# C1223 – Complex Analysis

Final Examination

## By Sébastien Boisgérault, MINES ParisTech, under CC BY-NC-SA 4.0

### January 29, 2018

# Problem R

### Preamble.

Let \(\Omega\) be an open subset of \(\mathbb{C}\). For any \(r>0\), we denote \(\Omega_r\) the set of points of \(\mathbb{C}\) whose distance to the complement of \(\Omega\) is larger than \(r\): \[ \Omega_r = \{z \in \mathbb{C} \; | \; d(z, \mathbb{C} \setminus \Omega) > r \}. \]Show that \(\Omega_r\) is an open subset of \(\Omega\) and that \(\Omega = \cup_{r>0} \Omega_r\).

Assume that \(\Omega\) is connected. Is \(\Omega_r\) necessarily connected? (Hint: consider for example \(\Omega = \{z \in \mathbb{C} \; | \; |\mathrm{Im} \, z| < |\mathrm{Re} \, z| +1 \}\) and \(r = 1\)).

Show that if \(z \in \mathbb{C} \setminus \Omega_r\), there is a \(w \in \mathbb{C} \setminus \Omega\) such that the segment \([w, z]\) is included in \(\mathbb{C} \setminus \Omega_r\). Deduce from this property that if \(\Omega\) is simply connected, then \(\Omega_r\) is also simply connected. Is the converse true?

Show that if \(\Omega\) is bounded and simply connected, \(\mathbb{C} \setminus \Omega\) is connected. (Hint: assume that \(\Omega\) is bounded but that \(\mathbb{C} \setminus \Omega\) is disconnected, then introduce a suitable dilation of this complement).

From now on, \(\Omega\) is a bounded open subset of \(\mathbb{C}\). Let \(\mathcal{F}\) be a class of holomorphic functions defined on \(\Omega\) (or a superset of \(\Omega\)). A holomorphic function \(f: \Omega \to \mathbb{C}\) has *uniform approximations* in \(\mathcal{F}\) if \[
\forall \, \epsilon > 0,
\exists \, \hat{f} \in \mathcal{F}, \;
\forall \, z \in \Omega, \; |f(z) - \hat{f}(z)| \leq \epsilon.
\]

Let \(f: \Omega \to \mathbb{C}\) be a holomorphic function and let \(a \in \mathbb{C} \setminus \Omega\). Assume that \(f\) has uniform approximations in the class of functions defined and holomorphic on \(\mathbb{C} \setminus \{a\}\). Show that if \(|a|\) is large enough, \(f\) has uniform approximations among polynomials.

Show that for any non-empty bounded open subset \(\Omega\) of \(\mathbb{C}\), there is a holomorphic function \(f :\Omega \to \mathbb{C}\) which doesn’t have uniform approximations among polynomials (Hint: consider \(z\mapsto 1/(z-a)\) for some suitable choice of \(a\)).

A holomorphic function \(f: \Omega \to \mathbb{C}\) has *locally uniform approximations* in \(\mathcal{F}\) if for any \(r>0\), its restriction to any \(\Omega_r\) has uniform approximations in \(\mathcal{F}\): \[
\forall \, \epsilon > 0, \forall \, r > 0,
\exists \, \hat{f} \in \mathcal{F}, \;
\forall \, z \in \Omega_r, \; |f(z) - \hat{f}(z)| \leq \epsilon.
\]

- Show that if \(\Omega\) is not simply connected, there is a holomorphic function \(f: \Omega \to \mathbb{C}\) which has no locally uniform approximations among polynomials (Hint: consider \(f:z\mapsto 1/(z-a)\) for some suitable choice of \(a\) then compare \[ \int_{\gamma} f(z) \, dz \; \mbox{ and } \; \int_{\gamma} \hat{f}(z) \, dz \] for some suitable closed rectifiable path \(\gamma\)).

From now on, we assume that \(\Omega\) is simply connected.

Let \(f:\Omega \to \mathbb{C}\) and let \(r>0\). We define the function \(\chi_r: \mathbb{C} \setminus \Omega \to \{0,1\}\) by:

\(\chi_r(z) = 1\) if for every \(\epsilon > 0\), there is a holomorphic function \(\hat{f}_z\) defined on \(\mathbb{C} \setminus \{z\}\) such that \(|f - \hat{f}_z| \leq \epsilon\) on \(\Omega_r\),

\(\chi_r(z) = 0\) otherwise.

Show that if some points \(z\) and \(w\) of \(\mathbb{C} \setminus \Omega\) satisfy \(\chi_r(z) = 1\) and \(|w - z| < r/2\) then \(\chi_r(w) = 1\) (Hint: first, prove that the open annulus \(A := A(w, r/2, +\infty)\) satisfies \(A \subset \mathbb{C} \setminus \{z\}\) and \(\overline{\Omega_r} \subset A\)).

Prove that \(\chi_r\) is locally constant then show that if \(\chi_r(a) = 1\) for some \(a \in \mathbb{C} \setminus \Omega\), then \(\chi_r(z) = 1\) for every \(z \in \mathbb{C} \setminus \Omega\).

Assume that \(f: \Omega \to \mathbb{C}\) has locally uniform approximations among holomorphic functions defined on \(\mathbb{C} \setminus \{a\}\) for some \(a \in \mathbb{C} \setminus \Omega\). Show that \(f\) has locally uniform approximations among polynomials.

Let \(\hat{f}: \mathbb{C} \setminus \{a_1,\dots, a_n\}\to \mathbb{C}\) be holomorphic (all the \(a_k\) are distincts). Show that there are holomorphic functions \(\hat{f}_k: \mathbb{C} \setminus \{a_k\} \to \mathbb{C}\) for \(k=1,\dots,n\) such that \[ \forall \, z \in \mathbb{C} \setminus \{a_1,\dots, a_n\}, \; \hat{f}(z) = \hat{f}_1(z) + \dots + \hat{f}_n(z). \]

Prove the following corollary: if a function \(f: \Omega \to \mathbb{C}\) has locally uniform approximations among holomorphic functions defined on \(\mathbb{C} \setminus \{a_1, \dots, a_n\}\) for some \(a_1, \dots, a_n \in \mathbb{C} \setminus \Omega\), then \(f\) has locally uniform approximations among polynomials.

