# Problem S

A function $$F: \Omega \mapsto \mathbb{C}^{m \times n}$$ where $$\Omega$$ is an open subset of $$\mathbb{C}$$ and $$m$$, $$n$$ are positive integers is a matrix-valued function of a complex variable. It is characterized by the collection of its element functions $$f_{kl}: \Omega \mapsto \mathbb{C}$$: $\forall \, z \in \Omega, \; F(z) = \left[ \begin{array}{cccc} f_{11}(z) & f_{12}(z) & \cdots & f_{1n}(z) \\ \vdots & \vdots & \cdots & \vdots \\ f_{m1}(z) & f_{12}(z) & \cdots & f_{mn}(z) \end{array} \right]$

Using element functions, we can readily adapt the tools of Complex Analysis:

• $$F$$ is holomorphic if all its element functions are holomorphic; its derivative $$F'$$ is defined element-wise: $\forall \, z \in \Omega, \; F'(z) = \left[ \begin{array}{cccc} f'_{11}(z) & f'_{12}(z) & \cdots & f'_{1n}(z) \\ \vdots & \vdots & \cdots & \vdots \\ f'_{m1}(z) & f'_{12}(z) & \cdots & f'_{mn}(z) \end{array} \right]$

• the integral of $$F$$ along some path (or path sequence) $$\gamma$$ is defined – if and only if it is defined for all its element functions – by $\forall \, z \in \Omega, \; \int_{\gamma} F(z) \, dz = \left[ \begin{array}{cccc} \int_{\gamma} f_{11}(z) \,dz& \int_{\gamma} f_{12}(z) \,dz & \cdots & \int_{\gamma}f_{1n}(z)\,dz \\ \vdots & \vdots & \cdots & \vdots \\ \int_{\gamma}f_{m1}(z) \, dz& \int_{\gamma}f_{12}(z) \, dz& \cdots & \int_{\gamma}f_{mn}(z) \, dz \end{array} \right].$

### Linear algebra reminder.

‌ The spectrum of a square matrix $$A$$ is the set $$\sigma(A) = \{z \in \mathbb{C} \; | \; \det(zI - A) = 0\}$$. The inverse matrix $$A^{-1}$$ of an invertible matrix $$A$$ depends continuously on the elements of $$A$$ (as long as $$A$$ stays invertible). Polynomials of a square matrix $$A$$ (e.g. $$I + 2A + A^2$$) commute with each other; rational functions of $$A$$ (e.g. $$(I + A)(I - A)^{-1}$$) also commute with each other whenever they are defined.

## Questions

1. Prove that for any holomorphic functions $$F: \Omega \to \mathbb{C}^{m\times n}$$ and $$G:\Omega \to \mathbb{C}^{n \times p}$$, the matrix function product $FG:z \in \Omega \mapsto F(z) G(z) \in \mathbb{C}^{m \times p}$ is holomorphic and $$(FG)' = F' G + F G'$$.

Let $$p \in \mathbb{N}^*$$ and $$A \in \mathbb{C}^{p \times p}$$. We denote $$\Omega$$ the set of complex numbers $$z$$ such that the matrix $$I - zA$$ is invertible and $$F$$ the matrix-valued function defined by $F: z \in \Omega \mapsto [I - zA]^{-1}.$

1. Show that $$\Omega$$ is open. Relate the set $$\mathbb{C} \setminus \Omega$$ of singularities of $$F$$ and the spectrum of $$A$$.

2. Show that the function $$F$$ is holomorphic and compute its derivative.

Hint: define the difference quotient $\epsilon(z, h) = \frac{F(z+h) - F(z)}{h}$ then compute $$\epsilon(z,h) [I - z A] [I - (z+h) A]$$ (optionally, consider first the case with $$p=1$$ and $$A = [a]$$ for some $$a \in \mathbb{C}$$ to guess the expression of the derivative in the general case).

3. Prove that the $$n$$-th order derivative $$F^{(n)}$$ of $$F$$ exists for every natural number $$n$$ and satisfies $\forall \, z \in \Omega, \; F^{(n)}(z) = n! A^n [I-zA]^{-n-1}.$ Hint: prove the result by induction on $$n$$.

