Proof If we can arrange (a) and (b), then it follows that we can also arrange (c) by Theorem 0 of [Reference Deutsch3] applied to the space $X_k=\{f\in C^k(\mathbb {R}):f{\sigma }=f\}$ with the norm $\|f\|=\sum _{i=0}^k\|D^if\|_\infty $ . (Cf. [Reference Deutsch3, Example (1), p. 1183, and Corollary 1.3]) So we need only arrange (a) and (b).

It is standard that in $X_0$ the functions $g(2\pi z/t)$ , where g is a trigonometric polynomial with real coefficients, are dense (e.g., see [Reference Deutsch3, Example (2), p. 1183] or [Reference Sohrab5, Remark 9.4.24 (2)]).

For the general case, we proceed by induction on k. When $k>0$ , fix $f\in X_k$ and apply the induction hypothesis to $Df$ to get an entire function $g_1$ satisfying $g_1(\mathbb {R})\subseteq \mathbb {R}$ , $g_1{\sigma }=g_1$ and $|D^{i-1}g_1(x)-D^if(x)|<\delta =\min ({\varepsilon }/2,{\varepsilon }/(2t))$ for $x\in \mathbb {R}$ and $i=1,\dots ,k$ . Define an entire function g by

$$ \begin{align*} g(z) = f(0)+\int_{0}^z g_1(s)\,ds - cz,\quad z\in\mathbb{C}, \end{align*} $$

where $c=(1/t)\int _0^t g_1(s)\,ds$ is the average value of $g_1$ over $[0,t]$ . Since $\int _0^tDf = 0$ by periodicity, we have

$$ \begin{align*} |c| = \Bigl|\frac1t\int_0^t (g_1(s)-Df(s))\,ds\Bigr| \leq \delta. \end{align*} $$

To verify that $g{\sigma }=g$ , it suffices to check that $g(x+t)=g(x)$ when x is real, and that follows by the following calculation.

$$ \begin{align*} g(x+t) = f(0)+\int_{0}^{x+t} g_1(s)\,ds - c(x+t) = g(x) + \int_{x}^{x+t} g_1(s)\,ds - ct = g(x). \end{align*} $$

For $x\in [0,t]$ , we also have

$$ \begin{align*} |g(x)-f(x)| & = \Bigl|\int_{0}^x (g_1(s)-Df(s))\,ds - cx\Bigr| \\ & \leq \int_{0}^x |g_1(s)-Df(s)|\,ds + |c|x \\ & \leq t\delta + t\delta = 2t\delta < {\varepsilon}, \end{align*} $$

and so, by periodicity, $|g(x)-f(x)| <{\varepsilon }$ for all $x\in \mathbb {R}$ . This takes care of the bound on $|D^ig(x)-D^if(x)|$ , $x\in \mathbb {R}$ , when $i=0$ . When $i=1$ we have

$$ \begin{align*} |Dg(x)-Df(x)| = |g_1(x)-c-Df(x)| \leq |g_1(x)-Df(x)|+|c| \leq 2\delta<{\varepsilon}, \end{align*} $$

and when $2\leq i\leq k$ , $|D^ig(x)-D^if(x)|=|D^{i-1}g_1(x)-D^if(x)|<\delta <{\varepsilon }$ . ▪

Proof For $r>0$ and $p\in \mathbb {R}$ , write $I_r(p)=(p-r,p+r)$ . For each $p\in F$ , choose $0<r_p<1$ , so that the intervals $I_{r_p}(p)$ are contained in $(0,t)$ when $0<p<t$ , $I_{r_t}(t) = I_{r_0}(0) + t$ if $0\in F$ , and the intervals $I_{r_p}(p)$ for $p\in F$ and have closures which are disjoint from each other. For each $p\in F$ , choose a $C^\infty $ bump function $\varphi _p$ with support equal to the closure of $I_{r_p}(p)$ , so that $0\leq \varphi _p\leq 1$ , $\varphi _p(p)=1$ , $\varphi _p$ has a flat point at p (i.e., $(D^i \varphi _p)(p)=0$ for all nonzero i), taking $\varphi _t(x)=\varphi _0(x-t)$ if $0\in F$ . Let $\theta _p$ be a $C^\infty $ function whose derivatives at p follow the requisite pattern, with $\theta _t(x)=\theta _0(x-t)$ if $0\in F$ , for example, take $\theta _p(x)=\sum _{i=0}^kk_{p,i}(x-p)^i$ . For suitably chosen positive constants $\lambda _p$ , we set $g(x) = \sum _{p\in F}\lambda _p\theta _p(x)\varphi _p(x)$ , $0\leq x\leq t$ , and extend g by periodicity to all of $\mathbb {R}$ so that $g{\sigma }=g$ . This function g is $C^\infty $ and its derivatives have the requisite pattern at the points $p\in F$ since for $i=0,\dots ,k$ we have $(D^i g)(p) = \lambda _p(D^i (\varphi _p\theta _p))(p) = \lambda _p D^{i}\theta _p(p)$ . Choose the constants $\lambda _p$ , $p\in F$ , small enough so that for each $i=0,\dots ,k$ ,