# Problem L

The *ray* with origin \(a\in \mathbb{C}\) and direction \(u \in \mathbb{C}^*\) is the function^{1} \[
\gamma: t \in \mathbb{R}_+ \mapsto a + t u.
\]

Let \(f:\Omega \mapsto \mathbb{C}\) be a holomorphic function defined on some open subset \(\Omega\) of \(\mathbb{C}\) that contains the image \(\gamma(\mathbb{R}_+)\) of the ray \(\gamma\). The Laplace transform \(\mathcal{L}_{\gamma}[f]\) of \(f\) along \(\gamma\) at \(s \in \mathbb{C}\) is given by \[ \mathcal{L}_{\gamma}[f](s) = \int_{\mathbb{R}_+} f(\gamma(t)) e^{-\gamma(t) s} \gamma'(t) \, dt \] (we consider that this integral is defined when its integrand is summable). This definition generalizes the classic Laplace transform \(\mathcal{L}[f]\) since \(\mathcal{L}_{\gamma}[f] = \mathcal{L}[f]\) when \(\gamma(t) = t\) (that is when \(a=0\) and \(u=1\)).

We assume that there are some \(\kappa > 0\) and \(\sigma \in \mathbb{R}\) such that \begin{equation} \label{bound} \forall \, z \in \gamma(\mathbb{R}_+), \; |f(z)| \leq \kappa e^{\sigma |z|}. \end{equation}Show that if \(\gamma(t) = a + t u\) and \(\mu(t) = a + t(\lambda u)\) for some \(\lambda > 0\), then \[ \mathcal{L}_{\mu}[f] = \mathcal{L}_{\gamma}[f] \] (Reminder: two functions are equal when the have the same domain of definition and the same values in this shared domain.)

Characterize geometrically the set \[ \Pi(u, \sigma) = \left\{ s \in \mathbb{C} \; \left| \; \mathrm{Re} \left( s u \right) > \sigma |u| \right. \right\} \] and show that \(\mathcal{L}_\gamma[f]\) is defined and holomorphic on \(\Pi(u, \sigma)\).

Let \(U\) be an open subset of \(\mathbb{C}^*\). We assume that bound is valid for every \(u \in U\) (for a given origin \(a\) and fixed values of \(\kappa\) and \(\sigma\)). Show that for any \(s\in \mathbb{C}\), the set \(U_{s}\) of directions \(u \in U\) such that \(s \in \Pi(u,\sigma)\) is open and that the function \(u \in U_s \mapsto \mathcal{L}_{\gamma}[f](s)\) is holomorphic (Hint: show that the complex-differentiation under the integral sign theorem is applicable).

Show that the derivative of \(\mathcal{L}_{\gamma}[f](s)\) with respect to \(u\) is zero (Hint: the result of question 1 may be used).

The exponential integral \(E_1(x)\) is defined for \(x>0\) by \[ E_1(x) = \int_x^{+\infty} \frac{e^{-t}}{t}dt. \]

- Compute the (classic) Laplace transform \(F\) of \[ t \in \mathbb{R}_+ \to \frac{1}{t+1} \] and give a formula for \(E_1(x)\) that depends on \(F(x)\). Prove that \(E_1\) has a unique holomorphic extension to the open right half-plane \(\{s \in \mathbb{C} \; | \; \mathrm{Re}(s) > 0\}\).

From now on, we study the case of \[ f: z \in \mathbb{C} \setminus \{-1\} \mapsto \frac{1}{z+1} \] with \(a=0\), \(\sigma=0\) and \(U =\{u \in\mathbb{C} \; | \; \mathrm{Re}(u)>0\}.\)

- Show that there is a \(\kappa > 0\) such that is valid for every \(u \in U\). Characterize geometrically the set \(U_s\); show that it is non-empty when \(s\in \mathbb{C}\setminus \mathbb{R}_-\).

Let \(s \in \mathbb{C} \setminus \mathbb{R}_-\). We define \[
G(s) := \mathcal{L}_{\gamma}[f](s)
\; \mbox{ if } \;
u \in U_s
\; \mbox{ and } \;
\gamma: t \in \mathbb{R}_+ \mapsto tu.
\] Note that this definition is *a priori* ambiguous since several \(u \in U_s\) exist for a given value of \(s\). For any \(u_0 \in \mathbb{C}^*\) and \(u_1 \in \mathbb{C}^*\) and for any \(\theta \in [0,1]\), we denote \[u_{\theta} = (1-\theta) u_0 + \theta u_1\] and whenever \(u_{\theta} \neq 0\), we denote \(\gamma_{\theta}\) the ray of origin \(a=0\) and direction \(u_{\theta}\).

Let \(s \in \mathbb{C}\setminus \mathbb{R}_-\). Show that if \(u_0 \in U_s\) and \(u_1 \in U_s\) then for every \(\theta \in [0,1]\), \(u_{\theta} \in U_s\) and that \[ \mathcal{L}_{\gamma_1}[f](s) - \mathcal{L}_{\gamma_0}[f](s) = \int_0^1 \frac{d}{d\theta} \mathcal{L}_{\gamma_{\theta}}[f](s) \, d\theta. \] Conclude that the definition of \(G\) is unambiguous.

We search for a new expression of the difference \(\mathcal{L}_{\gamma_1}[f](s) - \mathcal{L}_{\gamma_0}[f](s)\) to build an alternate proof for the conclusion of the previous question.