4. What is the (possibly infinite) radius $$r(A)$$ of the largest open disk centered on the origin and included in $$\Omega$$ ? Relate $$r(A)$$ and the spectral radius of $$A$$, defined as $\rho(A) = \max \, \{|\lambda| \; | \; \lambda \in \sigma(A)\}.$ Find a sequence of matrices $$A_n \in \mathbb{C}^{p \times p}$$ such that $\forall \, z \in D(0,r(A)), \; F(z) = [I - zA]^{-1} = \sum_{n=0}^{+\infty} A_n z^n.$ (the convergence of this matrix series should be interpreted element-wise).

5. Let $$n \in \mathbb{N}$$. Let $$z \in \mathbb{C} \setminus \sigma(A)$$ such that $$z \neq 0$$; write $$z^n[z I - A]^{-1}$$ as a function of $$z$$ and $$F(z^{-1})$$. Show that the function $z \in \mathbb{C} \setminus \sigma(A)\mapsto z^n [zI - A]^{-1}$ is complex-differentiable.

6. Compute the line integral $\frac{1}{i2\pi}\int_{\gamma} z^n [zI - A]^{-1} dz$ first when $$\gamma = r[\circlearrowleft]$$ with $$r > \rho(A)$$, then more generally for any sequence of rectifiable closed paths $$\gamma$$ of $$\mathbb{C} \setminus \sigma(A)$$ such that $$\forall \, \lambda \in \sigma(A)$$, $$\mathrm{ind}(\gamma, \lambda) = 1$$.

7. Let $A = \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right]$ and let $$\gamma$$ be a sequence of rectifiable closed paths of $$\mathbb{C} \setminus \mathbb{R}_-$$ such that $$\forall \, \lambda \in \sigma(A)$$, $$\mathrm{ind}(\gamma, \lambda) = 1$$. Compute $$[zI - A]^{-1}$$ and then the matrix $$\log A$$, defined as $\log A = \frac{1}{i2\pi} \int_{\gamma} (\log z) [z I - A]^{-1} \, dz.$

## Answers

1. For any $$z \in \Omega$$, $$j \in \{1,\dots,m\}$$ and $$l \in \{1,\dots, p\}$$, $FG(z)_{jl} = \sum_{k=1}^{n} F(z)_{jk} G(z)_{kl} = \sum_{k=1}^n f_{jk}(z) g_{kl}(z).$ By the sum and product rules for complex-valued holomorphic functions, every element of $$FG$$ is holomorphic and $\begin{split} (FG)'(z)_{jl} &= \sum_{k=1}^n f'_{jk}(z) g_{kl}(z) + \sum_{k=1}^n f_{jk}(z) g'_{kl}(z) \\ &= F'G(z)_{jl} + F G'(z)_{jl}. \end{split}$
2. The function $$d: z \in \mathbb{C} \mapsto \det (I - zA)$$ is a polynomial, thus it is continuous; the matrix $$I - z A$$ is invertible if and only if $$d(z) \neq 0$$, hence $$\Omega = d^{-1}(\mathbb{C}^*)$$ is open as the preimage of an open set by a continuous function.

Since the matrix $$I - z A$$ is invertible for $$z=0$$, we have $$0 \in \Omega$$; if $$z\neq 0$$, $I - z A = z (z^{-1}I - A)$ and it is invertible if and only if $$z^{-1}$$ is not an eigenvalue of $$A$$. Finally, $\mathbb{C} \setminus \Omega = \{\lambda^{-1} \; | \; \lambda \in \sigma(A) \setminus \{0\}\}.$