$$ \begin{align*} \lambda_p\|D^i(\varphi_p\theta_p)\|_\infty<{\varepsilon}/2. \end{align*} $$

Then g is a $C^\infty $ function which satisfies the conclusion in the place of f and with ${\varepsilon }$ replaced by ${\varepsilon }/2$ . Get the desired entire function f by applying Theorem 2.1 to g with ${\varepsilon }/2$ in the place of ${\varepsilon }$ . ▪

Proof Write $h(x,y)=(x,h^*(x,y))$ . The continuity of h is equivalent to that of $h^*$ . Also note that because h is one-to-one, so is $y\mapsto h^*(x,y)$ for each fixed x. Let $a\in X$ , $b,c\in \mathbb {R}$ with $(a,b)\in G$ and $h^*(a,b)=c$ . Fix an open neighborhood U of a and an open interval V containing b such that $U\times V\subseteq G$ . Choose an open interval W with $c\in W$ . Since $y\mapsto h^*(a,y)$ is continuous, there is a $\delta>0$ such that $|y-b|\leq \delta $ implies $y\in V$ and $h^*(a,y)\in W$ . The two functions $x\mapsto h^*(x,b\pm \delta )$ are continuous at a and hence there is an open neighborhood $U'\subseteq U$ of a such that $x\in U'$ implies $h^*(x,b\pm \delta )\in W$ . For $x\in U'$ , $y\mapsto h^*(x,y)$ is a continuous one-to-one function on the interval $[b-\delta ,b+\delta ]$ , so that the image of this interval is an interval with endpoints $h^*(x,b\pm \delta )$ , and hence is contained in W. Thus, $h^*(U'\times (b-\delta ,b+\delta ))\subseteq W$ and therefore $h^*$ is continuous at $(a,b)$ . ▪

Proof We may assume that $\bigcup _{i=0}^k E_i\not =\emptyset $ (otherwise take $f=0$ ) and that ${\varepsilon }\leq 1$ . Let $\mathscr {B}=\{U_r:r=1,2,\dots \}$ be a one-to-one enumeration of a base of bounded open intervals for $\mathbb {R}\setminus F$ . For each r, write $U_r=\bigcup _{n=1}^\infty U_{r,n}$ , the union of an increasing sequence of concentric intervals so that $\operatorname {cl} U_{r,n}\subseteq U_{r,n+1}$ . Let $\{(i_n,s_n):n=1,2,\dots \}$ list all pairs $(i,s)$ consisting of an $i=0,\dots ,k$ and a point $s\in E_i$ . Let $Q=\{h^*(p,q'):h\in \mathscr {H},\, p\in E_i,\, q'\in A_{p,i},\, (p,q')\in G^1_h,\, i=0,\dots ,k\}$ . List the quadruples $(h,j,q,V)$ consisting of an $h\in \mathscr {H}$ , a $j=0,\dots ,k$ and elements $q\in Q$ , $V\in \mathscr {B}$ as $\{(h_m,j_m,q_m,V_m):m=1,2,\dots \}$ with each quadruple listed infinitely many times. As with the $U_r$ above, we write $V_m=\bigcup _{n=1}^\infty V_{m,n}$ where if $V_m=U_r$ then $V_{m,n}=U_{r,n}$ . We will write $G^1_m$ , $G^2_m$ , $k_m$ for $G^1_{h_m}$ , $G^2_{h_m}$ , $k_{h_m}$ , respectively.

