Note that \(g\) is entire by Morera’s theorem, since \(0 = \int_T f_n \to \int_T g\) by uniform convergence and the \(f_n\) are holomorphic. By Cauchy’s theorem, up to a constant we have \begin{align*} f_n(z) = \oint_T {f_n(\xi) \over \xi - z}\,d\xi g(z) = \oint_T {g(\xi) \over \xi - z}\,d\xi ,\end{align*} Thus fixing \(K\) and \({\varepsilon}\), for any \(T \subseteq K\) containing \(z\), ` \begin{align*} {\left\lvert {f_n(z) - g(z)} \right\rvert} &= {\left\lvert { \oint_T {f_n(\xi) \over \xi - z},d\xi

• \oint_T {g(\xi) \over \xi - z},d\xi } \right\rvert} \ &\leq \oint_T {{\left\lvert {f_n(\xi) - g(\xi)} \right\rvert} \over \xi - z },d\xi\ &\leq \oint_T { \sup_{\xi\in T}{\left\lvert {f_n(\xi) - g(\xi)} \right\rvert} \over \xi - z },d\xi\ &\leq \oint_T { {\varepsilon}\over \xi - z },d\xi\ &= {\varepsilon}C \to 0 ,\end{align*} `{=html} where \(n = n({\varepsilon}, T)\) can be chosen to produce this \({\varepsilon}\) using that \(f_n\to g\) uniformly on \(T\). Taking a sup over the \(z\) enclosed by \(T\) on the LHS yields a bound on the open region enclosed by \(T\). Taking a union of all such \(T\) in \(K\) yields an open cover of \(K\), which by compactness has a finite subcover. This yields a finite collection \(\left\{{n = n({\varepsilon}, T_k)}\right\}_{k\leq N}\), and taking the maximum such \(n\) yields a uniform bound for all of \(K\).

Write \(g(z) \coloneqq\lim_{n\to\infty}f_n(z)\) for the pointwise limit, then \(g(z) = f(z)\) on a set with a limit point. By the identity principle, \(g\equiv f\) on \(\Omega\), making \(f\) the pointwise limit of the \(f_n\).

By Montel, locally uniformly bounded implies normal and locally equicontinuous. So \(\left\{{f_n}\right\}\) is normal, and thus has a locally uniformly convergent subsequence \(\left\{{f_{n_k}}\right\}\). Since singletons \(\left\{{z}\right\}\) are compact, \(f_{n_k}(z) \to g(z)\) pointwise, and by uniqueness of limits, \(\lim_{k\to\infty } f_{n_k} = g = f\) on any compact \(K \subseteq \Omega\).

It remains to show that the original sequence \(\left\{{f_n}\right\}\) converges locally uniformly to \(f\), not just the subsequence. Suppose not, then there exists a compact \(K \subseteq \Omega\) and \({\varepsilon}>0\) so that \({\left\lVert {f_n - f} \right\rVert}_{K, \infty} > {\varepsilon}\) for infinitely many \(n\). This produces a subsequence \(\left\{{f_{n_j}}\right\}\) with \({\left\lVert {f_{n_j} - f} \right\rVert} > {\varepsilon}\) for all \(j\). However, since \({\mathcal{F}}\) was normal, every subsequence has a locally uniformly convergent subsequence, so this has a further subsequence \(f_{n_{j'}}\) uniformly converging to \(f\), a contradiction.

Let \(K\) be compact, where \(z\in K\implies {\left\lvert {z} \right\rvert} \leq R\) for some constant \(R\). For the remainder of the problem, we only work in \(K\).

Why these claims imply the result:

If \(f_n(z)\to z\) uniformly, both are uniformly bounded, and \(e^z\) is uniformly continuous, then \(e^{f(z)}\to e^z\) uniformly.

\(\implies\):

• Fix \(r \in (0, 1)\) and let \(\gamma = \left\{{{\left\lvert {z} \right\rvert} = r}\right\}\). This is compact, so \(f_n\to f\) uniformly on \(\gamma\): \begin{align*} \int_\gamma {\left\lvert {f_n(z) - f(z) } \right\rvert} \,dz &\leq\int_\gamma \sup_{w\in \gamma } {\left\lvert {f_n(w) - f(w) } \right\rvert} \,dz\\ &\leq\int_\gamma {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \,dz\\ &= {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \int_\gamma \,dz\\ &= {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \mathop{\mathrm{length}}(\gamma) \\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}

\(\impliedby\):

• Let \(K\) be compact, then choose \(\gamma\) enclosing but not intersecting \(K\).