3. For any pair $$(k,l) \in \{1,\dots,p\}^2$$, any $$z\in\Omega$$ and $$h \in \mathbb{C}$$ small enough, $\epsilon(z,h)_{kl} = \frac{f_{kl}(z+h) - f_{kl}(z)}{h}$ is the difference quotient of $$f_{kl}$$ at $$z$$. Therefore, $$F$$ is complex-differentiable at $$z$$ if (and only if) $$\epsilon(z, h)$$ has a limit element-wise when $$h \to 0$$ and $$F'(z)$$ is this limit. Now, we have $\begin{split} \epsilon(z,h) & = \frac{[I - (z+h) A]^{-1} - [I - zA]^{-1}}{h}\\ \end{split}$ The matrices $$I - (z+h)A$$ and $$I - zA$$ commute, hence $\epsilon(z,h) [I -(z+h) A][I - zA] = \frac{[I - zA] - [I - (z+h)A]}{h} = A,$ thus $$\epsilon(z, h) = A [I - z A]^{-1} [I - (z+h)A]^{-1}$$. Since matrix inversion is continuous $F'(z) = \lim_{h \to 0} \epsilon(z,h) = A [I - z A]^{-2}.$

4. The function $$F$$ is holomorphic; all of its elements are holomorphic, thus they have derivatives at all orders and so does $$F$$. Assume that the $$n+1$$-th order derivative of $$F$$ satisfies $F^{(n+1)}(z) = n A [I-zA]^{-1} F^{(n)}(z).$ This property certainly holds for $$n=0$$, as $$F'(z) = A [I-zA]^{-1} F(z)$$. Now, by the product rule, as $$A$$ and $$F(z) = [I - zA]^{-1}$$ commute, $\begin{split} F^{(n+2)}(z) &= n A F'(z) F^{(n)}(z) + n A F(z) F^{(n+1)}(z) \\ &= A F(z) (n A F(z) F^{(n)}(z)) + n A F(z) F^{(n+1)}(z) \\ &= A F(z) F^{(n+1)}(z) + n A F(z) F^{(n+1)}(z) \\ &= (n+1) A F(z) F^{(n+1)}(z) \\ \end{split}$ Consequently, the formula holds for all $$n$$, which leads to $F^{(n)}(z) = n! A^n [I-zA]^{-n-1}.$

5. The radius $$r(A)$$ of the largest disk centered on the origin and included in $$\Omega$$ is $r(A) = \min \{|\lambda^{-1}| \; | \; \lambda \in \sigma(A) \setminus \{0\}\} \\$ if $$\sigma(A) \setminus \{0\}$$ is non-empty and $$r(A) = +\infty$$ otherwise. If we adopt the convention $$0^{-1} = +\infty$$, we can rewrite this as $r(A) = (\max \{|\lambda| \; | \; \lambda \in \sigma(A)\})^{-1} = \rho(A)^{-1}.$ The function $$F$$ is holomorphic in $$\Omega$$, thus every element $$f_{kl}$$ is holomorphic in $$\Omega$$, which has a Taylor series expansion in $$D(0,r)$$: $\forall \, z \in D(0,r), \; f_{kl}(z) = \sum_{n=0}^{+\infty} \frac{f^{(n)}_{kl}(0)}{n!} z^n.$ Consequently, $\forall \, z \in D(0,r), \; F(z) = \sum_{n=0}^{+\infty} \frac{F^{(n)}(0)}{n!} z^n = \sum_{n=0}^{+\infty} A^n z^n.$

6. If $$z \in \mathbb{C} \setminus \sigma(A)$$ and $$z\neq 0$$, then $z^n[z I - A]^{-1} = z^{n-1}[I - z^{-1}A]^{-1} = z^{n-1} F(z^{-1}).$ Consequently, $$z \in \mathbb{C} \setminus \sigma(A) \mapsto z^n[z I - A]^{-1}$$ is complex-differentiable at $$z$$ – by the product and chain rules – if $$z\neq 0$$, which is all we need unless $$0 \not \in \sigma(A)$$. In this case, $$A$$ is invertible and by continuity of the matrix inversion, the function $$z\mapsto z^n[zI-A]^{-1}$$ is continuous at $$z=0$$. Hence, $$0$$ is a removable singularity of $$z\mapsto z^{n-1}F(z^{-1})$$ and the function $$z \in \mathbb{C} \setminus \sigma(A) \mapsto z^n[zI-A]^{-1}$$ is also complex-differentiable at $$z=0$$.

