# derivative continuity versus analyticity

While presenting Cauchy's theorem in their well-known textbook, the authors chose to state a stronger condition on the analytical function $$f(z)$$ that require it to be not only analytic within the simple domain $$D$$ and on its contour $$C$$, but also that the derivative $$f'(z)$$ is continuous on the same complex domain. They admitted that this condition is stronger than needed (only analyticity of $$f(z)$$ is actually needed) for the theorem, and that they gave such version for simpler proof within their book's scope.

However, this is a bit confusing to me, because the authors are here clearly distinguishing between being "analytic" versus being "of continuous derivative $$f'(z)$$" on all points of the domain under question.

But I had thought that being analytic would automatically mean that the derivative function $$f'(z)$$ [as a limit by definition, $$\lim\limits_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}$$] exists and that it exists uniquely (regardless of the direction in which $$h$$ approaches zero) at each point. And if I recall the basic requirement for a function to be continuous at any point, it is that the limit $$\lim\limits_{x\rightarrow x_{0}}f(x)$$ exists (same value from all directions) and is equal to the function's value there [$$\lim\limits_{x\rightarrow x_{0}}f(x)=f(x_{0})$$]. Similarly, a function $$f(x)$$ is differentiable if it has its derivative $$f'(x)$$ continuous using the same definiton just mentioned (again as existent limit from all directions)...

Furthermore, more modern definitions of the term "analytic" (e.g. here and here) refers to being representable as Taylor series, which again implies having existent higher derivatives everywhere in the domain, which therefore must mean that these derivatives are themselves continuous functions.

So drawing a distinction between being "analytic" and being "of continuous derivative" is somewhat confusing here. Any help to clarify this would be helpful.

• They define analytic as having complex derivatives, not as a power series. At the beginning it is not known if its derivatives are continuous. There are plenty of examples of functions that are (real) differentiable with discontinuous derivatives. So, who knows, maybe it also happens with complex derivatives. And they are doing the Green's theorem proof. That is why the assumed continuity of derivatives. They are writing a book for engineers. I guess that is why they opt for a proof not as general but fast that follows form a computation and a formula. – conditionalMethod Nov 15 '19 at 19:16
• @conditionalMethod "they define analytic as having complex derivatives, not a power series" - I think this is all the more a reason to the confusion I mentioned. Also, I think being differentiable over a domain's points disallows having discontinuous derivatives there (if not, the limit at each point wouldn't exist in the first place). – user135626 Nov 15 '19 at 19:29
• Having continuous derivatives has to be proven and what you are saying is not a proof of it. One of the proofs is using Cauchy's theorem. At that moment of proving Cauchy's theorem, the continuity of the derivatives is still not known. – conditionalMethod Nov 15 '19 at 19:45

In the case of functions on the real line, differentiability does not imply continuity of the derivative. A standard example is $$f(x) = x^2 \sin(1/x)$$ If $$x \ne 0$$, $$f'(x) = 2 x \sin(1/x) - \cos(1/x)$$. This does not have a limit as $$x \to 0$$. On the other hand, $$f'(0)$$ does exist and is $$0$$, as can be seen from the limit definition of derivative and the fact that $$|f(x)| \le x^2$$.
I think your confusion arises from the fact that there are two quite different limits here: $$\lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = f'(z)\ \text{the definition of f'(z)}$$ and $$\lim_{h \to 0} f'(z+h) = f'(z) \ \text{for its continuity}$$ In the case of functions on the reals, you can have one without the other.
• Many thanks for this answer. However, may I please ask about how could we find $f'(0)$ to exist and be 0 from the limit definition and $|f|\leq x^{2}$ in the method you indicated (I am actually having difficulty working this out)? – user135626 Nov 15 '19 at 20:38