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# 3.5.1 Appendix: Real Analytic Functions

### Multi-index notation

The following multi-index notation simplifies many presentations of formulas. Let $$x=(x_1,\ldots,x_n)$$ and
$$u:\ \Omega\subset\mathbb{R}\mapsto \mathbb{R}^1\ \ (\mbox{or}\ \mathbb{R}^m\ \mbox{for systems}).$$
The n-tuple of non-negative integers (including zero)
$$\alpha=(\alpha_1,\ldots,\alpha_n)$$
is called multi-index. Set
\begin{eqnarray*}
|\alpha|&=&\alpha_1+\ldots+\alpha_n\\
\alpha!&=&\alpha_1!\alpha_2!\cdot\ldots\cdot\alpha_n!\\
x^\alpha&=&x_1^{\alpha_1}x_2^{\alpha_2}\cdot\ldots\cdot x_n^{\alpha_n}\ \ (\mbox{for a monom})\\
D_k&=&\frac{\partial}{\partial x_k}\\
D&=&(D_1,\ldots,D_n)\\
D^\alpha&=&D_1^{\alpha_1}D_2^{\alpha_2}\cdot\ldots\cdot D_n^{\alpha_n}\equiv\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\partial x_2^{\alpha_2}\ldots\partial x_n^{\alpha_n}}.
\end{eqnarray*}
Define a partial order by
$$\alpha\ge\beta\ \ \mbox{if and only if}\ \ \alpha_i\ge\beta_i\ \ \mbox{for all}\ i.$$
Sometimes we use the notations
$${\bf 0}=(0,0\ldots,0), \ \ \bf{ 1}=(1,1\ldots,1),$$
where
$${\bf 0},\ {\bf 1}\in\mathbb{R}.$$

Using this multi-index notion, we have

1.
$$(x+y)^\alpha=\sum_{\begin{array}{c}\beta,\gamma\\ \beta+\gamma=\alpha\end{array}}\frac{\alpha!}{\beta!\gamma!}x^\beta y^\gamma,$$
where $$x,\ y\in\mathbb{R}$$ and $$\alpha,\ \beta,\ \gamma$$ are multi-indices.

2. Taylor expansion for a polynomial $$f(x)$$ of degree $$m$$:
$$f(x)=\sum_{|\alpha|\le m}\frac{1}{\alpha!}\left(D^\alpha f(0)\right) x^\alpha,$$
here is $$D^\alpha f(0):=\left(D^\alpha f(x)\right)|_{x=0}$$.

3. Let $$x=(x_1,\ldots,x_n)$$ and $$m\ge0$$ an integer, then
$$(x_1+\ldots+x_n)^m=\sum_{|\alpha|=m}\frac{m!}{\alpha!}x^\alpha.$$

4.
$$\alpha!\le|\alpha|!\le n^{|\alpha|}\alpha!.$$

5. Leibniz's rule:
$$D^\alpha(fg)= \sum_{\begin{array}{c}\beta,\gamma\\ \beta+\gamma=\alpha\end{array}}\frac{\alpha!}{\beta!\gamma!}(D^\beta f)(D^\gamma g).$$

6.
\begin{eqnarray*}
D^\beta x^\alpha&=&\frac{\alpha!}{(\alpha-\beta)!}x^{\alpha-\beta}\ \ \mbox{if}\ \alpha\ge\beta,\\
D^\beta x^\alpha&=&0\ \ \mbox{otherwise}.
\end{eqnarray*}

7. Directional derivative:
$$\frac{d^m}{dt^m}f(x+ty)=\sum_{|\alpha|=m}\frac{|\alpha|!}{\alpha!}\left(D^\alpha f(x+ty)\right)y^\alpha,$$
where $$x,\ y\in\mathbb{R}$$ and $$t\in\mathbb{R}^1$$.