7. If $$|z|>\rho(A)$$, $$z \in \mathbb{C} \setminus \sigma(A)$$, $$|z^{-1}| < r(A)$$ and $z^n[z I - A]^{-1} = z^{n-1}[I - z^{-1}A]^{-1} = z^{n-1} \sum_{l=0}^{+\infty} A^{l} z^{-l} = \sum_{l=0}^{+\infty} A^{l} z^{n-1-l}.$ This convergence if uniform (element-wise) on compact subsets of the annulus, therefore $\begin{split} \frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^n [zI - A]^{-1} dz &= \frac{1}{i2\pi} \int_{r[\circlearrowleft]} \sum_{l=0}^{+\infty} A^{l} z^{n-1-l} dz \\ &= \sum_{l=0}^{+\infty} A^{l} \left[\frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^{n-1-l} dz\right] \\ \end{split}$ The integral $$\int_{r[\circlearrowleft]} z^{p} \, dz$$ is zero unless the integer $$p$$ is $$-1$$, in which case it is equal to $$i2\pi$$, hence $\frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^n [zI - A]^{-1} dz = A^{n}$

Now, If $$\gamma$$ is a sequence of rectifiable closed paths of $$\mathbb{C} \setminus \sigma(A)$$ such that $$\forall \, \lambda \in \sigma(A)$$, $$\mathrm{ind}(\gamma, \lambda) = 1$$ and $$r > \rho(A)$$, the sequence of paths $$\mu =r[\circlearrowleft] \, | \, \gamma^{\leftarrow}$$ satisfies $$\forall \, z \in \sigma(A)$$, $$\mathrm{ind}(\mu,z) = 0$$. Cauchy’s integral theorem provides $$\int_{\mu} z^n[zI-A]^{-1} \, dz = 0$$, or equivalently, $\frac{1}{i2\pi} \int_{\gamma} z^n[zI-A]^{-1} \, dz = \frac{1}{i2\pi} \int_{r[\circlearrowleft]} z^n[zI-A]^{-1} \, dz = A^n.$

8. We have $z I - A = \left[ \begin{array}{rr} z & +1 \\ -1 & z \end{array} \right]$ hence $$\det (zI -A)= z^2 +1$$. The spectrum of $$A$$ is $$\sigma(A) = \{-i, +i\}$$ and $[zI - A]^{-1} = \frac{1}{z^2+1} \left[ \begin{array}{rr} z & -1 \\ +1 & z \end{array} \right].$ Consequently, $\frac{1}{i2\pi} \int_{\gamma} (\log z) [z I - A]^{-1} \, dz = \frac{1}{i2\pi} \int_{\gamma} \left[ \begin{array}{rr} f(z) & -g(z) \\ g(z) & f(z) \end{array} \right]\, dz$ with $f(z) = \frac{z\log z}{z^2+1}, \; g(z) = \frac{\log z}{z^2+1}.$ Let $$\Omega = \mathbb{C} \setminus \mathbb{R}_- \setminus \sigma(A)$$; this set is open. Both functions $$f$$ and $$g$$ are holomorphic on $$\Omega$$, have isolated singularities in $$\sigma(A)$$ and are the quotient of holomorphic functions $$a$$ and $$b$$ defined on $$\Omega \cup \sigma(A)$$ such that when $$\lambda \in \sigma(A)$$, $$a(\lambda) \neq 0$$, $$b(\lambda) = 0$$ and $$b'(\lambda) \neq 0$$. Thus, we have $\sum_{z \in \sigma(A)} \mathrm{ind}(\gamma, z) \mathrm{res}(f, z) = \frac{(-i)\log (-i)}{2(-i)} + \frac{i\log i}{2i} = 0$ and $\sum_{z \in \sigma(A)} \mathrm{ind}(\gamma, z) \mathrm{res}(g, z) = \frac{\log (-i)}{2(-i)} + \frac{\log i}{2i} = \frac{\pi}{2}.$ Since the open set $\Omega \cup \sigma(A) = \mathbb{C} \setminus \mathbb{R}_-$ is simply connected (it is star-shaped), $$\mathrm{Int} \, \gamma \subset \Omega \cup \sigma(A)$$. Finally, the residue theorem provides $\log A = \frac{1}{i2\pi} \int_{\gamma} (\log z) [z I - A]^{-1} \, dz = \frac{\pi}{2} \left[ \begin{array}{rr} 0 & -1 \\ +1 & 0 \end{array} \right]$

