# 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 $$r$$: $\Omega_r = \{z \in \mathbb{C} \; | \; d(z, \mathbb{C} \setminus \Omega) > r \}.$
1. Show that $$\Omega_r$$ is an open subset of $$\Omega$$ and that $$\Omega = \cup_{r>0} \Omega_r$$.

2. 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 $$r = 1$$).

3. 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?

4. 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 $\forall \, \epsilon > 0, \exists \, \hat{f} \in \mathcal{F}, \; \forall \, z \in \Omega, \; |f(z) - \hat{f}(z)| \leq \epsilon.$

1. 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.

2. 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 $$\mathcal{F}$$: $\forall \, \epsilon > 0, \forall \, r > 0, \exists \, \hat{f} \in \mathcal{F}, \; \forall \, z \in \Omega_r, \; |f(z) - \hat{f}(z)| \leq \epsilon.$

1. 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.

1. 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 $$\overline{\Omega_r} \subset A$$).

2. 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 $$z \in \mathbb{C} \setminus \Omega$$.

3. 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.

4. 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 function1 $\gamma: t \in \mathbb{R}_+ \mapsto a + t u.$

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}
1. 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.)

2. 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 $$\Pi(u, \sigma)$$.

3. 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).

4. 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 $E_1(x) = \int_x^{+\infty} \frac{e^{-t}}{t}dt.$

1. 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 $$U =\{u \in\mathbb{C} \; | \; \mathrm{Re}(u)>0\}.$$

1. 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 $$u_{\theta}$$.

1. 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.

2. 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).

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

1. (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, $$\Omega = \cup_{r>0} \Omega_r$$.

2. (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.

3. (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.

4. (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.

5. (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, $\forall \, z \in \Omega, \; |f(z) - p(z)| \leq |f(z) - \hat{f}(z)| + |\hat{f}(z) - p(z)| \leq \epsilon.$

6. (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 $$\Omega$$.

7. (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.

8. (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 $$\chi(w) = 1$$.

9. (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 $$\mathbb{C} \setminus \Omega$$.

10. (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.

11. (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.

1. (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 $$\mathcal{L}_{\mu}[f](s) = \mathcal{L}_{\lambda}[f](s)$$).

2. (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.

3. (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 $$u$$.

4. (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 $\frac{d}{du}\mathcal{L}_{\mu}[f](s) = \int_{\mathbb{R}_+} \frac{\partial \psi}{\partial u} (u, t) \, dt = \lim_{r \to +\infty} \int_{0}^r \frac{\partial \psi}{\partial u} (u, t) \, dt = 0.$

5. (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.

6. (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.

7. (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.

8. (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.

9. (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).