Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck with the following exercise. I'm trying to show that the functional $J$ on the set of holomorphic functions with $J(f)=f^{(p)}(z_0)$ is continuous. But I have no approach on the continuity of a function of functions. If we change a function $f$ slightly, how do we know that it will give only a small change in $f(z_0)$?

Thanks for help!

share|cite|improve this question
Holomorphic functions are from $\mathbb{C}$ to $\mathbb{C}$, but a topology is not mentioned explicitly; and yes, that $\Delta$ does not work too well here. – Marie. P. Nov 23 '11 at 0:01
I'm guessing the topology is uniform convergence on all compacts. that result is a consequence of a classical convergence theorem, I think it's called weierstrass theorem (unsure) – Glougloubarbaki Nov 23 '11 at 0:06
up vote 1 down vote accepted

We can define a metric $d$ on the set $\mathcal O(\mathbb C)$ of holomorphic functions: $$d(f,g):=\sum_{n=1}^{+\infty}\frac 1{2^n}\min\left(1,\sup_{z\in B_n}|f(z)-g(z)|\right),$$ where $B_n=\{z\in\mathbb C, |z|\leq n\}$. The topology given by this metric is equivalent to the topology of the uniform convergence on compact sets. Indeed, if $\{f_k\}$ is a sequence of holomorphic functions on $\mathbb C$ which converges to $f$ on each compact sets, then fix $\varepsilon>0$ and pick $n_0$ such that $\sum_{n\geq n_0+1}2^{-n}\leq\frac{\varepsilon}2$. Since $\{f_k\}$ converges uniformly to $f$ on $B_{n_0}$, we can pick for each $n\leq n_0$ an integer $j_n$ such that for all $k\geq j_n$, $2^{-n}\min\left(1,\sup_{z\in B_n}|f_k(z)-f(z)|\right)\leq \frac{\varepsilon}{2n_0}$. Hence for all $k\geq k_0:=\max_{1\leq n\leq n_0}j_n$, we have $d(f_k,f)\leq\varepsilon$. Conversely, if $d(f_k,f)$ converges to $0$, and $K\subset\mathbb C$ is compact, then take $n$ such that $K\subset B_n$. We have $$d(f_k,f)\geq 2^{-n}\min\left(1,\sup_{z\in B_n}|f_k(z)-f(z)|\right)=2^{-n}\sup_{z\in B_n}|f_k(z)-f(z)|$$ for $k$ large enough, hence $f_k\to f$ uniformly on $K$.

Now, we prove the sequential continuity of $J$. Let $\{f_k\}\subset\mathcal O(\mathbb C)$ and $f\in \mathcal O(\mathbb C)$ such that $d(f_k,f)\to 0$. Thanks to the Cauchy integral formula we have, if $B(z_0,1)\subset B_N$ $$| J(f_k)-J(f)|=|(f_k-f)^{(p)}(z_0)|=\left|\frac {p!}{2\pi i}\int_{C(z_0,1)}\frac{f_k(z)-f(z)}{(z-z_0)^{p+1}}dz\right|\leq p!\sup_{z\in B_N}|f_k(z)-f(z)|,$$ hence $$\min(1, | J(f_k)-J(f)|)\leq p! 2^Nd(f_k,f).$$ Since $d(f_k,f)\to 0$, we have $| J(f_k)-J(f)|\leq 2^{-1}$ for $k$ large enough (say $k\geq k_0$) therefore $$\forall k\geq k_0\quad | J(f_k)-J(f)|\leq p!2^Nd(f_k,f),$$ which shows that $\lim_{k\to\infty}| J(f_k)-J(f)|=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.