8. Taylor's theorem: Let $$u\in C^{m+1}$$ in a neighborhood $$N(y)$$ of $$y$$, then, if $$x\in N(y)$$,
$$u(x)=\sum_{|\alpha|\le m}\frac{1}{\alpha!}\left(D^\alpha u(y)\right)(x-y)^\alpha+R_m,$$
where
$$R_m=\sum_{|\alpha|=m+1}\frac{1}{\alpha!}\left(D^\alpha u(y+\delta(x-y))\right)x^\alpha,\ 0<\delta<1,$$
$$\delta=\delta(u,m,x,y)$$,
or
$$R_m=\frac{1}{m!}\int_0^1\ (1-t)^m\Phi^{(m+1)}(t)\ dt,$$
where $$\Phi(t)=u(y+t(x-y))$$. It follows from 7. that
$$R_m=(m+1)\sum_{|\alpha|=m+1}\frac{1}{\alpha!}\left(\int_0^1\ (1-t)D^\alpha u(y+t(x-y))\ dt\right)(x-y)^\alpha.$$

9. Using multi-index notation, the general linear partial differential equation of order $$m$$  can be written as
$$\sum_{|\alpha|\le m}a_\alpha (x)D^\alpha u=f(x)\ \ \mbox{in}\ \Omega\subset\mathbb{R}.$$

### Power series

Here we collect some definitions and results for power series in $$\mathbb{R}$$.

Definition. Let $$c_\alpha\in\mathbb{R}^1\ (\mbox{or}\ \in\mathbb{R}^m)$$. The series
$$\sum_\alpha c_\alpha\equiv\sum_{m=0}^\infty\left(\sum_{|\alpha|=m}c_\alpha\right)$$
is said to be convergent if
$$\sum_\alpha |c_\alpha|\equiv\sum_{m=0}^\infty\left(\sum_{|\alpha|=m}|c_\alpha|\right)$$
is convergent.

Remark. According to the above definition, a convergent series is absolutely convergent. It follows that we can rearrange the order of summation.

Using the above multi-index notation and keeping in mind that we can rearrange convergent series, we have

10. Let $$x\in\mathbb{R}$$, then
\begin{eqnarray*}
\sum_\alpha x^\alpha&=&\prod_{i=1}^n\left(\sum_{\alpha_i=0}^\infty x_i^{\alpha_i}\right)\\
&=&\frac{1}{(1-x_1)(1-x_2)\cdot\ldots\cdot(1-x_n)}\\
&=&\frac{1}{({\bf 1}-x)^{\bf 1}},
\end{eqnarray*}
provided $$|x_i|<1$$ is satisfied for each $$i$$.

11. Assume $$x\in\mathbb{R}$$ and $$|x_1|+|x_2|+\ldots+|x_n|<1$$, then
\begin{eqnarray*}
\sum_\alpha\frac{|\alpha|!}{\alpha!}x^\alpha&=&\sum_{j=0}^\infty\sum_{|\alpha|=j}\frac{|\alpha|!}{\alpha!}x^\alpha\\
&=&\sum_{j=0}^\infty(x_1+\ldots+x_n)^j\\
&=&\frac{1}{1-(x_1+\ldots+x_n)}.
\end{eqnarray*}

12. Let $$x\in\mathbb{R}$$, $$|x_i|<1$$ for all $$i$$, and $$\beta$$ is a given multi-index. Then
\begin{eqnarray*}
\sum_{\alpha\ge\beta}\frac{\alpha!}{(\alpha-\beta)!}x^{\alpha-\beta}&=&D^\beta\frac{1}{({\bf 1}-x)^1}\\
&=&\frac{\beta!}{({\bf 1}-x)^{1+\beta}}\
\end{eqnarray*}

13. Let $$x\in\mathbb{R}$$ and $$|x_1|+\ldots+|x_n|<1$$. Then
\begin{eqnarray*}
\sum_{\alpha\ge\beta}\frac{|\alpha|!}{(\alpha-\beta)!}x^{\alpha-\beta}&=&D^\beta\frac{1}{1-x_1-\ldots-x_n}\\
&=&\frac{|\beta|!}{(1-x_1-\ldots-x_n)^{1+|\beta|}}\ .
\end{eqnarray*}

Consider the power series

\label{power}\tag{3.34}
\sum_\alpha c_\alpha x^\alpha

and assume this series is convergent for a $$z\in\mathbb{R}$$. Then, by definition,
$$\mu:=\sum_\alpha|c_\alpha||z^\alpha|<\infty$$
and the series (\ref{power}) is uniformly convergent for all $$x\in Q(z)$$, where
$$Q(z):\ \ |x_i|\le|z_i|\ \ \mbox{for all}\ \ i.$$

Figure 3.5.1.1: Definition of $$D\in Q(z)$$

Thus the power series (\ref{power}) defines a continuous function defined on $$Q(z)$$, according to a theorem of Weierstrass.

