# Exercises

## Taylor Series of a Rational Function

### Questions

1. Show that the function $f: x \in \mathbb{R} \mapsto \frac{1}{1+x^2}$ is analytic.

2. Determine for any $$x_0 \in \mathbb{R}$$ the open interval of convergence of its Taylor series expansion at $$x_0$$.

1. The function $$f$$ is the restriction of the holomorphic function $f^*: z \in \mathbb{C} \setminus \{i, -i\} \mapsto \frac{1}{1+z^2}.$

2. For any $$x_0 \in \mathbb{R}$$, the disk $D(x_0) = D(x_0, \sqrt{1+x_0^2})$ is included in $$\mathbb{C} \setminus \{i, -i\}$$. The radius of the disk of convergence of the Taylor expansion of $$f^*$$ at $$x_0$$ is therefore at least $$\sqrt{1+x_0^2}$$; it cannot exceed this threshold: otherwise, the sum $$g(z)$$ of its Taylor series would be defined and holomorphic in an open set that contains $$\overline{D}_0$$ and therefore bounded on $$D_0$$; but $$g$$ and $$f^*$$ are identical on $$D_0$$ where $$f^*$$ is unbounded. Finally, as the Taylor expansion of $$f$$ at $$x_0$$ has the same coefficient as the Taylor expansion of $$f^*$$ at $$x_0$$, the open interval of convergence of $$f^*$$ at $$x_0$$ is $\left]x_0 - \sqrt{1+ x_0^2}, x_0 + \sqrt{1+x_0^2} \right[.$

## Analytic Functions of a Real Variable

### Questions

1. Show that the function $$f: \mathbb{R} \to \mathbb{C}$$ defined by $f(x) = \left| \begin{array}{rl} e^{-1/x} & \mbox{if } \, x > 0 \\ 0 & \mbox{otherwise.} \end{array} \right.$ is smooth but is not analytic.

2. Let $$K$$ be an compact interval of $$\mathbb{R}$$ and $$f: K \to \mathbb{C}$$ be a smooth function. Show that $$f$$ is analytic if and only if there are positive constants $$\alpha > 0$$ and $$r>0$$ such that $\forall \, x \in K, \forall \, n \in \mathbb{N}, \; |f^{(n)}(x)| \leq \alpha \, r^n n!$

1. By induction, for any $$x>0$$, $$f^{(n)}(x) = g_n(x) e^{-1/x}$$ where $$g_n$$ is a rational function, defined for $$x>0$$ by $g_0(x) = 1 \; \wedge \; \forall \, n \in \mathbb{N}, \; g_{n+1}(x) = g'_n(x) + \frac{g_n(x)}{x^2}.$ On the other hand, for $$x \leq 0$$, the $$n$$-th order derivative (left-derivative at $$x=0$$) of $$f$$ at $$x$$ is defined and equal to $$0$$. To prove that $$f$$ is smooth, we now have to prove that the right, $$n$$-th order derivative of $$f$$ at $$0$$ exists and is equal to its left derivative, that is zero. We may proceed by induction: suppose that $$f^{(n)}(0)$$ exists and is zero; then $\frac{f^{(n)}(h)-f^{(n)}(0)}{h} = \frac{1}{h}\int_0^h f^{(n+1)}(x) dx.$ But for any $$n$$, we have $\lim_{x \to 0^+} f^{(n+1)}(x) = \lim_{x \to 0^+} g_{n+1}(x) e^{-1/x} = 0,$ hence the right-hand side of the equation tends to zero when $$h\to 0^+$$: the $$n+1$$-th order right derivative of $$f$$ exists at $$x=0$$ and is equal to zero.

Now, the function $$f$$ cannot be analytic: given that its derivatives at $$x=0$$ are zero at all order, its Taylor series expansion at the origin is zero. The function $$f$$ would be zero in a neighbourhood of the origin, and this property does not hold.