We will build the required function as $f=\sum _{n=1}^\infty \lambda _n u_n$ , where for each $n\in {\mathbb N}$ , $\lambda _n\in \mathbb {R}$ satisfies $|\lambda _n|\leq 1$ and $u_n\colon \mathbb {C}\to \mathbb {C}$ is an entire function such that $u_n(\mathbb {R})\subseteq \mathbb {R}$ . We recursively define the following.

1. (i) $\lambda _n$ and $u_n$ .

2. (ii) An increasing sequence of finite sets $\emptyset =K_0\subseteq K_1\subseteq \dots $ of pairs $(w,p)$ consisting of a $w=0,\dots ,k$ and a point $p\in E_w$ .

3. (iii) A decreasing sequence of positive numbers $1=\delta _0>\delta _1>\dots $ .

We will arrange that the following properties hold for $n=1,2,\dots $ . In this list, $f_n$ denotes the sum $\sum _{k=1}^n \lambda _k u_k$ , and $f_0=0$ .

1. (1) $u_n{\sigma }=u_n$ and $|(D^{i}u_n)(x)| < 2^{-n-1}\delta _{n-1}{\varepsilon }$ for $x\in \mathbb {R}$ , $i=0,\dots ,k$ .

2. (2) $(D^{w}u_n)(p) = 0$ for $(w,p)\in K_{n-1}$ .

3. (3) $(D^{i}u_n)(x) = 0$ whenever $x\in F$ , $i=0,\dots ,k$ .

4. (4) $|u_n(z)| < 2^{-n}$ , for $|z|\leq n$ .

5. (5) For each $(w,p)\in K_n$ , $(D^w f_n)(p)\in A_{p,w}$ .

6. (6) If n is odd then $K_n = K_{n-1}\cup \{(i_\ell ,s_\ell )\}$ for the least $\ell $ such that $(i_\ell ,s_\ell )\not \in K_{n-1}$ . We have $(D^{i_\ell }u_n)(s_\ell ) \not = 0$ .

7. (7) If $n=2m$ is even, we have the following. Suppose that

   $$ \begin{align*} (x,(D^{j_m}f_{n-1})(x))\in G^1_{m}\ \text{and}\ (x,q_m)\in G^2_{m}\ \text{whenever}\ x\in V_m. \end{align*} $$

   1. (a) If $q_m=h^*_m(x,(D^{j_m}f_{n-1})(x))$ for some $x\in V_m\cap \operatorname {cl} Y_{h_m,q_m,j_m}$ then for some $p\in V_m\cap Y_{h_m,q_m,j_m}$ , $h^*_m(p,(D^{j_m}f_{n})(p))=q_m$ and $K_n = K_{n-1}\cup \{(j_m,p)\}$ .

   2. (b) If $q_m\not =h^*_m(x,(D^{j_m}f_{n-1})(x))$ for all $x\in V_m\cap \operatorname {cl} Y_{h_m,q_m,j_m}$ then $u_n=0$ , $K_n = K_{n-1}$ and $\delta _n<\inf \{|k^*_m(x,q_m) - (D^{j_m}f_{n-1})(x)|:x\in V_{m,n}\cap \operatorname {cl} Y_{h_m,q_m,j_m}\}$ .

We now explain how to carry out the construction at odd stages and at even stages.

First suppose n is odd. This stage includes the initial step $n=1$ . Set $K_n = K_{n-1}\cup \{(i_\ell ,s_\ell )\}$ for the least $\ell $ such that $(i_\ell ,s_\ell )\not \in K_{n-1}$ . Apply Lemma 2.2 to get an entire function $u_n\colon \mathbb {C}\to \mathbb {C}$ satisfying (1), (2), (3), and (6). Arrange (4) by replacing $u_n$ by a smaller positive multiple of itself if necessary. Since $(D^{i_\ell }u_n)(s_\ell )\not =0$ by (6), and $A_{s_\ell ,i_\ell }$ is dense in $\mathbb {R}$ , we may choose $\lambda _n$ so that $0<\lambda _n<1$ and $(D^{i_\ell }f_n)(s_\ell ) = (D^{i_\ell }f_{n-1})(s_\ell ) + \lambda _n (D^{i_\ell }u_n)(s_\ell )\in A_{s_\ell ,i_\ell }$ . This gives (5) because if $(w,p)\in K_{n-1}$ then $(D^w u_n)(p)=0$ by (2), so $(D^w f_n)(p) = (D^w f_{n-1})(p)$ which by (5) for $n-1$ belongs to $A_{p,w}$ . For $\delta _n$ , choose any positive number satisfying $\delta _n<\delta _{n-1}$ .

