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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!

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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

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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 $\lim_{k\to\infty}| J(f_k)-J(f)|=0$.

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