The interior of $$Q(z)$$ is not empty if and only if $$z_i\not=0$$ for all $$i$$, see Figure 3.5.1.1.
For given $$x$$ in a fixed compact subset $$D$$ of $$Q(z)$$ there is a $$q$$, $$0<q<1$$, such that
$$|x_i|\le q|z_i|\ \ \mbox{for all}\ i.$$
Set
$$f(x)=\sum_\alpha c_\alpha x^\alpha.$$

Proposition A1. (i) In every compact subset $$D$$ of $$Q(z)$$ one has $$f\in C^\infty(D)$$ and
the formal differentiate series, that is $$\sum_\alpha D^\beta c_\alpha x^\alpha$$, is uniformly convergent on the closure of $$D$$ and is equal to $$D^\beta f$$.
}

(ii)
$$|D^\beta f(x)|\le M|\beta|!r^{-|\beta|}\ \ \mbox{in}\ D,$$
where
$$M=\frac{\mu}{(1-q)^n},\qquad \qquad r=(1-q)\min_i|z_i|.$$

Proof. See F. John [10], p. 64. Or an exercise. Hint: Use formula 12. where $$x$$ is replaced by $$(q,\ldots,q)$$.

Remark. From the proposition above it follows
$$c_\alpha=\frac{1}{\alpha!}D^\alpha f(0).$$

Definition. Assume $$f$$ is defined on a domain $$\Omega\subset\mathbb{R}$$, then $$f$$ is said to be {\it real analytic in $$y\in\Omega$$} if there are $$c_\alpha\in\mathbb{R}^1$$ and if there is a neighborhood $$N(y)$$ of $$y$$ such that
$$f(x)=\sum_\alpha c_\alpha(x-y)^\alpha$$
for all $$x\in N(y)$$, and the series converges (absolutely) for each $$x\in N(y)$$.
A function $$f$$ is called {\it real analytic in $$\Omega$$} if it is real analytic for each $$y\in\Omega$$.
We will write $$f\in C^\omega(\Omega)$$ in the case that $$f$$ is real analytic in the domain $$\Omega$$.
A vector valued function $$f(x)=(f_1(x),\ldots,f_m)$$ is called real analytic if each coordinate is real analytic.

Proposition A2. (i) Let $$f\in C^\omega(\Omega)$$. Then $$f\in C^\infty(\Omega)$$.}

(ii)
Assume $$f\in C^\omega(\Omega)$$. Then for each $$y\in \Omega$$ there exists a neighborhood $$N(y)$$ and positive constants $$M$$, $$r$$ such that
$$f(x)=\sum_\alpha\frac{1}{\alpha!}(D^\alpha f(y))(x-y)^\alpha$$
for  all $$x\in N(y)$$, and the series converges (absolutely) for each $$x\in N(y)$$, and
$$|D^\beta f(x)|\le M|\beta|!r^{-|\beta|}.$$

The proof follows from Proposition A1.

An open set $$\Omega\in\mathbb{R}$$ is called connected if $$\Omega$$ is not a union of two nonempty
open sets with empty intersection. An open set $$\Omega\in\mathbb{R}$$ is connected if and only if its path connected, see [11], pp. 38, for example. We say that $$\Omega$$ is path connected if for any $$x,y\in\Omega$$ there is a continuous curve $$\gamma(t)\in\Omega$$, $$0\le t\le1$$, with $$\gamma(0)=x$$ and $$\gamma(1)=y$$.  From the theory of one complex variable we know that a continuation of an analytic function is uniquely determined. The same is true  for real analytic functions.

Proposition A3. Assume $$f\in C^\omega(\Omega)$$ and $$\Omega$$ is connected. Then
$$f$$ is uniquely determined if for one $$z\in\Omega$$ all $$D^\alpha f(z)$$ are known.

Proof. See F. John [10], p. 65. Suppose $$g, h\in C^\omega(\Omega)$$ and
$$D^\alpha g(z)=D^\alpha h(z)$$ for every $$\alpha$$. Set $$f=g-h$$ and
\begin{eqnarray*}
\Omega_1&=&\{x\in\Omega:\ D^\alpha f(x)=0\ \ \mbox{for all}\ \alpha\},\\
\Omega_2&=&\{x\in\Omega:\ D^\alpha f(x)\not=0\ \ \mbox{for at least one}\ \alpha\}.
\end{eqnarray*}
The set $$\Omega_2$$ is open since $$D^\alpha f$$ are continuous in $$\Omega$$. The set $$\Omega_1$$ is also open since $$f(x)=0$$ in a neighbourhood of $$y\in\Omega_1$$. This follows from
$$f(x)=\sum_\alpha \frac{1}{\alpha!}(D^\alpha f(y))(x-y)^\alpha.$$
Since $$z\in\Omega_1$$, i. e., $$\Omega_1\not=\emptyset$$, it follows $$\Omega_2=\emptyset$$.

$$\Box$$

It was shown in Proposition A2 that derivatives of a real analytic function satisfy estimates.
On the other hand it follows, see the next proposition, that a function $$f\in C^\infty$$ is real analytic if these estimates are satisfied.