Let again \(s \in \mathbb{C}\setminus \mathbb{R}_-\), \(u_0 \in U_s\) and \(u_1 \in U_s\); let \(r \geq 0\) and \(\gamma_0^r\) and \(\gamma_1^r\) be the paths defined by \[ \gamma_{0}^r: t \in [0,1] \mapsto \gamma_0(tr) \; \mbox{ and } \; \gamma_{1}^r: t \in [0,1] \mapsto \gamma_1(tr) \] Show that \[ \mathcal{L}_{\gamma_1}[f](s) - \mathcal{L}_{\gamma_0}[f](s) = \lim_{r \to +\infty} \int_{\mu_r} f(z) e^{-sz} \, dz \; \; \mathrm{ where } \; \mu_r = (\gamma_0^r)^{\leftarrow} | (\gamma_1^r) \] Conclude again that the definition of \(G\) is unambiguous (Hint: “close” the path \(\mu_r\) and use Cauchy’s integral theorem).

Prove that \(E_1\) has a unique holomorphic extension to \(\mathbb{C} \setminus \mathbb{R}_-\).

# Problem R – Answers

**(1.5pt)**If \(z \in \Omega_r\), then \(d(z, \mathbb{C} \setminus \Omega) > r > 0\). Since any point \(z\) of \(\mathbb{C} \setminus \Omega\) satisfies \(d(z, \mathbb{C} \setminus \Omega) = 0\), we have \(\mathbb{C} \setminus \Omega \subset \mathbb{C} \setminus \Omega_r\) and thus \(\Omega_r \subset \Omega\).For any \(z \in \Omega_r\), the number \(\epsilon = d(z, \mathbb{C} \setminus \Omega) - r\) is positive. If \(|w - z| < \epsilon\) then for any \(v \in \mathbb{C} \setminus \Omega\), \(|z - v| \geq d(z, \mathbb{C} \setminus \Omega)\) and \[ |w - v| \geq |z - v| - |w - z| > d(z, \mathbb{C} \setminus \Omega) - \epsilon = r. \] Consequently \(D(z, \epsilon) \subset \Omega\) and \(\Omega\) is open.

Finally, if \(z \in \Omega\), since \(\Omega\) is open, \(d = d(z, \mathbb{C} \setminus \Omega) > 0\), thus if \(r = d/2\), the point \(z\) belongs to \(\Omega_r\). Consequently, \(\Omega = \cup_{r>0} \Omega_r\).

**(1pt)**Let \(\Omega = \{z \in \mathbb{C} \; | \; |\mathrm{Im} \, z| < |\mathrm{Re} \, z| +1 \}\). This set is open: the function \[ \phi: z \in \mathbb{C} \to |\mathrm{Re} \, z| + 1 - |\mathrm{Im} \, z| \in \mathbb{R} \] is continuous and \(\Omega\) is the pre-image of the open set \(\mathbb{R}_{+}^*\) by \(\phi\). Now, for any purely imaginary number \(iy\) of \(\Omega\), since \(|iy - i| \leq 1\) or \(|iy + i| \leq 1\), we have \(d(iy, \mathbb{C} \setminus \Omega) \leq 1\). Therefore, no such point belongs to \(\Omega_1\). On the other hand, \(d(-2, \mathbb{C} \setminus \Omega) = d(-2, \mathbb{C} \setminus \Omega) = 3\sqrt{2}/2 > 1\) and hence \(-2 \in \Omega_1\) and \(2 \in \Omega_1\).Assume that \(\Omega_1\) is connected. Since it is open, it is path-connected: there is a continuous function \(\gamma:[0,1] \to \Omega_{1}\) that joins \(-2\) and \(2\). By the intermediate value theorem, there is a \(t\in \left]0,1\right[\) such that \(\mathrm{Re} \, \gamma(t) = 0\). Since \(\gamma(t) \in \Omega_1\), we have a contradiction. Consequently, \(\Omega_1\) is not connected.

**(2.5pt)**By definition of \(\Omega_r\), its complement satisfies \[ \mathbb{C} \setminus \Omega_r = \{ z \in \mathbb{C} \; | \; d(z, \mathbb{C} \setminus \Omega) \leq r \}. \] Thus, if \(z \in \mathbb{C} \setminus \Omega_r\), since \(\mathbb{C} \setminus \Omega\) is closed, there is a \(w \in \mathbb{C} \setminus \Omega\) such that \(|z - w| \leq r\). Since for any \(\lambda \in [0,1]\), the point \(z_{\lambda} = \lambda w + (1-\lambda) z\) satisfies \(|z_{\lambda} - w| = (1-\lambda) |z - w| \leq r\), we also have \(z_{\lambda} \in \mathbb{C} \setminus \Omega_r\). Hence, \([w, z] \subset \mathbb{C} \setminus \Omega_r\).Assume that \(\Omega\) is simply connected. Let \(z \in \mathbb{C} \setminus \Omega_r\) and let \(\gamma\) be a closed path of \(\Omega_r\); since \(\Omega_r \subset \Omega\), \(\gamma\) is also a closed path of \(\Omega\). Let \(w \in \mathbb{C} \setminus \Omega\) such that \([w, z] \subset \mathbb{C} \setminus \Omega_r\). Since \([w, z]\) is connected and the function \(\xi \in \mathbb{C} \setminus \Omega_r \mapsto \mathrm{ind}(\gamma, \xi)\) is locally constant, \(\mathrm{ind}(\gamma, z) = \mathrm{ind}(\gamma, w)\). Since \(\Omega\) is simply connected, \(\mathrm{ind}(\gamma, w)= 0\). Hence, \(\Omega_r\) is simply connected.