Now suppose that $n=2m$ is even. If the assumption of (7) fails, take $u_n=0$ , $\lambda _n=0$ , $K_n=K_{n-1}$ , and let $\delta _n$ be any number such that $0<\delta _n<\delta _{n-1}$ . Clearly (1)–(7) hold. Now suppose that the assumption is satisfied, i.e., $(x,(D^{j_m}f_{n-1})(x))\in G^1_{m}$ and $(x,q_m)\in G^2_{m}$ whenever $x\in V_m$ .

Case 1. $q_m\not = h^*_m(x,(D^{j_m}f_{n-1})(x))$ for all $x\in V_m\cap \operatorname {cl} Y_{h_m,q_m,j_m} $ .

Let $u_n=0$ , $\lambda _n=0$ , $K_n=K_{n-1}$ . Clearly (1)– (5) hold. For (7), choose any positive number $\delta _n<\delta _{n-1}$ so that $\delta _n<\inf \{|k^*_m(x,q_m) - (D^{j_m}f_{n-1})(x)|:x\in V_{m,n}\cap \operatorname {cl} Y_{h_m,q_m,j_m}\}$ where the right-hand side is positive because $\operatorname {cl}(V_{m,n})\cap \operatorname {cl} (Y_{h_m,q_m,j_m})$ is compact and contained in $V_m\cap \operatorname {cl} Y_{h_m,q_m,j_m}$ .

Case 2. $h^*_m(p,(D^{j_m}f_{n-1})(p))=q_m$ for some $p\in V_m\cap Y_{h_m,q_m,j_m}$ .

By definition, $p\in Y_{h_m,q_m,j_m}$ means that $p\in E_{j_m}$ and there is a (unique) $q'\in A_{p,j_m}$ such that $(p,q')\in G^1_m$ and $q_m=h^*_m(p,q')$ . Since the vertical sections of $h^*_m$ are one-to-one, we must have $(D^{j_m}f_{n-1})(p)=q'$ and hence $(D^{j_m}f_{n-1})(p)\in A_{p,j_m}$ . Let $u_n=0$ , $\lambda _n=0$ , $K_n=K_{n-1}\cup \{(j_m,p)\}$ . Clearly (1)–(5) and (7) hold. For $\delta _n$ , choose any positive number satisfying $\delta _n<\delta _{n-1}$ .

Case 3. $h^*_m(x,(D^{j_m}f_{n-1})(x))\not =q_m$ for all $x\in V_m\cap Y_{h_m,q_m,j_m}$ , but

$$ \begin{align*} h^*_m(p,(D^{j_m}f_{n-1})(p))=q_m \end{align*} $$

for some $p\in V_m\cap \operatorname {cl} Y_{h_m,q_m,j_m}$ .

The assumptions give $h^*_m(p,(D^{j_m}f_{n-1})(p)) = q_m$ and $p\not \in V_m\cap Y_{h_m,q_m,j_m}$ , and since $p\in V_m$ this gives $p\not \in Y_{h_m,q_m,j_m}$ . It follows from (5) that $(j_m,p)\not \in K_{n-1}$ because if $(j_m,p)\in K_{n-1}$ then $p\in E_{j_m}$ and by (5) we have

$$ \begin{align*} (D^{j_m}f_{n-1})(p)\in A_{p,j_m}, \end{align*} $$

which together with $h^*_m(p,(D^{j_m}f_{n-1})(p))=q_m$ gives $p\in Y_{h_m,q_m,j_m}$ , contradiction. Apply Lemma 2.2 to get an entire function $u_n\colon \mathbb {C}\to \mathbb {C}$ satisfying (1), (2), (3) and $(D^{j_m} u_n)(p)>0$ . Arrange (4) by replacing $u_n$ by a smaller positive multiple of itself if necessary. The function $x\mapsto (k^*_m(x,q_m) - (D^{j_m}f_{n-1})(x))/(D^{j_m}u_n)(x)$ is continuous at p with value $0$ there, and $p\in V_m\cap \operatorname {cl} Y_{h_m,q_m,j_m}$ , so we can pick an element $p'$ of $V_m\cap Y_{h_m,q_m,j_m}$ so close to p that the number

