Is $\frac{d}{dz}\left(f\left(z\right)\right)$ continuous at z?

Answer & Explanation

Linda Birchfield

Expert

2022-01-18Added 39 answers

Intuitively, you would maybe think so but the answer is no.

As a counterexample, consider

Then the derivative is

Note that

zesponderyd

Expert

2022-01-19Added 42 answers

If you are thinking to functions over the complex numbers, it is a celebrated result of Goursat that if the derivative exists over an open set, then the function is analytic over the open set; in particular its derivative is analytic as well, in particular differentiable and, therefore, continuous.

However, the derivative might not exist over an open set. An example is the function$f\left(z\right)={\left|z\right|}^{2}=z\stackrel{\u2015}{z}$ , which is only differentiable at zero; indeed, the Wirtinger derivative is

$\frac{\partial f}{\partial \stackrel{\u2015}{z}}=z$

and a function can be differentiable (over the complex numbers) only where the Wirtinger derivative vanishes (it’s just the Cauchy-RIemann condition expressed in a different way). The derivative actually exists at zero, because

$\underset{z\to 0}{lim}\frac{f\left(z\right)-f\left(0\right)}{z-0}=\underset{z\to 0}{lim}\stackrel{\u2015}{z}=0$

The derivative only exists at a point so it is continuous over its domain, but I guess it’s not the notion of continuity you were thinking to.

However, the derivative might not exist over an open set. An example is the function

and a function can be differentiable (over the complex numbers) only where the Wirtinger derivative vanishes (it’s just the Cauchy-RIemann condition expressed in a different way). The derivative actually exists at zero, because

The derivative only exists at a point so it is continuous over its domain, but I guess it’s not the notion of continuity you were thinking to.

Most Popular Questions