Alternatively, assume that \(\Omega_r\) is multiply connected. Let \(\mathbb{C} \setminus \Omega_r = K \cup L\) where \(K\) is bounded and non-empty and \(d(K, L) > 0\). Since \(\Omega_r \subset \Omega\), \(\mathbb{C} \setminus \Omega \subset \mathbb{C} \setminus \Omega_r\). Let \(K' = K \cap (\mathbb{C} \setminus \Omega)\) and \(L' = L \cap (\mathbb{C} \setminus \Omega)\). Clearly, \(\mathbb{C} \setminus \Omega = K' \cup L'\), \(K'\) is bounded and \(d(K', L') > 0\). Now, let \(z \in K\) and let \(w \in \mathbb{C} \setminus \Omega\) such that \([w, z] \subset \mathbb{C} \setminus \Omega_r = K \cup L\). Every connected \(C\) subset of \(K \cup L\) that contains a point of \(K\) is included in \(K\) since for any \(r \leq d(K,L)\), the dilation \(C + B(0,r)\) doesn’t intersect \(L\). Since \([w,z]\) is a connected subset of \(K \cup L\) and \(z \in K\), we have \([w,z] \subset K\), thus \(K'\) is also non-empty. Finally, \(\Omega\) is multiply connected.

The converse result is false: for any open subset \(\Omega\) which is bounded and not simply connected, for \(r\) large enough, \(\Omega_r = \varnothing\) which is simply connected.

**(2.5pt)**Assume that \(\mathbb{C} \setminus \Omega\) is disconnected. Since \(\Omega\) is bounded, for \(r\) large enough, the annulus \(A(0,r+\infty)\), which is connected, is a subset of \(\mathbb{C} \setminus \Omega\). Consider a dilation of \(\mathbb{C} \setminus \Omega\) which is not (path-)connected; let \(V_{\infty}\) be the component of this dilation that contains the annulus above, while \(V\) is the (non-empty) union of the other components of the dilation. By construction, \(V\) and \(V_{\infty}\) are open and disjoints and \(V\) is bounded. The set \[K = (\mathbb{C} \setminus \Omega) \cap V = (\mathbb{C} \setminus \Omega) \setminus V_{\infty}\] is closed, non-empty and bounded; with \[ L = (\mathbb{C} \setminus \Omega) \cap V_{\infty} = (\mathbb{C} \setminus \Omega) \setminus V, \] which is closed, we have \(\mathbb{C} \setminus \Omega = K \cup L\). Thus \(\mathbb{C} \setminus \Omega\) is multiply connected.**(1.5pt)**Since \(\Omega\) is bounded, there is a \(r>0\) such that \(\Omega \subset K = \overline{D(0, r)}\). Let \(a \in \mathbb{C}\) such that \(|a|>r\); let \(\epsilon > 0\) and \(\hat{f}: \mathbb{C} \setminus \{a\}\) be a holomorphic function such that \(|f - \hat{f}| \leq \epsilon/2\) on \(\Omega\). Since \(K\) is a compact subset of \(D(0, |a|) \subset \mathbb{C} \setminus \{a\}\), the Taylor series expansion \(\sum a_n (z-c)^n\) of \(\hat{f}\) in \(D(0, |a|)\) is uniformly convergent in \(K\), thus there is a \(m \in \mathbb{N}\) such that the polynomial \[ p(z) = \sum_{n=0}^m a_n (z-c)^n \] satisfies \(|\hat{f} - p| \leq \epsilon/2\) in \(K\) and hence in \(\Omega\). Consequently, \[ \forall \, z \in \Omega, \; |f(z) - p(z)| \leq |f(z) - \hat{f}(z)| + |\hat{f}(z) - p(z)| \leq \epsilon. \]**(1.5pt)**Assume that there is a polynomial \(p\) such that \(|f - p| < 1\) on \(\Omega\). Since \(\Omega\) is bounded, the set \(K = \overline{\Omega}\) is compact and since \(p\) is continuous on \(K\), it is bounded on \(\Omega\). Given that \[ |f(z)| \leq |f(z) - p(z)| + |p(z)| \] the function \(f\) is necessarily bounded on \(\Omega\).Now all we have to do is to find a holomorphic function on \(\Omega\) which is not bounded. Consider \(f: z\mapsto 1/(z-a)\) where \(a\) is a point of the boundary of \(\Omega\) (such a point exists since \(\Omega \neq \varnothing\) and \(\Omega \neq \mathbb{C}\)); since \(\Omega\) is open, \(a \not \in \Omega\) and the function \(f\) is defined and holomorphic on \(\Omega\). By construction there is a sequence \(z_n \in \Omega\) such that \(z_n \to a\) and thus \[|f(z_n)| = \left|\frac{1}{z_n - a}\right| \to +\infty,\] thus \(f\) is not bounded on \(\Omega\).

