Suppose ddz(f(z)) exists at a point z. Is ddz(f(z)) continuous at z?

Reginald Metcalf

Reginald Metcalf

Answered

2022-01-17

Suppose ddz(f(z)) exists at a point z. Is ddz(f(z)) continuous at z?

Answer & Explanation

Linda Birchfield

Linda Birchfield

Expert

2022-01-18Added 39 answers

Intuitively, you would maybe think so but the answer is no.
As a counterexample, consider
g(x):={x2sin(1/x),x00x=0
Then the derivative is
g(x)={2x2sin(1/x)cos(1/x),x00x=0
Note that limx0g(x) does not exist, even though g(0)=0.

zesponderyd

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(z)=|z|2=zz, which is only differentiable at zero; indeed, the Wirtinger derivative is
fz=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
limz0f(z)f(0)z0=limz0z=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.

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