Cauchy Estimates

In mathematics, specifically in complex analysis, Cauchy's estimate gives local bounds for the derivatives of a holomorphic function. These bounds are optimal.

Cauchy's estimate is also called Cauchy's inequality, but must not be confused with
the Cauchy–Schwarz inequality.

Statement and consequence

Let $f$ be a holomorphic function on the open ball $B(a, r)$ in $\mathbb C$. If $M$ is the sup of $, f,$ over $B(a, r)$, then Cauchy's estimate says: for each integer $n > 0$,
:$, f^(a), \le \frac M$
where $f^$ is the ''n''-th complex derivative of $f$; i.e., $f' = \frac$ and $f^ = (f^)^'$ (see ).

Moreover, taking $f(z) = z^n, a = 0, r = 1$ shows the above estimate cannot be improved.

As a corollary, for example, we obtain Liouville's theorem, which says a bounded entire function is constant (indeed, let $r \to \infty$ in the estimate.) Slightly more generally, if $f$ is an entire function bounded by $A + B, z, ^k$ for some constants $A, B$ and some integer $k > 0$, then $f$ is a polynomial.

Proof

We start with Cauchy's integral formula applied to $f$, which gives for $z$ with $, z - a , < r'$,
:$f(z) = \frac \int_ \frac \, dw$
where $r' < r$. By the differentiation under the integral sign (in the complex variable), we get:
:$f^(z) = \frac \int_ \frac \, dw.$
Thus,
:$, f^(a), \le \frac \int_ \frac = \frac.$
Letting $r' \to r$ finishes the proof. $\square$

(The proof shows it is not necessary to take $M$ to be the sup over the whole open disk, but because of the maximal principle, restricting the sup to the near boundary would not change $M$.)

Related estimate

Here is a somehow more general but less precise estimate. It says: given an open subset $U \subset \mathbb$, a compact subset $K \subset U$ and an integer $n > 0$, there is a constant $C$ such that for every holomorphic function $f$ on $U$,
:$\sup_ , f^, \le C \int_U , f, \, d\mu$
where $d\mu$ is the Lebesgue measure.

This estimate follows from Cauchy's integral formula (in the general form) applied to $u =\psi f$ where $\psi$ is a smooth function that is $=1$ on a neighborhood of $K$ and whose support is contained in $U$. Indeed, shrinking $U$, assume $U$ is bounded and the boundary of it is piecewise-smooth. Then, since $\partial u / \partial \overline = f \partial \psi / \partial \overline$, by the integral formula,
:$u(z) = \frac \int_ \frac \, dw + \frac \int_U \frac \, dw \wedge d\overline$
for $z$ in $U$ (since $K$ can be a point, we cannot assume $z$ is in $K$). Here, the first term on the right is zero since the support of $u$ lies in $U$. Also, the support of $\partial \psi/\partial \overline$ is contained in $U - K$. Thus, after the differentiation under the integral sign, the claimed estimate follows.

As an application of the above estimate, we can obtain the Stieltjes–Vitali theorem, which says that a sequence of holomorphic functions on an open subset $U \subset \mathbb$ that is bounded on each compact subset has a subsequence converging on each compact subset (necessarily to a holomorphic function since the limit satisfies the Cauchy–Riemann equations). Indeed, the estimate implies such a sequence is equicontinuous on each compact subset; thus, Ascoli's theorem and the diagonal argument give a claimed subsequence.

In several variables

Cauchy's estimate is also valid for holomorphic functions in several variables. Namely, for a holomorphic function $f$ on a polydisc $U = \prod_^n B(a_j, r_j) \subset \mathbb^n$, we have: for each multiindex $\alpha \in \mathbb^n$,
:$\left , \left(\frac^ f\right) (a) \ \le \frac \sup_U , f,$
where $a = (a_1, \dots, a_n)$, $\alpha! = \prod _j!$ and $r^ = \prod r_j^$.

As in the one variable case, this follows from Cauchy's integral formula in polydiscs. and its consequence also continue to be valid in several variables with the same proofs.

See also

*Taylor's theorem

References

*
*

Further reading

* https://math.stackexchange.com/questions/114349/how-is-cauchys-estimate-derived/114363

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Complex analysis