# Problem L

A function $$f: \mathbb{C} \to \mathbb{C}$$ is of exponential type if $\exists \, \sigma \in \mathbb{R}, \; \exists \, \kappa > 0, \forall \, z \in \mathbb{C}, \; |f(z)| \leq \kappa e^{\sigma |z|}.$ We define $$\sigma(f)$$ as the infimum of the exponential bounds $$\sigma$$: $\sigma(f) = \inf \, \{ \sigma \in \mathbb{R} \; | \; \exists \, \kappa > 0, \forall \, z \in \mathbb{C}, \; |f(z)| \leq \kappa e^{\sigma |z|} \}.$

## Questions

We suppose that $$f$$ is complex-differentiable on $$\mathbb{C}$$ and denote $f(z) = \sum_{n=0}^{+\infty} a_n z^n$ its Taylor expansion at $$0$$.

1. Show that if $$f$$ is of exponential type, $$\sigma(f) \geq 0$$ unless $$f$$ is identically zero.

2. Show that the function $f_0: z \in \mathbb{C} \to e^{-z}$ is of exponential type and compute $$\sigma(f_0)$$.

3. Prove that if for some $$\kappa > 0$$ and $$\sigma \geq 0$$, $$\forall \, n \in \mathbb{N}$$, $$|n! a_n| \leq \kappa \sigma^n$$, then $$f$$ is of exponential type and $$\sigma(f) \leq \sigma$$.

4. Prove that if $$\forall \, z \in \mathbb{C}, \; |f(z)| \leq \kappa e^{\sigma |z|}$$ then $\forall \, n \in \mathbb{N}, \; \forall \, r > 0, \; |a_n| \leq \kappa \frac{e^{\sigma r}}{r^n}.$ Let $$n \in \mathbb{N}$$; what is the value of $$r$$ that provides the tightest upper bound on $$|a_n|$$? Compute this bound.

5. Prove that if $$f$$ is of exponential type and $$\sigma > \sigma(f)$$, there is a $$\kappa > 0$$ such that $$\forall \, n \in \mathbb{N}$$, $$|n! a_n| \leq \kappa \sigma^n$$. Reminder – Stirling’s formula: $n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n, \; \mbox{ that is } \; \lim_{n \to + \infty} \frac{n!e^n}{\sqrt{2\pi n} n^n} = 1.$

6. Suppose that $$f$$ is of exponential type; show that the function $$g:\mathbb{C} \to \mathbb{C}$$ defined by $g(z) = \sum_{n=0}^{+\infty} |a_n| z^n$ is of exponential type and $$\sigma(g) = \sigma(f)$$.

7. What is the domain of the Laplace transform of the function $$f_0$$? Compute $$\mathcal{L}[f_0](s)$$ for any $$s$$ in this domain. Prove that in general the domain of the Laplace transform of a function of exponential type $$f$$ contains the set $$\{s \in \mathbb{C} \; | \; \mathrm{Re} \, \, s > \sigma(f)\}$$.

8. Show that whenever $$\mathrm{Re} \, \, s >\sigma(f)$$, $\mathcal{L}[f](s) = \int_{\mathbb{R}_+} f(t) e^{-st} \, dt = \sum_{n=0}^{+\infty} \frac{n! a_n}{s^{n+1}}.$ Prove that $$\mathcal{L}[f]$$ has a unique holomorphic extension to the open set $\Omega = \mathrm{dom} \, \mathcal{L}[f] \cup \{s\in \mathbb{C} \; | \; |s| > \sigma(f)\}.$ We denote $$\overline{\mathcal{L}}[f]$$ this extension. Compute $$\overline{\mathcal{L}}[f_0]$$.