2. If the function $$f$$ is smooth, for any real numbers $$c$$ and $$y$$ in $$K$$, the Taylor formula with integral remainder is applicable at any order $$n$$: $f(y) = \sum_{p=0}^n \frac{f^{(p)}(c)}{p!} (y-c)^p + \int_c^y \frac{f^{(n+1)}(x)}{n!}(x-c)^n dx.$ If there exist $$\alpha>0$$ and $$r>0$$ such that the inequality $\forall \, x \in K, \; |f^{(n)}(x)| \leq \alpha \, r^n n!$ holds, the remainder satisfies $\begin{split} \left|\int_c^y \frac{f^{(n+1)}(x)}{n!}(x-c)^n dx \right| & \leq \alpha \, r^{n+1} \frac{(n+1)!}{n!} \left| \int_c^y (x - c)^n \, dx \right| \\ & = \alpha \, (r |y-c|)^{n+1} \end{split}.$ Thus, if $$|y-c| < 1/r$$, the Taylor expansion of $$f$$ at $$y$$ centered on $$c$$ is convergent. As $$c$$ is an arbitrary point of $$I$$, the function $$f$$ is analytic.

Conversely, if $$f$$ is analytic, it has a holomorphic extension – that we may still denote $$f$$ – to some open neighbourhood $$U$$ of $$K$$. The distance $$d$$ between $$K$$ and $$\mathbb{C} \setminus U$$ is positive: for any $$c \in K$$, the disk $$D(c,d)$$ is included in $$U$$. Let $$r$$ be a positive radius smaller than $$d$$ and $$\alpha$$ be an upper bound of $$|f|$$ on $$K + \overline{D}(0,r)$$; for any natural number $$n$$, we have $\left|\frac{f^{(n)}(c)}{n!}\right| = \left|\frac{1}{i2\pi}\int_{r[\circlearrowleft]+c} \frac{f(z)}{(z-c)^{n+1}} dz \right| \leq \alpha \left(r^{-1}\right)^n$ which concludes the proof.

## Periodic Analytic Functions

### Notations.

‌ In this exercise, $$\mathbb{U}$$ is the unit circle centered at the origin: $\mathbb{U} = \{z \in \mathbb{C} \; | \; |z| = 1\}.$ For any radius $$0 \leq r < 1$$, we define the annulus $A_r = A(0,r, 1/r) = \{z \in \mathbb{C} \; | \; r < |z| <1/r \}$ (with the convention that $$A_0 = \mathbb{C}^*$$). For any $$0 < \epsilon \leq +\infty$$, the notation $$\Omega_{\epsilon}$$ refers to the horizontal strip $\Omega_{\epsilon} = \{z \in \mathbb{C} \; | \; |\mathrm{Im} \,\, z| < \epsilon\}.$

### Questions

Let $$f:\mathbb{U} \to \mathbb{C}$$ be a function with an analytic extension in some open neighbourhood $$U$$ of $$\mathbb{U}$$.
1. Prove that there is an annulus $$A_r$$ such that $${A}_r \subset U$$.

2. Let $$g: t \in \mathbb{R} \mapsto f(e^{it})$$; show that $$g$$ is a $$2\pi$$-periodic analytic function.

Conversely, let $$g: t \in \mathbb{R} \to \mathbb{C}$$ be a $$2\pi$$-periodic analytic function.

1. Show there is an analytic extension $$g^*$$ of $$g$$ on some strip $$\Omega_{\epsilon}$$, such that $\forall \, z \in \Omega_{\epsilon}, \; g^*(z+2\pi) = g^*(z).$

2. Show that there exist a function $$f:\mathbb{U} \to \mathbb{C}$$ with an analytic extension in some open neighbourhood of $$\mathbb{U}$$ such that $\forall \, t \in \mathbb{R}, \; g(t)= f(e^{it}).$

1. The set $$\mathbb{U}$$ is compact and the set $$\mathbb{C} \setminus U$$ is closed; their intersection is empty, thus the distance $$d = d(\mathbb{U}, \mathbb{C} \setminus U)$$ is positive (it may be $$+\infty$$ if $$U = \mathbb{C}$$). On the other hand, for any $$r<1$$, $d(\mathbb{U}, \mathbb{C} \setminus A_r) = \min(1- r, 1/r - 1) = 1 - r.$ Thus, for any $$r$$ such that $$1 - r < d$$, the annulus $${A}_r$$ is a subset of $$U$$.