Definition. Let $$y\in\Omega$$ and $$M,\ r$$ positive constants. Then $$f$$ is said to be in the class  $$C_{M,r}(y)$$ if $$f\in C^\infty$$ in a neighbourhood of $$y$$ and if
$$|D^\beta f(y)|\le M|\beta|!r^{-|\beta|}$$
for all $$\beta$$.

Proposition A4. $$f\in C^\omega(\Omega)$$ if and only if $$f\in C^\infty(\Omega)$$ and for every compact subset $$S\subset\Omega$$ there are positive constants $$M,\;r$$ such that
$$f\in C_{M,r}(y)\ \ \mbox{for all}\ y\in S.$$

Proof. See F. John [10], pp. 65-66. We will prove the local version of the proposition, that is, we show it for each fixed $$y\in\Omega$$. The general version follows from Heine-Borel theorem. Because of Proposition A3 it remains to show that the Taylor series
$$\sum_\alpha\frac{1}{\alpha!}D^\alpha f(y)(x-y)^\alpha$$
converges (absolutely) in a neighborhood of $$y$$ and that this series is equal to $$f(x)$$.

Define a neighborhood of $$y$$ by
$$N_d(y)=\{x\in\Omega:\ \ |x_1-y_1|+\ldots+|x_n-y_n|<d\},$$
where $$d$$ is a sufficiently small positive constant. Set $$\Phi(t)=f(y+t(x-y))$$. The one-dimensional Taylor theorem says
$$f(x)=\Phi(1)=\sum_{k=0}^{j-1}\frac{1}{k!}\Phi^{(k)}(0)+r_j,$$
where
$$r_j=\frac{1}{(j-1)!}\int_0^1\ (1-t)^{j-1}\Phi^{(j)}(t)\ dt.$$
From formula 7. for directional derivatives it follows for $$x\in N_d(y)$$ that
$$\frac{1}{j!}\frac{d^j}{dt^j}\Phi(t)=\sum_{|\alpha|=j}\frac{1}{\alpha!}D^\alpha f(y+t(x-y))(x-y)^\alpha.$$
From the assumption and the multinomial formula 3. we get for $$0\le t\le 1$$
\begin{eqnarray*}
\left|\frac{1}{j!}\frac{d^j}{dt^j}\Phi(t)\right|&\le&M\sum_{|\alpha|=j}\frac{|\alpha|!}{\alpha!}r^{-|\alpha|}\left|(x-y)^\alpha\right|\\
&=& Mr^{-j}\left(|x_1-y_1|+\ldots +|x_n-y_n|\right)^j\\
&\le&M\left(\frac{d}{r}\right)^j.
\end{eqnarray*}
Choose $$d>0$$ such that $$d<r$$, then the Taylor series converges (absolutely) in $$N_d(y)$$ and it is equal to $$f(x)$$ since the remainder satisfies, see the above estimate,
$$|r_j|=\left|\frac{1}{(j-1)!}\int_0^1\ (1-t)^{j-1}\Phi^j(t)\ dt\right|\le M\left(\frac{d}{r}\right)^j.$$

$$\Box$$

We recall that the notation $$f<<F$$ ($$f$$ is majorized by $$F$$) was defined in the previous section.

Proposition A5. (i) $$f=(f_1,\ldots,f_m)\in C_{M,r}(0)$$ if and only if $$f<<(\Phi,\ldots,\Phi)$$, where
$$\Phi(x)=\frac{Mr}{r-x_1-\ldots-x_n}\ .$$
}

(ii) $$f\in C_{M,r}(0)$$ and $$f(0)=0$$ if and only if
$$f<<(\Phi-M,\ldots,\Phi-M),$$
where
$$\Phi(x)=\frac{M(x_1+\ldots+x_n)}{r-x_1-\ldots-x_n}\ .$$

Proof.
$$D^\alpha\Phi(0)=M|\alpha|!r^{-|\alpha|}.$$

$$\Box$$

Remark. The definition of $$f<<F$$ implies, trivially, that $$D^\alpha f<<D^\alpha F$$.