9. Compute for any $$t\geq 0$$ and any $$r> \sigma(f)$$ the line integral $\frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds.$

## Answers

1. If $$\sigma(f) < 0$$ then $$\sigma = 0$$ is an exponential bound of $$f$$: there is a $$\kappa > 0$$ such that $$\forall\, z \in \mathbb{C}$$, $$|f(z)| \leq \kappa$$. The function $$f$$ is entire and bounded, thus by Liouville’s theorem, it is identically zero.

2. For any complex number $$z$$, $|f_0(z)| = |e^{-z}| = e^{-\mathrm{Re}(z)} \leq e^{|z|}$ hence $$\sigma = 1$$ is an exponential bound of $$f_0$$. Conversely, if $$|f_0(z)| \leq \kappa e^{\sigma|z|}$$ holds for any complex number $$z$$, then $\forall \, x \geq 0, \; f_0(-x)=e^{x} \leq \kappa e^{\sigma |-x|} = \kappa e^{\sigma x},$ thus $$\sigma \geq 1$$. Finally, $$\sigma(f_0) = 1$$.

3. If for some $$\kappa > 0$$ and $$\sigma \geq 0$$, $$\forall \, n \in \mathbb{N}$$, $$|n! a_n| \leq \kappa \sigma^n$$, then $|f(z)| \leq \kappa \sum_{n=0}^{+\infty} \frac{(\sigma |z|)^n}{n!} = \kappa e^{\sigma |z|}$ hence $$f$$ is of exponential type and $$\sigma(f) \leq \sigma$$.

4. For any $$n \in \mathbb{N}$$ and $$r>0$$, $a_n = \frac{1}{i2\pi} \int_{r[\circlearrowleft]}\frac{f(z)}{z^{n+1}},$ hence if $$\forall \, z \in \mathbb{C}$$, $$|f(z)| \leq \kappa e^{\sigma |z|}$$, by the M-L inequality, $|a_n| \leq \frac{1}{2\pi} \kappa \frac{e^{\sigma r}}{r^{n+1}}\times 2\pi r = \kappa \frac{e^{\sigma r}}{r^n}.$ If $$n=0$$, the infimum of the right-hand side with respect to $$r>0$$ is $$\kappa$$. Otherwise, the right-hand side is a differentiable function of $$r>0$$ that tends to $$+\infty$$ when $$r\to 0$$ or $$r\to +\infty$$. Therefore, it has a minimum at a $$r>0$$ that satisfies $\frac{d}{dr} \ln \left(\kappa \frac{e^{\sigma r}}{r^n}\right) = 0.$ Given that $\frac{d}{dr} \ln \left(\kappa \frac{e^{\sigma r}}{r^n}\right) = \frac{d}{dr} \left(\ln \kappa + \sigma r - n \ln r\right) =\sigma - \frac{n}{r},$ we have $|a_n| \leq \min_{r>0} \kappa \frac{e^{\sigma r}}{r^n} = \kappa \left(\frac{e \sigma}{n}\right)^n.$

5. Let $$\sigma > \sigma(f)$$. Let $$\lambda < 1$$ such that $$\lambda \sigma > \sigma(f)$$; there is a $$\kappa > 0$$ such that $\forall \, z \in \mathbb{C}, \; |f(z)| \leq \kappa e^{\lambda\sigma|z|}.$ The result of the previous question yields for any $$n>0$$ $|a_n| \leq \kappa \left(\frac{e \lambda \sigma}{n}\right)^n$ and $$|a_0| \leq \kappa$$, therefore $\forall \, n \in \mathbb{N}^*, \; |n ! a_n| \leq \kappa \left[ {\sqrt{2\pi n}} \lambda^n(1+\epsilon_n) \right] \sigma^n \; \; \mathrm{ with } \lim_{n \to +\infty} \epsilon_n = 1.$ The sequence $\left|\kappa \left[ {\sqrt{2\pi n}} \lambda^n(1+\epsilon_n) \right] \right|, \; n \in \mathbb{N}^*$ converges to $$0$$ when $$n$$ tends to $$+\infty$$ and thus has some upper bound $$\kappa'$$ that we may select greater than $$\kappa$$. Finally, $$|n! a_n| \leq \kappa' \sigma^n$$ for any $$n\in \mathbb{N}$$.