$$ \begin{align*} \lambda_n = \frac{k^*_m(p',q_m) - (D^{j_m}f_{n-1})(p')}{(D^{j_m}u_n)(p')} \end{align*} $$

satisfies $|\lambda _n|<1$ . With this value of $\lambda _n$ , we have $k^*_m(p',q_m)=(D^{j_m}f_n)(p')$ and therefore

$$ \begin{align*} h^*_m(p',(D^{j_m}f_{n})(p')) = q_m, \end{align*} $$

and hence (7) holds if we take $K_n=K_{n-1}\cup \{(j_m,p')\}$ . Note that (5) is satisfied. For $\delta _n$ , choose any positive number satisfying $\delta _n<\delta _{n-1}$ .

This completes the construction. Property (4) ensures that the formula $f = \sum _{n=1}^\infty \lambda _nu_n$ defines an entire function and that

$$ \begin{align*} (D^{i}f)(z) = \sum_{n=1}^\infty \lambda_n(D^{i}u_n)(z) \end{align*} $$

for all $i=0,\dots ,k$ and $z\in \mathbb {C}$ . Clearly $f(\mathbb {R})\subseteq \mathbb {R}$ and by (1), we have $f{\sigma }=\sum _{n=1}^\infty \lambda _nu_n{\sigma }=f$ . We now verify (a)–(c).

(a) When $i=0,\dots ,k$ , $x\in \mathbb {R}$ , we have using (1) that $|(D^{i}f)(x)| \leq \sum _{i=1}^\infty |(D^{i}u_i)(x)| < \sum _{i=1}^\infty 2^{-i-1}{\varepsilon } = {\varepsilon }$ and, using (3), when $x\in F$ we have $(D^{i}f)(x) = 0$ .

(b) When $p\in E_i$ , $i=0,\dots ,k$ , (6) ensures that $(i,p)\in K_n$ if n is sufficiently large. Then (5) gives $(D^{i}f_n)(p) \in A_{p,i}$ . From (2), we get that $(D^{i}u_j)(p)=0$ , $j\geq n+1$ , and hence $(D^{i}f)(p) = (D^{i}f_n)(p) \in A_{p,i}$ .

(c) Suppose $i=0,\dots ,k$ , $q\in \mathbb {R}$ , $U\subseteq \mathbb {R}\setminus F$ is an open interval, and we have $(p,(D^i f)(p))\in G^1_h$ , and $q = h^*(p,(D^i f)(p))$ for some $p\in U\cap \operatorname {cl} Y_{h,q,i}$ . We wish to show that $q = h^*(\bar {p},(D^i f)(y))$ for some $y\in U\cap E_i$ . We may assume that $U\in \mathscr {B}$ and for some open intervals $W_1,W_2\subseteq \mathbb {R}$ , for all $x\in U$ ,

• • $(D^i f)(x)\in W_1\subseteq \operatorname {cl} W_1\subseteq W_2$ , and $U\times W_2\subseteq G^1_h$ and

• • $(x,q)\in G^2_h$ .

Since $D^i f_n\to D^i f$ uniformly on compact sets, for large enough n we have for $x\in U$ that $(D^i f_n)(x)\in W_2$ and hence $(x,(D^i f_n)(x))\in G^1_h$ . By assumption, $p\in \operatorname {cl} Y_{h,q,i}$ , so $Y_{h,q,i}\not =\emptyset $ and hence $q\in Q$ . Fix r such that $U=U_r$ . Choose an even $n=2m$ with $(h_m,j_m,q_m,V_m)=(h,i,q,U_r)$ , n large enough so that $p\in U_{r,n}=V_{m,n}$ and $(x,(D^i f_{n-1})(x))\in G^1_h$ for all $x\in U$ .

Claim. $q = h^*(x,(D^i f_{n-1})(x))$ for some $x\in U\cap \operatorname {cl} Y_{h,q,i}$ .