• Since \(\gamma, K\) are disjoint compact sets, define \(M \coloneqq\inf \left\{{{\left\lvert {z-\xi} \right\rvert} {~\mathrel{\Big\vert}~}z\in K, \xi\in \gamma}\right\}\), the \(0<M<\infty\).

• Apply Cauchy’s formula to the function \(F_n(z) \coloneqq f_n(z) - f(z)\), where we want to show \({\left\lvert {F_n(z)} \right\rvert} < {\varepsilon}\): \begin{align*} F_n(z) &= {1\over 2\pi i} \int_\gamma { F_n(\xi) \over z-\xi} \,d\xi\\ \implies {\left\lvert {f_n(z) - f(z) } \right\rvert} &\leq {1\over 2\pi }\int_\gamma {\left\lvert {f_n(\xi) - f(\xi) \over z-\xi} \right\rvert} \,d\xi\\ &\leq {1\over 2\pi} \int_\gamma {{\left\lvert { f_n(\xi) - f(\xi) } \right\rvert} \over M} \,d\xi\\ &\leq {1\over 2\pi M} \int_\gamma {{\left\lvert { f_n(\xi) - f(\xi) } \right\rvert} } {\left\lvert {\,d\xi} \right\rvert} \\ ,\end{align*} where by hypothesis we can bound this integral by an \({\varepsilon}\). So given \({\varepsilon}\), choose \(n\) large enough to bound the integral as above by some \({\varepsilon}\) depending only on \(n\) and not on \(z\). Taking \(\sup\) of both sides yields \({\left\lVert {f_n - f} \right\rVert}_{\infty, K} \leq {{\varepsilon}\over 2\pi M}\), so \(f_n\to f\) uniformly on \(K\).

No: polynomials are holomorphic and the uniform limit of holomorphic functions is holomorphic. However, \(f(z) \coloneqq\overline{z}\) is continuous on \(\overline{{\mathbb{D}}}\) but not holomorphic, so can not be uniformly approximated by any sequence of polynomials.

The key insight: \begin{align*} \int_\gamma A \,dz &= \int_0^b A \cdot i \,dt&& z=x+it,\, \,dz= i\,dt\\ &=iA \int_0^b \,dt\\ &= iAb .\end{align*}

So now estimate the difference: \begin{align*} {\left\lvert { \int_{\gamma} f(z) \,dz- iAb } \right\rvert} &= {\left\lvert { \int_\gamma f(z) \,dz- \int_\gamma A \,dz} \right\rvert} \\ &= {\left\lvert { \int_\gamma \qty{ f(z) - A } \,dz} \right\rvert} \\ &\leq\int_\gamma {\left\lvert { f(z) - A } \right\rvert} \,dz\\ &\leq \sup_{z = x+iy\in \gamma} {\left\lvert {f(x+iy) - A} \right\rvert} \cdot \mathop{\mathrm{length}}(\gamma_x) \\ &\overset{x\to \infty}\longrightarrow 0 ,\end{align*} using that \(\mathop{\mathrm{length}}(\gamma_x) = b\) is constant.

Key observation: \begin{align*} iA\alpha = \int_\gamma {A\over z}\,dz .\end{align*} Why this is true: \begin{align*} \int_\gamma {A\over z}\,dz= \int_0^\alpha {1\over Re^{it}} iRe^{it}dt = \int_0^\alpha iA \,dt= iA\alpha .\end{align*}

Now estimate the difference:

\begin{align*} {\left\lvert { \int_\gamma f(z) \,dz- iA\alpha } \right\rvert} &= {\left\lvert { \int_\gamma f(z) \,dz- \int_\gamma {A\over z} \,dz} \right\rvert}\\ &= {\left\lvert {\int_\gamma f(z) - {A\over z} \,dz} \right\rvert} \\ &= {\left\lvert {\int_\gamma{zf(z) - A \over z} \,dz} \right\rvert} \\ &\leq \int_\gamma {\left\lvert {zf(z) - A \over z} \right\rvert} \,dz\\ &= \int_\gamma { {\left\lvert {zf(z) - A} \right\rvert} \over R} \,dz\\ &\leq {1\over R } \int_\gamma {\left\lVert {zf(z) - A} \right\rVert}_{\infty, \gamma} \,dz\\ &= {{\varepsilon}\over R}\cdot \mathop{\mathrm{length}}(\gamma) \\ &= {{\varepsilon}\over R} \cdot R\alpha \\ &= {\varepsilon}\alpha \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}