Reginald Metcalf

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?

Linda Birchfield

Expert

Intuitively, you would maybe think so but the answer is no.
As a counterexample, consider
$g\left(x\right):=\left\{\begin{array}{ll}{x}^{2}\mathrm{sin}\left(1/x\right),& x\ne 0\\ 0& x=0\end{array}$
Then the derivative is
${g}^{\prime }\left(x\right)=\left\{\begin{array}{ll}2{x}^{2}\mathrm{sin}\left(1/x\right)-\mathrm{cos}\left(1/x\right),& x\ne 0\\ 0& x=0\end{array}$
Note that $\underset{x\to 0}{lim}{g}^{\prime }\left(x\right)$ does not exist, even though ${g}^{\prime }\left(0\right)=0$.

zesponderyd

Expert

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 $f\left(z\right)={|z|}^{2}=z\stackrel{―}{z}$, which is only differentiable at zero; indeed, the Wirtinger derivative is
$\frac{\partial f}{\partial \stackrel{―}{z}}=z$
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
$\underset{z\to 0}{lim}\frac{f\left(z\right)-f\left(0\right)}{z-0}=\underset{z\to 0}{lim}\stackrel{―}{z}=0$
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.

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