6. The sum $$\sum_{n=0}^{+\infty} |a_n| z^n$$ is convergent for any $$z \in \mathbb{C}$$ because the Taylor expansion of $$f$$ is absolutely convergent on the whole complex plane; the function $g: z \in \mathbb{C} \mapsto \sum_{n=0}^{+\infty} |a_n| z^n$ is actually an entire function.

For any $$\sigma$$ such that $$\sigma(f) < \sigma$$, by question 5, there is a $$\kappa > 0$$ such that $$|n! a_n| \leq \kappa \sigma^n$$ and thus $$|n! |a_n|| \leq \kappa \sigma^n$$. By question 3, the function $$g$$ is of exponential type and $$\sigma(g) \leq \sigma$$; thus $$\sigma(g) \leq \sigma(f)$$. The converse inequality may be proved by the same kind of argument (or directly as $$|f(z)| \leq g(|z|)$$).

7. For any $$\sigma \in \mathbb{R}$$, $$e^{-\sigma t} f_0(t) = e^{-(\sigma + 1)t},$$ hence the function $$t \in \mathbb{R}_+ \mapsto e^{-\sigma t} f_0(t)$$ is integrable if and only if $$\sigma > -1$$. Therefore, the Laplace transform of the function $$f_0$$ is defined exactly on $$\{s \in \mathbb{C} \; | \; \mathrm{Re} \, \, s > -1\}.$$ For any such $$s$$, we have $\mathcal{L}[f_0](s) = \int_{0}^{+\infty} e^{-t} e^{-st} \, dt = \left[-\frac{e^{-(s+1)t}}{s+1}\right]^{+\infty}_0 = \frac{1}{s+1}.$ In general, if $$f$$ is an entire function of exponential type, then for any complex number $$s$$ such that $$\sigma = \mathrm{Re}(s) > \sigma(f)$$, there is a $$\epsilon > 0$$ such that $$\sigma - \epsilon > \sigma(f)$$, hence there is some $$\kappa >0$$ such that $\forall \,t > 0, \; |f(t) e^{-s t}| = |f(t)|e^{-\sigma t} \leq \kappa e^{-\epsilon t},$ thus the function $$t \in \mathbb{R}_+ \mapsto f(t) e^{-st}$$ is integrable.

8. If $$\mathrm{Re}(s) > \sigma(f)$$, we have $\mathcal{L}[f](s) = \int_{\mathbb{R}_+} \left[ \sum_{n=0}^{+\infty} a_n t^n \right] e^{-st} \, dt.$ For any $$m \in \mathbb{N}$$, $\left| \left[ \sum_{n=0}^{m} a_n t^n \right] e^{-st} \right| \leq \left[ \sum_{n=0}^{+\infty} |a_n| t^n \right] e^{-(\mathrm{Re}(s)) t} = g(t) e^{-(\mathrm{Re}(s)) t}.$ Therefore, as $$\sigma(g) = \sigma(f)$$, for any $$\sigma>0$$ such that $$\sigma(f) < \sigma < \mathrm{Re}(s)$$, there is a $$\kappa > 0$$ such that $\left| \left[ \sum_{n=0}^{m} |a_n| t^n \right] e^{-st} \right| \leq \kappa e^{\sigma t} e^{-(\mathrm{Re}(s))t} = \kappa e^{-(\mathrm{Re}(s) - \sigma) t}.$ The right-hand side of this inequality is an integrable function of $$t$$; by the dominated convergence theorem $\mathcal{L}[f](s) = \int_{\mathbb{R}_+} \left[ \sum_{n=0}^{+\infty} a_n t^n \right] e^{-st} \, dt = \sum_{n=0}^{+\infty} a_n \left[ \int_{\mathbb{R}_+} t^n e^{-st} \, dt \right].$ For any natural number $$n$$, we have $\int_{\mathbb{R}_+} t^n e^{-st} \, dt = \mathcal{L}[t\mapsto t^n](s) = \frac{n!}{s^{n+1}}$ and therefore $\mathcal{L}[f](s) = \sum_{n=0}^{+\infty} \frac{n! a_n}{s^{n+1}}.$

Now, the right-hand side of this equation is a Laurent series centered on the origin which is convergent whenever $$\mathrm{Re}(s) > \sigma(f)$$. Consequently, the sum $$g(s)$$ is actually convergent and holomorphic in the annulus $$A(0, \sigma(f), +\infty)$$.