**(2pt)**If \(\Omega\) is not simply connected, there is a closed rectifiable path \(\gamma\) of \(\Omega\) and a point \(a \in \mathbb{C} \setminus \Omega\) which is in the interior of \(\gamma\), that is \(\mathrm{ind}(\gamma, a)\) is a non-zero integer. The function \(z \mapsto 1/(z-a)\) is defined and holomorphic on \(\Omega\). Additionally \[ \frac{1}{2\pi} \int_{\gamma} f(z) dz = \mathrm{ind}(\gamma, a) \in \mathbb{Z}^*. \] Since the distance between \(\gamma([0,1])\) and the complement of \(\Omega\) is positive, there is a \(r>0\) such that \(\gamma([0,1]) \subset \Omega_r\). Now if \(\hat{f}\) is a polynomial such that \(|f - \hat{f}| \leq \epsilon\) on \(\Omega_r\), \[ \left| \frac{1}{2\pi} \int_{\gamma} f(z) \, dz \right| \leq \left| \frac{1}{2\pi} \int_{\gamma} \hat{f}(z) \, dz \right| + \left| \frac{1}{2\pi} \int_{\gamma} (f-\hat{f})(z) \, dz \right|. \] By Cauchy’s theorem (the local version, in \(\mathbb{C}\)), the first integral in the right-hand side is zero. By the M-L inequality, the second one is dominated by \((\epsilon/2\pi) \times \ell(\gamma)\). Thus for \(\epsilon < 2\pi / \ell(\gamma)\), we would have \[ \left| \frac{1}{2\pi} \int_{\gamma} f(z) \, dz \right| < 1. \] Consequently \(z\mapsto 1/(z-a)\) is holomorphic on \(\Omega\) but cannot be locally uniformly approximated by polynomials.**(3pt)**The condition \(|w - z| <r/2\) yields \[ \{z\} \subset \overline{D(w, r/2)} = \mathbb{C} \setminus A(w, r/2, +\infty). \] or equivalently, \(A = A(w, r/2, +\infty) \subset \mathbb{C} \setminus \{z\}\). Additionally, any \(v \in \overline{\Omega}_r\) satisfies \(d(v, \mathbb{C} \setminus \Omega) \geq r\) and in particular \(|v - z| \geq r\). Since \(|w - z| < r/2\), \[ |v - w| \geq |v - z| - |w - z| > r - r/2 = r/2, \] hence \(v \in A(w, r/2, +\infty)\). Consequently, \(\overline{\Omega_r} \subset A\).Since \(\chi_r(z) = 1\), for any \(\epsilon > 0\), there is a function \(\hat{f}_z\) holomorphic in \(\mathbb{C} \setminus \{z\}\) such such that \(|f - \hat{f}| \leq \epsilon / 2\) on \(\Omega_r\). Now, since the annulus \(A\) is included in the domain of definition of \(\hat{f}_z\), we have the Laurent series expansion \[ \forall \, v \in A, \; \hat{f}_z(v) = \sum_{n=-\infty}^{+\infty} a_n (v - w)^n. \] This expansion is locally uniformly convergent; since \(\overline{\Omega_r}\) is compact and included in \(A\), there is a natural number \(m\) such that the function \(\hat{f}_w(v) :=\sum_{n=-m}^m a_n (v - w)^n\) – which is holomorphic on \(\mathbb{C} \setminus \{w\}\) – satisfies \(|\hat{f}_z - \hat{f}_w| \leq \epsilon/2\) on \(\overline{\Omega}_r\). Finally, on \(\Omega_r\), we have \[ |f - \hat{f}_w| \leq |f - \hat{f}_z| + |\hat{f}_z - \hat{f}_w| \leq \epsilon/2 + \epsilon/2 = \epsilon. \] and thus \(\chi(w) = 1\).

**(1pt)**We know that if \(\chi(z) = 1\) and \(|w - z| < r/2\), then \(\chi(w) = 1\). Now, by contraposition of this property, if \(\chi(w) = 0\) and \(|w-z| < r/2\), we have \(\chi(z) = 0\). Therefore \(\chi\) is locally constant. Now since \(\Omega\) is simply connected and bounded, by question 4, \(\mathbb{C} \setminus \Omega\) is connected. Since \(\chi: \mathbb{C}\setminus \Omega \to \{0,1\}\) is locally constant, if \(\chi(a) = 1\) for some \(a \in \mathbb{C}\setminus \Omega\), \(\chi = 1\) on \(\mathbb{C} \setminus \Omega\).**(1pt)**If \(f: \Omega \to \mathbb{C}\) has locally uniform approximations among the holomorphic functions defined on \(\mathbb{C} \setminus \{a\}\), then for any \(r>0\) \(\chi_r(a)=1\). By the previous question, \(\chi_r(b)=1\) for any \(b \in \mathbb{C} \setminus \Omega\) and thus \(f\) has locally uniform approximations among the holomorphic functions defined on \(\mathbb{C} \setminus \{b\}\). We can select a \(b\) that is arbitrarily large, thus by question 5, \(f\) has locally uniform approximations among polynomials.**(4pt)**We proceed by induction. The result is plain if \(n=1\); now assume that the result holds for a given \(n \geq 1\) and let \(\hat{f}: \mathbb{C} \setminus \{a_1,\dots, a_n, a_{n+1}\}\to \mathbb{C}\) be holomorphic. Since \(a_{n+1}\) is an isolated singularity of \(f\), there is a \(r>0\) such that \(A(a_{n+1}, 0, r)\) is a non-empty annulus included in the domain of definition of \(f\). Let \(\sum_{k={-\infty}}^{+\infty} b_k (z-a_{n+1})^k\) be the Laurent series expansion of \(\hat{f}\) in this annulus and define \[ \hat{f}_{n+1}(z) = \sum_{n = -{\infty}}^{-1} b_k (z - a_{n+1})^k \] Since there are only negative powers, the sum is convergent (and holomorphic) in \(\mathbb{C} \setminus \{a_{n+1}\}\). Now since in the annulus \[ \hat{f}(z) - \hat{f}_{n+1}(z) = \sum_{n=0}^{+\infty} b_k (z - a_{n+1})^k \] the point \(a_{n+1}\) is a removable singularity of the function \(\hat{f} - \hat{f}_k\) that may thus be extended holomorphically to \(\mathbb{C} \setminus \{a_1,\dots, a_n\}\). We may apply the induction hypothesis to this extension; we get holomorphic functions \(\hat{f}_k: \mathbb{C} \setminus \{a_k\} \to \mathbb{C}\) for \(k=1,\dots,n\) such that \[ \forall \, z \in \mathbb{C} \setminus \{a_1,\dots, a_{n+1}\}, \; \hat{f}(z) - \hat{f}_{n+1}(z)= \hat{f}_1(z) + \dots + \hat{f}_n(z) \] which is the induction hypothesis at stage \(n+1\).Now the corollary: assume that for any \(r>0\) and any \(\epsilon>0\), there is a holomorphic function \(\hat{f}: \mathbb{C} \setminus \{a_1, \dots, a_n\} \to \mathbb{C}\) such that \(|f - \hat{f}| \leq \epsilon / 2\) in \(\Omega_r\). Let \(\hat{f}_k: \mathbb{C} \setminus \{a_k\} \to \mathbb{C}\) for \(k=1,\dots,n\) be such that \[ \forall \, z \in \mathbb{C} \setminus \{a_1,\dots, a_n\}, \; \hat{f}(z) = \hat{f}_1(z) + \dots + \hat{f}_n(z). \] Since the restriction of every \(\hat{f}_k\) to \(\Omega\) is locally uniformly approximated by holomorphic functions on \(\mathbb{C} \setminus \{a_k\}\), by question 9, it is locally uniformly approximated by polynomials, so there is a polynomial \(p_k\) such that \(|\hat{f}_k - p_k| \leq \epsilon/2n\) on \(\Omega_r\). Let \(p = \sum_{k=1}^n p_k\); on \(\Omega_r\), we have \[ |f - p| \leq |f - \hat{f}| + \sum_{k=1}^n |\hat{f}_k - p_k| \leq \epsilon. \] Thus, \(f\) is locally uniformly approximated by polynomials.