If not, then by (7), $\delta _n<\inf \{|k^*_m(x,q_m) - (D^{j_m}f_{n-1})(x)|:x\in V_{m,n}\cap \operatorname {cl} Y_{h_m,q_m,j_m}\}$ and $u_n=0$ , so that $f_n=f_{n-1}$ . We have that for $j>n$ , $|(D^{i}u_j)(p)| < 2^{-j-1}\delta _{n}$ . This gives

$$ \begin{align*} |(D^i f)(p) - (D^i f_{n-1})(p)|\leq \sum_{j=n}^{\infty} 2^{-j-1}\delta_{n} < \delta_{n} < |k^*_m(p,q) - (D^{i}f_{n-1})(p)|. \end{align*} $$

This contradicts $h^*(p,(D^i f)(p))=q$ (which is equivalent to $k^*_h(p,q)=(D^i f)(p)$ ). This completes the proof of the claim.

The claim says that $q_m=h^*_m(x,(D^{j_m}f_{n-1})(x))$ for some $x\in V_m\cap \operatorname {cl} Y_{h_m,q_m,j_m}$ , so by (7), for some $y\in V_m\cap Y_{h_m,q_m,j_m}$ , $h^*_m(y,(D^{j_m}f_{n})(y))=q_m$ and $K_n = K_{n-1}\cup \{(j_m,y)\}$ . But then by (2), $(D^{j_m}u_k)(y) = 0$ if $k>n$ , so $h^*_m(y,(D^{j_m}f)(y)) = h^*_m(y,(D^{j_m}f_{n})(y))=q_m$ . ▪

Proof (A) Using Theorem 2.1, we can approximate g by an entire function $g_0$ so that $g_0(\mathbb {R})\subseteq \mathbb {R}$ , $g_0{\sigma }=g_0$ , and for $x\in \mathbb {R}$ , $i=0,\dots ,k$ , $|D^ig_0(x)-D^ig(x)|<{\varepsilon }/2$ and if $x\in F$ then $D^ig_0(x)=D^ig(x)$ . If we approximate $g_0$ as in Corollary 3.5, replacing ${\varepsilon }$ by ${\varepsilon }/2$ , we get the desired function f. Hence, we may assume that g is entire.

For $p\in E_i$ , define $B_{p,i}=A_{p,i}-(D^i g)(p)$ . Writing $\xi _i$ for the fiber-preserving homeomorphism of $\mathbb {R}^{2}$ given by $\xi _i(x,y)=(x,y+(D^i g)(x))$ , take $\mathscr {H}_i=\{h\circ \xi _i:h\in \mathscr {H}\}$ and apply Theorem 3.3 with $A_{p,i}$ replaced by $B_{p,i}$ and $\mathscr {H}$ replaced by $\overline {\mathscr {H}}=\bigcup _i\mathscr {H}_i$ . (Here, $h\circ \xi _i\colon \xi _i^{-1}(G^1_h)\to G^2_h$ .) If $f_1$ is the resulting entire function, the function $f=g+f_1$ is as desired. We have $f{\sigma }=g{\sigma }+f_1{\sigma }=g+f_1=f$ . Clauses (a) and (b) are immediate from the corresponding clauses of Theorem 3.3 and the definition of $B_{p,i}$ .

For (c), fix an $i=0,\dots ,k$ , $q\in \mathbb {R}$ , $h\in \mathscr {H}$ , an open interval $U\subseteq \mathbb {R}\setminus F$ , and assume that $(x,(D^i f)(x))\in G^1_h$ and $q = h^*(x,(D^i f)(x))$ for some $x\in U\cap \operatorname {cl} Y_{h,q,i}$ . Then $\xi _i(x,(D^i f_1)(x))\in G^1_h$ , so $(x,(D^i f_1)(x))\in \xi _i^{-1}(G^1_h)$ , and $q = (h^*\circ \xi _i)(x,(D^i f_1)(x))$ .

Claim. $Y_{h,q,i}\subseteq Y_{h\circ \xi _i,q,i}$ where the set $Y_{h\circ \xi _i,q,i}$ is defined using the $B_{p,i}$ instead of the $A_{p,i}$ .