The next proposition shows that compositions majorize  if the involved functions majorize. More precisely, we have

Proposition A6. Let $$f,\ F:\ \mathbb{R}\mapsto\mathbb{R}^m$$ and $$g,\ G$$ maps a neighborhood of $$0\in\mathbb{R}^m$$ into $${\mathbb R}^p$$. Assume all functions $$f(x),\ F(x),\ g(u),\ G(u)$$ are in $$C^\infty$$, $$f(0)=F(0)=0$$, $$f<<F$$ and $$g<<G$$. Then
$$g(f(x))<<G(F(x))$$.
}

Proof. See F. John [10], p. 68. Set
$$h(x)=g(f(x)),\ \ \ H(x)=G(F(x)).$$
For each coordinate $$h_k$$ of $$h$$ we have, according to the chain rule,
$$D^\alpha h_k(0)=P_\alpha(\delta^\beta g_l(0),D^\gamma f_j(0)),$$
where $$P_\alpha$$ are polynomials with non-negative integers as coefficients, $$P_\alpha$$
are independent on $$g$$ or $$f$$ and $$\delta:=(\partial/\partial u_1,\ldots,\partial/\partial u_m)$$. Thus,
\begin{eqnarray*}
|D^\alpha h_k(0)|&\le&P_\alpha(|\delta^\beta g_l(0)|,|D^\gamma f_j(0)|)\\
&\le&P_\alpha(\delta^\beta G_l(0),D^\gamma F_j(0))\\
&=&D^\alpha H_k(0).
\end{eqnarray*}
$$\Box$$

Using this result and Proposition A4,  which characterizes real analytic functions,  it follows that compositions of real analytic functions are real analytic functions again.

Proposition A7. Assume $$f(x)$$ and $$g(u)$$ are real analytic, then $$g(f(x))$$ is real analytic if $$f(x)$$ is in the domain of definition of $$g$$.

Proof. See F. John [10], p. 68. Assume that $$f$$ maps a neighborhood of $$y\in\mathbb{R}$$ in $$\mathbb{R}^m$$ and $$g$$ maps a neighborhood of $$v=f(y)$$ in ${\mathbb R}^m$. Then $$f\in C_{M,r}(y)$$ and $$g\in C_{\mu,\rho}(v)$$ implies
$$h(x):=g(f(x))\in C_{\mu,\rho r/(mM+\rho)}(y).$$
Once one has shown this inclusion, the proposition follows from Proposition~A4. To show the inclusion, we set
$$h(y+x):=g(f(y+x))\equiv g(v+f(y+x)-f(x))=:g^*(f^*(x)),$$
where $$v=f(y)$$ and
\begin{eqnarray*}
g^*(u):&=&g(v+u)\in C_{\mu,\rho}(0)\\
f^*(x):&=&f(y+x)-f(y)\in C_{M,r}(0).
\end{eqnarray*}
In the above formulas $$v,\ y$$ are considered as fixed parameters. From Proposition~A5 it follows
\begin{eqnarray*}
f^*(x)&<<&(\Phi-M,\ldots,\Phi-M)=:F\\
g^*(u)&<<&(\Psi,\ldots,\Psi)=:G,
\end{eqnarray*}
where
\begin{eqnarray*}
\Phi(x)&=&\frac{Mr}{r-x_1-x_2-\ldots-x_n}\\
\Psi(u)&=&\frac{\mu\rho}{\rho-x_1-x_2-\ldots-x_n}.
\end{eqnarray*}
From Proposition A6 we get
$$h(y+x)<<(\chi(x),\ldots,\chi(x))\equiv G(F),$$
where
\begin{eqnarray*}
\chi(x)&=&\frac{\mu\rho}{\rho-m(\Phi(x)-M)}\\
&=&\frac{\mu\rho(r-x_1-\ldots-x_n)}{\rho r-(\rho+mM)(x_1+\ldots+x_n)}\\
&<<&\frac{\mu\rho r}{\rho r-(\rho+mM)(x_1+\ldots+x_n)}\\
&=&\frac{\mu\rho r/(\rho+mM)}{\rho r/(\rho+mM)-(x_1+\ldots x_n)}.
\end{eqnarray*}
See an exercise for the ''$$<<$$''-inequality.

$$\Box$$

### Contributors:

• Integrated by Justin Marshall.