# Problem L – Answers

**(1pt)**Since \[ \mathcal{L}_{\mu}[f](s) = \int_{\mathbb{R}_+} f(a + t\lambda u) e^{-s(a+ t\lambda u)} (\lambda u) \, dt, \] if \(\mathcal{L}_{\mu}[f](s)\) is defined, the change of variable \(\tau = \lambda t\) yields \[ \mathcal{L}_{\mu}[f](s) = \int_{\mathbb{R}_+} f(a + \tau u) e^{-s(a+ \tau u)} u \, d\tau , \] thus \(\mathcal{L}_{\lambda}[f](s)\) is defined and \(\mathcal{L}_{\mu}[f](s) = \mathcal{L}_{\lambda}[f](s)\). Conversely, if \(\mathcal{L}_{\lambda}[f](s)\) is defined, the same argument with \(1/\lambda\) instead of \(\lambda\) shows that \(\mathcal{L}_{\mu}[f](s)\) is also defined (and obviously \(\mathcal{L}_{\mu}[f](s) = \mathcal{L}_{\lambda}[f](s)\)).**(2pt)**The set \(\Pi(u, \sigma)\) is an open half-plane: if \(u = |u|e^{i\theta}\) and \(s = x + iy\),

then the condition \(\mathrm{Re}(s u) > \sigma |u|\) is equivalent to \[ x \cos \theta - y \sin \theta > \sigma. \]The Laplace transform of \(f\) along \(\gamma\) satisfies \[ \begin{split} \mathcal{L}_{\gamma}[f](s) & = \int_{\mathbb{R}_+} f(a+tu) e^{-s(a+tu)}u\, dt \\ & = (e^{-sa}u) \int_{\mathbb{R}_+} f(a+tu) e^{-(su)t}\, dt \end{split} \] and thus \[ \mathcal{L}_{\gamma}[f](s) = (e^{-sa} u) \mathcal{L}[t \mapsto f(a+tu)](su). \] The function \(t \in \mathbb{R}_+ \mapsto f(a+tu)\) is locally integrable (it is continuous since \(f\) is holomorphic) and since for any \(t\geq 0\) \[ -|a| + t |u| \leq |a + tu| \leq |a| + t |u|, \] we have \[ |f(a+tu)| \leq \kappa e^{\sigma |a+tu|} \leq \kappa \max(e^{-\sigma|a|},e^{\sigma|a|}) e^{(\sigma|u|) t} \] therefore \(t \geq 0 \mapsto |f(a+tu)|e^{-\sigma^+ t}\) is integrable whenever \(\sigma^+ > \sigma |u|\). Hence the Laplace transform \(\mathcal{L}[t \mapsto f(a+tu)]\) is defined and holomorphic on the set \(\Pi(u, \sigma) =\{s \in \mathbb{C} \; | \; \mathrm{Re}(s) > \sigma |u|\}\) and \(\mathcal{L}_{\gamma}[f]\) as well, as a composition and product of holomorphic functions.

**(3pt)**For a \(s \in \mathbb{C}\), the set of \(u\) such that \(s \in \Pi(u, \sigma)\) is open as the preimage of the open set \(\left]0, + \infty\right[\) by the continuous function \(u \in U \mapsto \mathrm{Re}(s u) - \sigma|u|\).Let \(\psi: U_s \times \mathbb{R}_+ \to \mathbb{C}\) be defined as \[ \psi(u, t) = f(a+tu) e^{-(a+tu)s} u. \] Since \[ |f(a+tu)| \leq \kappa e^{\sigma|a+tu|} \leq \kappa_1 e^{t|u| \sigma} \] with \(\kappa_1 = \kappa \max(e^{-\sigma|a|},e^{\sigma|a|})\) and \[ |e^{-(a+tu)s}| = e^{-\mathrm{Re}((a+tu)s)} = e^{-\mathrm{Re}(as)}e^{-t\mathrm{Re}(su)}, \] we end up with \[ |\psi(u, t)| \leq (\kappa_1 e^{-\mathrm{Re}(as)}|u|) e^{t(\sigma |u|- \mathrm{Re}(s u))}. \]

Since for any \(u \in U_s\) \[ \mathcal{L}_{\gamma}[f](s) = \int_{\mathbb{R}_+} \psi(u, t) \, dt \] we check the assumptions of the complex-differentiation under the integral sign theorem:

For every \(u \in U_s\), the function \(t \in \mathbb{R}_+ \mapsto \psi(u, t)\) is Lebesgue measurable (it is actually continuous since \(f\) is holomorphic).

If \(s \in \Pi(u_0, \sigma)\), since \(\epsilon := \mathrm{Re}(s u_0) - \sigma |u_0|> 0\), by continuity there is a \(0<r<|u_0|\) such that \(|u-u_0| \leq r\) ensures \(\mathrm{Re}(s u) - \sigma |u| \geq \epsilon/2\). Let \(\kappa_2\) be an upper bound of \(\kappa_1 e^{-\mathrm{Re}(as)}|u|\) when \(|u-u_0| \leq r\). The bound on \(\psi\) that we have derived above yields \[ |\psi(u,t)| \leq \kappa_2 e^{-t \epsilon/2} \] and the right-hand side of this inequality is a Lebesgue integrable function of \(t\).