Let $p\in Y_{h,q,i}$ . Then $p\in E_i$ and for some $q'\in A_{p,i}$ , $(p,q')\in G^1_h$ and $q=h^*(p,q')$ . We then have $q'-(D^i g)(p)\in B_{p,i}$ , $(p,q'-(D^i g)(p)) = \xi _i^{-1}(p,q')\in \xi _i^{-1}(G^1_h)$ and $q=(h^*\circ \xi _i)(p,q'-(D^i g)(p)) = (h\circ \xi _i)^*(p,q'-(D^i g)(p))$ . Thus, $p\in Y_{h\circ \xi _i,q,i}$ , which proves the claim.

By the claim, $x\in \operatorname {cl} Y_{h\circ \xi _i,q,i}$ and hence by (c) of Theorem 3.3, $q = (h\circ \xi _i)^*(p,(D^i f_1)(p)) = (h^*\circ \xi _i)(p,(D^i f_1)(p))$ for some $p\in U\cap E_i$ . Then $h^*(p,(D^i f)(p)) = (h^*\circ \xi _i)(p,(D^i f_1)(p))=q$ .

(B) This part follows from (A) by an argument similar to that used for (A). Given $g\in C^k(\mathbb {R})$ satisfying $g{\sigma }={\sigma } g$ , write $g=g_1+\operatorname {id}$ . As pointed out above, we have $g_1{\sigma }=g_1$ . For $p\in E_i$ , define $B_{p,i}=A_{p,i}-(D^i \operatorname {id})(p)$ . (So $B_{p,0}=A_{p,0}-p$ , $B_{p,1}=A_{p,1}-1$ , $B_{p,i}=A_{p,i} (i>1)$ .) Writing $\xi _i$ for the fiber-preserving homeomorphism of $\mathbb {R}^{2}$ given by $\xi _i(x,y)=(x,y+(D^i \operatorname {id})(x))$ , take $\mathscr {H}_i=\{h\circ \xi _i:h\in \mathscr {H}\}$ and apply part (A) to $g_1$ with $A_{p,i}$ replaced by $B_{p,i}$ and $\mathscr {H}$ replaced by $\overline {\mathscr {H}}=\bigcup _i\mathscr {H}_i$ . (Here, $h\circ \xi _i\colon \xi _i^{-1}(G^1_h)\to G^2_h$ .) If $f_1$ is the resulting entire function, the function $f=\operatorname {id}+f_1$ is as desired. Since $f_1$ satisfies $f_1{\sigma }=f_1$ , it follows that f satisfies $f{\sigma }={\sigma } f$ . Clauses (a) and (b) of part (B) are immediate from the corresponding clauses of part (A) and the definition of $B_{p,i}$ . For (c), as in part (A), fix an $i=0,\dots ,k$ , $q\in \mathbb {R}$ , $h\in \mathscr {H}$ , an open interval $U\subseteq \mathbb {R}\setminus F$ , and assume that $(x,(D^i f)(x))\in G^1_h$ and $q = h^*(x,(D^i f)(x))$ for some $x\in U\cap \operatorname {cl} Y_{h,q,i}$ . Then $\xi _i(x,(D^i f_1)(x))\in G^1_h$ , so $(x,(D^i f_1)(x))\in \xi _i^{-1}(G^1_h)$ , and $q = (h^*\circ \xi _i)(x,(D^i f_1)(x))$ . As in the proof of (A), we get $Y_{h,q,i}\subseteq Y_{h\circ \xi _i,q,i}$ where the set $Y_{h\circ \xi _i,q,i}$ is defined using the $B_{p,i}$ instead of the $A_{p,i}$ . (In the proof of the claim in part (A), replace the four g’s by $\operatorname {id}$ .) Then finish exactly as in part (A), reading “by (c) of part (A)” instead of “by (c) of Theorem 3.3.” ▪