2. The $$2\pi$$-periodicity of $$g$$ is clear: for any $$t\in \mathbb{R}$$, $g(t + 2\pi) = f(e^{i(t+2\pi)}) = f(e^{it} e^{i2\pi}) = f(e^{it}) = g(t).$

The assumption on $$f$$ and the result of the previous question provide a holomorphic extension $$f^*:A_r \to \mathbb{C}$$ to $$f:\mathbb{U} \to \mathbb{C}$$ for some $$r<1$$. Now, $|e^{iz}| = e^{\mathrm{Re} \,\, iz} = e^{-\mathrm{Im} \,\, z},$ thus if $$|\mathrm{Im} \,\, z| < \ln 1/r$$, then $$\ln r < -\mathrm{Im} \,\, z < \ln 1/r$$ which yields $$r <|e^{iz}|<1/r$$. Therefore, if we set $$\epsilon = \ln 1/r > 0$$, we have $\forall \, z \in \mathbb{C}, \; (z \in \Omega_{\epsilon} \, \Rightarrow \, e^{iz} \in A_r).$

Consequently, setting $$g^*(z) = f^*(e^{iz})$$ defines a function $$g^*$$ on $$\Omega_{\epsilon}$$; it is an extension of $$g:\mathbb{R} \to \mathbb{C}$$ and it is holomorphic as the composition of holomorphic functions. Therefore, the function $$g:\mathbb{R} \to \mathbb{C}$$ is analytic.

Alternate proof. Consider the Laurent series expansion of $$f^*$$ in $$A_r$$: $f^*(z) = \sum_{n = -\infty}^{+\infty} a_n z^n.$ For any real numbers $$t_0$$ and $$t$$, we have $(e^{it})^n = e^{int} = e^{int_0} e^{in(t-t_0)} = e^{int_0} \sum_{m =0}^{+\infty} \frac{1}{m!} i^m n^m (t-t_0)^m,$ therefore $g(t) = f(e^{it}) = \sum_{n = -\infty}^{+\infty} a_n e^{int_0} \left[ \sum_{m =0}^{+\infty} \frac{1}{m!} i^m n^m (t-t_0)^m \right].$ We can change the order of the summation in this double series if $\sum_{(m,n) \in \mathbb{N} \times \mathbb{Z}} \left| a_n e^{int_0} \frac{1}{m!} i^m n^m (t-t_0)^m \right| < +\infty$ The general term of this double series satisfies $\left|a_n e^{int_0} \frac{1}{m!} i^m n^m (t-t_0)^m \right| \leq |a_n| \frac{1}{m!}|n|^m |t-t_0|^m$ hence the sum is bounded by $\sum_{n=-\infty}^{+\infty} |a_n| (e^{|t-t_0|})^{|n|} \leq \sum_{n=-\infty}^{+\infty} |a_n| (e^{t-t_0})^{n} + \sum_{n=-\infty}^{+\infty} |a_n| (e^{-(t-t_0)})^{n}.$ The Laurent series expansion of $$f^*$$ is absolutely convergent in $$A_r$$, hence the sums in the right-hand side of this inequality are finite if $r < e^{t-t_0} < 1/r \; \mbox{ and } \; r < e^{-(t-t_0)} < 1/r$ that is if $$|t-t_0| < \epsilon = \ln 1/r$$. After the change in the order of the summation, we end up with: $\forall \, t \in \mathbb{R}, \; |t - t_0| < \epsilon \; \Rightarrow \; g(t) = \sum_{m = 0}^{+\infty} b_m (t-t_0)^m$ where $b_m = \left[ \sum_{n =-\infty}^{+\infty} a_n e^{int_0} \frac{1}{m!} i^m n^m \right],$ hence the function $$g$$ is analytic.

3. The function $$g$$ is analytic; let $$g^{0}$$ be an analytic extension of $$g$$ in some open neighbourhood $$\Omega$$ of $$\mathbb{R}$$. However, if the distance between $$\mathbb{R}$$ and $$\mathbb{C} \setminus \Omega$$ is equal to zero – it may happen as both sets but neither of them is compact – then $$\Omega$$ contains no strip $$\Omega_{\epsilon}$$.