For every \(t \in \mathbb{R}_+\), it is plain that the function \(u \in U_s \mapsto \psi(u, t)\) is holomorphic.

Consequently \(\mathcal{L}_{\gamma}[f](s)\) is a complex-differentiable function of \(u\).

**(1.5pt)**First method: since we know that the complex-derivative of \(\mathcal{L}_{\gamma}[f](s)\) exists with respect to \(u\), we can apply the chain rule to the differentiable function \[ \chi: \lambda \in \mathbb{R}_+^{*} \to \mathcal{L}_{\mu}[f](s) \; \mbox{ with } \; \mu(t) = a + t(\lambda u) \] at \(t=1\): it yields \[ \frac{d\chi}{d\lambda} (1) = \frac{d}{du}\mathcal{L}_{\mu}[f](s) \times \frac{d(\lambda u)}{d\lambda} (1) = \frac{d}{du}\mathcal{L}_{\mu}[f](s) \times u. \] Since by question 1 we know that the function \(\chi\) is constant, we conclude that \(\frac{d}{du}\mathcal{L}_{\mu}[f](s) = 0\).Second method: we use the result of the differentiation under the integral sign. It provides \[ \frac{d}{du}\mathcal{L}_{\mu}[f](s) = \int_{\mathbb{R}_+} \frac{\partial \psi}{\partial u} (u, t) \, dt. \] Since \(\psi(u, t) = g(tu) u\) with \(g(z) = f(a+z) e^{-s(a+z)}\), we have \[ \begin{split} \frac{\partial \psi}{\partial u} (u, t) &= \frac{d}{du} (g(tu) u) = g'(tu) tu + g(tu) = \frac{d g(tu)}{dt} t + g(tu) \\ &= \frac{d}{dt} (g(tu) t) \end{split} \] and thus \[ \begin{split} \int_0^r \frac{\partial \psi}{\partial u} (u, t) \, dt &= \int_0^r \frac{d}{dt} (g(tu) t) \, dt = [g(tu) t]^r_0 \\ &= f(a+ ru) e^{-s(a+ru)} r \end{split} \] Since \(\epsilon := \mathrm{Re}(s u) - \sigma|u| > 0\) \[ |f(a+ ru) e^{-s(a+ru)} r| \leq |\psi(u, r)r / u| \leq (\kappa_1 e^{-\mathrm{Re}(as)}) e^{-\epsilon r} r. \] The right-hand side of this inequality converges to \(0\) when \(r\to +\infty\). Therefore, by the dominated convergence theorem \[ \frac{d}{du}\mathcal{L}_{\mu}[f](s) = \int_{\mathbb{R}_+} \frac{\partial \psi}{\partial u} (u, t) \, dt = \lim_{r \to +\infty} \int_{0}^r \frac{\partial \psi}{\partial u} (u, t) \, dt = 0. \]

**(2pt)**Since the Laplace transform of \(t \in \mathbb{R}_+ \to 1/(t+1)\) is given by \[ F(s) = \int_{0}^{+\infty} \frac{e^{-s t}}{t+1} \, dt, \] the change of variable \(t+1 \to t\) yields \[ F(s) = \int_1^{+\infty} \frac{e^{-s (t-1)}}{t} \, dt = e^s \int_1^{+\infty} \frac{e^{-s t}}{t} \, dt. \] If \(s\) is equal to the real number \(x>0\), the change of variable \(xt \to t\) then provides \[ F(x) = e^x \int_1^{+\infty} \frac{e^{-xt}}{xt} \, d(xt) = e^x \int_x^{+\infty} \frac{e^{-t}}{t} \, dt \] and thus \(E_1(x) = e^{-x} F(x)\). Since \(t \in \mathbb{R}_+ \to 1/(t+1) e^{-\sigma^+t}\) is integrable whenever \(\sigma^+ > 0\) the Laplace transform \(F\) of \(t \mapsto 1/(t+1)\) is defined and holomorphic on \(\{s \in \mathbb{C} \; | \; \mathrm{Re}(s) > 0\}\). Thus, the function \(G: s \mapsto e^{-s} F(s)\) extends holomorphically \(E_1\) on this open right-hand plane. Any other function \(\tilde{G}\) with the same property would be equal to \(G\) on \(\mathbb{R}_+^*\) and thus every positive real number \(x\) would be a non-isolated zero of \(G - \tilde{G}\). By the isolated zeros theorem, since the open right-hand plane is connected, \(G\) and \(\tilde{G}\) are necessarily equal.**(2pt)**For any \(u \in U\), since \(\mathrm{Re} \, u > 0\), for any \(t \geq 0\), \(\mathrm{Re}(\gamma(t)) = t \, \mathrm{Re} \, u \geq 0\). Since for any \(z\) such that \(\mathrm{Re} \, z \geq 0\), \(|z + 1| \geq 1\), we have \(|f(z)| = 1 / |z + 1| \leq 1\) and thus \[ \forall \, z \in \gamma(\mathbb{R}_+), \, |f(z)| = \kappa e^{\sigma |z|} \] with \(\kappa = 1\) and \(\sigma=0\).By definition, for any \(s \in \mathbb{C} \setminus \mathbb{R}_-\), \(U_s\) is the set of all directions \(u\) such that \(\mathrm{Re}(u) > 0\) and \(\mathrm{Re}(s u) > 0\). To be more explicit, if \(\alpha\) denotes the argument of \(u\) in \(\left[-\pi, \pi\right[\) and \(\beta\) the argument of \(s\) in \(\left]-\pi,\pi \right[\), this is equivalent to \[ -\pi/2 < \alpha < \pi/2 \; \mbox{ and } \; - \pi/2 < \alpha + \beta < \pi/2 \] The points \(u = r e^{i\alpha}\) that satisfy these constraints form an open sector: \(r>0\) is arbitrary and if \(\beta \geq 0\), \(\alpha\) is subject to \[ -\frac{\pi}{2} < \alpha < \frac{\pi}{2} - \beta \] and if \(\beta < 0\) \[ -\frac{\pi}{2} - \beta < \alpha < \frac{\pi}{2}. \] In any case, since \(|\beta| < \pi\), the set of admissible arguments \(\alpha\) is not empty.