Now, the set $\Omega = \mathrm{dom} \, \mathcal{L}[f] \cup A(0,\sigma(f), +\infty).$ is open and connected as the union of two connected sets with a non-empty intersection. We may define on $$\Omega$$ the function $\overline{\mathcal{L}}[f](s) = \left| \begin{array}{cl} \mathcal{L}[f](s) & \mbox{ if } \, s\in \mathrm{dom} \, \mathcal{L}[f], \\ g(s) & \mbox{ if } |s| > \sigma(f). \end{array} \right.$ This definition is non-ambiguous: since $$\mathcal{L}[f](s)$$ and $$g$$ are holomorphic and identical on the non-empty set $$\{s \in \mathbb{C} \; | \; \mathrm{Re} \, s > \sigma(f)\}$$, by the isolated zeros theorem, they have the same values on the connected open set $\mathrm{dom} \, \mathcal{L}[f] \cap A(0,\sigma(f), +\infty).$

The function $$\overline{\mathcal{L}}[f]$$ clearly is a holomorphic extension of $$\mathcal{L}[f]$$ to $$\Omega$$. This extension is unique: the set $$\Omega$$ is connected as the union of two connected sets with a non-empty intersection, therefore the isolated zeros theorem also applies.

For $$f=f_0$$, we have $\mathrm{dom} \, \mathcal{L}[f] = \{ s \in \mathbb{C} \; | \; \mathrm{Re}(s) > -1 \}$ and as $$\sigma(f_0) = 1$$, $A(0, \sigma(f_0), +\infty) = \{s \in \mathbb{C} \; | \; |s| > 1\},$ thus $$\Omega = \mathbb{C} \setminus \{-1\}$$. The function $s \in \mathbb{C} \setminus \{-1\} \mapsto \frac{1}{s+1}$ is holomorphic. As it extends the original Laplace transform of $$f_0$$, this is the extended Laplace transform of $$f_0$$.

9. For any $$t\geq 0$$ and any $$r> \sigma(f)$$, $\frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds = \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \left[ \sum_{n=0}^{+\infty} \frac{n! a_n}{s^{n+1}} \right] e^{st} \, ds.$ The series $$\sum_{n=0}^{+\infty} {n! a_n}/{s^{n+1}}$$ is convergent, uniformly with respect to $$s$$ on the circle of radius $$r$$ centered on the origin, hence $\frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds = \sum_{n=0}^{+\infty} n! a_n \left[ \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \frac{e^{st}}{s^{n+1}} \, ds \right].$ The function $$s \in \mathbb{C}^* \mapsto {e^{st}}/{s^{n+1}}$$ is complex-differentiable, therefore $\frac{1}{i2\pi}\int_{r[\circlearrowleft]} \frac{e^{st}}{s^{n+1}} \, ds = \mathrm{res} \left(s \mapsto \frac{e^{st}}{s^{n+1}}, 0\right).$ Since $\lim_{s \to 0} s^{n+1} \frac{e^{st}}{s^{n+1}} = 1,$ the origin is a pole of order $$n+1$$ of this function and $\mathrm{res} \left(s \mapsto \frac{e^{st}}{s^{n+1}}, 0\right) = \lim_{s \to 0} \frac{1}{n!}\frac{d^n}{ds^n}e^{st} = \frac{t^n}{n!}.$ Finally, $\frac{1}{i2\pi}\int_{r[\circlearrowleft]} \overline{\mathcal{L}}[f] e^{st} \, ds = \sum_{n=0}^{+\infty} n! a_n \frac{t^n}{n!} = \sum_{n=0}^{+\infty} a_n t^n = f(t).$