Let’s build a new analytic extension $$g^*$$ on such a strip from $$g^0$$. First, the set $$\Omega$$ contains some open tubular neighbourhood $$V_{\epsilon}$$ of $$[0,2\pi]$$ for any $$\epsilon > 0$$ small enough: $V_{\epsilon} = \{z \in \mathbb{C} \; | \; d(z, [0,2\pi]) < \epsilon\} \subset \Omega.$ Indeed, $$[0,2\pi]$$ is compact, $$\mathbb{C} \setminus \Omega$$ is closed and their intersection is empty, hence $$d(\mathbb{C} \setminus \Omega, [0,2\pi]) > 0$$; any $$\epsilon$$ smaller than (or equal to) this distance is admissible.

Consider the function $$g^*$$ defined on $$\Omega_{\epsilon}$$ by $g^*(z) = g^0(z+2\pi k) \; \mbox{ if } \; k \in \mathbb{Z} \; \mbox{ and } \; z + 2\pi k \in V_{\epsilon}.$ It is plain that $$g^*$$ is analytic and extends $$g$$ to $$\Omega_{\epsilon}$$; by construction it also satisfies the property $\forall \, z \in \Omega_{\epsilon}, \; g^*(z+2\pi) = g^*(z).$ The only point to check is that this definition is unambiguous, as we may have for some $$z$$ several integers $$k$$ and $$\ell$$ such that $$z_k = z + 2\pi k \in V_{\epsilon}$$ and $$z_{\ell} = z + 2\pi \ell \in V_{\epsilon}$$. Assume for example that $$k < \ell$$; in this case $$z_k \in D(0,\epsilon)$$ and $$\ell = k + 1$$, i.e. $$z_{\ell} = z_k + 2\pi$$. The functions $w \in D(0,\epsilon) \mapsto g^{0}(w) \; \mbox{ and } \; w \in D(0,\epsilon) \mapsto g^{0}(w+2\pi)$ are holomorphic and identical on $$\left]-\epsilon,\epsilon\right[$$; by the isolated zeros theorem, they are identical on $$D(0,\epsilon)$$ (which is connected) and in particular $$g(z_k) = g(z_{\ell})$$. The definition of $$g^*$$ is actually unambiguous.

4. To answer the question, we exhibit an analytic function $$f^*: A_r \to \mathbb{C}$$ with $$\epsilon = \ln 1/r$$ (or equivalently $$r = e^{-\epsilon}$$) such that $\forall \, z \in A_r, \; f^*(e^{iz}) = g^*(z).$ For any $$w \in \mathbb{C}^*$$, there is a solution $$z_0$$ to the equation $e^{iz} = w, \; z \in \mathbb{C}$ and the other solutionss are $$z_0 + 2\pi k$$, for $$k \in \mathbb{Z}$$. Additionally, if $$w \in A_r$$, then $$z \in \Omega_{\epsilon}$$ with $$\epsilon = \ln 1/r$$. We may define $$f^*: A_r \to \mathbb{C}$$ by $f^*(w) = g^{*}(z), \; e^{iz} = w.$ This definition is unambiguous: two $$z$$ that correspond to the same $$w$$ differ from a multiple of $$2\pi$$, but $$g^*$$ is $$2\pi$$-periodic hence the right-hand sides of this definition are equal.

Let’s prove that $$f^*$$ is analytic. Let $$w_0$$ in $$A_r$$ and $$z_0$$ such that $$e^{iz_0} = w_0$$, the expression $\phi(w) = -i \log \frac{w}{w_0} + z_0$ defines an analytic function $$\phi$$ in an neighbourhood of $$w_0$$. It satisfies $$e^{i(\phi(w)-z_0)} = w/w_0$$, thus $e^{i\phi(w)} = w.$ Consequently, in a neighbourhood of $$w_0$$, $f^*(w) = g^*(\phi(w))$ and $$f^*$$ is holomorphic – locally everywhere – as a composition of holomorphic functions.