**(3pt)**It is plain that \[ \mathrm{Re}(u_{\theta}) = \mathrm{Re}((1-\theta) u_0 + \theta u_1) = (1-\theta) \mathrm{Re}(u_0) + \theta \mathrm{Re} (u_1). \] and that for any \(s \in \mathbb{C}\) \[ \mathrm{Re}(su_{\theta}) = \mathrm{Re}(s((1-\theta) u_0 + \theta u_1)) = (1-\theta) \mathrm{Re}(su_0) + \theta \mathrm{Re} (su_1). \] Since \(u \in U_s\) if and only if \(\mathrm{Re}(u) > 0\) and \(\mathrm{Re}(su) > 0\), if \(u_0 \in U_s\) and \(u_1 \in U_s\) then \(u_{\theta} \in U_s\) for any \(\theta \in [0,1]\).The function \(u \in U_s \mapsto \mathcal{L}_{\gamma} [f](s)\) is holomorphic, hence the composition of \[ \theta \in [0,1] \mapsto u_{\theta} \in U_s \; \mbox{ and } \; u \in U_s \mapsto \mathcal{L}_{\gamma} [f](s) \] is continuously differentiable, \[ \mathcal{L}_{\gamma_0}[f](s) - \mathcal{L}_{\gamma_1}[f](s) = \int_0^1 \frac{d}{d\theta} \mathcal{L}_{\gamma_{\theta}}[f](s) \, d\theta \] and \[ \frac{d}{d\theta} \mathcal{L}_{\gamma_{\theta}}[f](s) = \frac{d \mathcal{L}_{\gamma}[f](s)}{du} |_{u=u_{\theta}} \times \frac{d u_{\theta}}{d\theta} \] which is zero by question 4. Hence \(\mathcal{L}_{\gamma_0}[f](s) = \mathcal{L}_{\gamma_1}[f](s)\); the definition of \(G\) is unambiguous.

**(4pt)**For \(k \in \{0,1\}\) and any \(r\geq 0\), \[ \int_{\gamma_k^r} f(z) e^{-sz} \, dz = \int_0^1 f((tr) u_k) e^{-s((tr)u_k)} r u_k \, dt = \int_0^r f(t u_k) e^{-s(t u_k)} u_k \, dt \] thus by the dominated convergence theorem \[ \mathcal{L}_{\gamma_1}[f](s) - \mathcal{L}_{\gamma_0}[f](s) = \lim_{r\to +\infty} \left( \int_{\gamma_1^r} f(z) e^{-sz} dz - \int_{\gamma_0^r} f(z) e^{-sz} dz \right). \] Since \((\gamma_0^r)^{\leftarrow}\) and \(\gamma_1^r\) are consecutive, with the path \(\mu_r = (\gamma_0^r)^{\leftarrow} \,|\, \gamma_1^r\), this is equivalent to \[ \mathcal{L}_{\gamma_1}[f](s) - \mathcal{L}_{\gamma_0}[f](s) = \lim_{r\to +\infty} \int_{\mu_r} f(z) e^{-sz} dz. \]Now, the path \(\nu_r\) defined by \(\nu_r(\theta) = (1-\theta) r u_0 + \theta r u_1\) is such that \(\mu_r \, | \, \nu_r^{\leftarrow}\) is a closed rectifiable path in \(U_s\) which is simply connected. Therefore by Cauchy’s integral theorem \[ \int_{\mu_r} f(z) e^{-sz} dz = \int_{\nu_r} f(z) e^{-sz} dz. \]

For any \(\theta \in [0,1]\), we have \(\mathrm{Re}(s u_\theta) = (1 - \theta) \mathrm{Re}(s u_0) + \theta \mathrm{Re}(s u_1)\), thus \[ \epsilon := \min_{\theta \in [0,1]} \mathrm{Re}(s u_{\theta}) > 0 \] and \[ |f(\nu_r(\theta)) e^{-s\nu_r(\theta)}| \leq \kappa e^{- \mathrm{Re}(s ru_{\theta})} \leq \kappa e^{-\epsilon r}. \] Given that \(\ell(\nu_r) = r |u_1 - u_0|\), the M-L inequality provides \[ \left| \int_{\nu_r} f(z) e^{-sz} dz \right| \leq \kappa e^{-\epsilon r}r |u_1 - u_0| \] and thus \[ \lim_{r \to +\infty} \int_{\nu_r} f(z) e^{-sz} dz = 0. \]

Finally, \(\mathcal{L}_{\gamma_1}[f](s) = \mathcal{L}_{\gamma_0}[f](s)\): the definition of \(G\) is unambiguous.**(1pt)**By construction the function \[s \in \mathbb{C} \setminus \mathbb{R}_- \mapsto e^{-s} G(s)\] is holomorphic (since every \(\mathcal{L}_{\gamma}\) is holomorphic, \(G\) is holomorphic); It extends \(s \mapsto e^{-s}F(s)\), thus it also extends \(E_1\). Since \(\mathbb{C} \setminus \mathbb{R}_-\) is connected, by the isolated zeros theorem, this extension is unique (the argument is identical to the one used in question 6).

# Notes

This notation emphasizes that the complex number \(\gamma(t)\) depends on \(t\); however it also depends implicitly on some \(a\) and \(u\) that are usually clear from the context. Feel free to use a more explicit notation if you feel that it is beneficial.↩