Proof For $i=0,\dots ,k$ , let $E_i=[0,t]\cap \bigcup _{n=1}^\infty A^i_n$ . For pairs $(p,i)$ with $p\in E_i$ , for some unique n we have $p\in A^i_n$ . Define $A_{p,i}=B^i_n$ . If $0,t\in E_i$ , then for some n we have $0,t\in A^i_n$ . Then $A_{0,i}=A_{t,i}=B^i_n$ and therefore also ${\sigma }(A_{0,i})=A_{t,i}$ by invariance of $B^i_n$ . The orbit of each point has at most two points in $[0,t]$ , so $F_1=[0,t]\cap \bigcup _{\ell \in {\mathbb Z}}{\sigma }^\ell (F)$ is finite. We have $E_i\cap F_1=\emptyset $ for each i, because $E_i\cap F_1 \subseteq (\bigcup _{n=1}^\infty A^i_n)\cap (\bigcup _{\ell =1}^\infty {\sigma }^\ell (F))$ and for all $n,\ell $ , $A^i_n\cap {\sigma }^\ell (F) = {\sigma }^{\ell }(A^i_n\cap F) = \emptyset $ . Similarly, $g(F_1)\cap B^0_n =\emptyset $ because $g({\sigma }^\ell (F))\cap B^0_n = {\sigma }^{\ell }(g(F)\cap B^0_n) = \emptyset $ for all $n,\ell $ . Take ${\varepsilon }'=\min ({\varepsilon },\inf \{Dg(x):x\in \mathbb {R}\})$ which is positive by periodicity of $Dg$ and since $Dg>0$ . Corollary 3.5 (B) with $\mathscr {H}=\{\operatorname {id}_{\mathbb {R}^2}\}$ gives $f $ such that $f (\mathbb {R})\subseteq \mathbb {R}$ , $f {\sigma }={\sigma } f $ and for all $x\in \mathbb {R}$ and $i=0,\dots ,k$ ,

1. (i) $|(D^{i}f )(x)-(D^{i}g )(x)|<{\varepsilon }'$ , and if $x\in F_1$ then $D^if (x)=D^ig (x)$ .

2. (ii) for each $p\in E_i$ , $(D^i f )(p)\in A_{p,i}$ .

3. (iii) for any $q\in \mathbb {R}$ , if $q = f(x)$ for some $x\in \operatorname {cl} \{p\in E_0: q\in A_{p,0}\}$ , $x\notin F_1$ , then $q = f(p)$ for some $p\in E_0$ . (Since f is injective, necessarily $x=p$ .)

From (i) we get (b). From (i) we also get for $x\in \mathbb {R}$ , $|Df(x)-Dg(x)| <{\varepsilon }'\leq Dg(x)$ , so $Df(x)>0$ giving (a). For (c), fix $i=0,\dots ,k$ , $n\in {\mathbb N}$ . We want to show that $D^if(A^i_n)\subseteq B^i_n$ , and when $i=0$ , $f(A^0_n)=B^0_n$ . For $p\in A^i_n$ we have for some $\ell $ that ${\sigma }^\ell (p)\in [0,t]$ . When $i=0$ , we get, using (ii), $f (p) = {\sigma }^{-\ell }f ({\sigma }^\ell (p))\in {\sigma }^{-\ell }A_{{\sigma }^\ell (p),0} = B^0_n$ . When $i>0$ , we get similarly $D^if (p) = D^if ({\sigma }^\ell (p)) \in A_{{\sigma }^\ell (p),i} = B^i_n$ .

There remains to show that $B^0_n\subseteq f(A^0_n)$ . Let $q\in B^0_n$ . Since $f(t)=f(0)+t$ , for some $\ell $ , ${\sigma }^\ell (q)\in [f(0),f(t)]$ and therefore, for some $x\in [0,t]$ , $f(x)={\sigma }^\ell (q)$ . We have $x\notin F_1$ because otherwise $g(x)=f(x)={\sigma }^\ell (q)\in B^0_n$ , contradicting the fact that $g(F_1)\cap B^0_n = \emptyset $ . Since $A^0_n$ is dense, arbitrarily close to x there are $p\in A^0_n\cap [0,t]$ and these satisfy ${\sigma }^\ell (q)\in B^0_n = A_{p,0}$ , so by (iii), there is a $p\in E_0$ such that ${\sigma }^\ell (q) = f(p)$ and therefore $f({\sigma }^{-\ell }(p))=q$ . This gives $q\in f(A^0_n)$ as long as $p\in A^0_n$ . Let m be such that $p\in A^0_m$ . Then ${\sigma }^\ell (q)=f (p)\in f (A^0_m)\subseteq B^0_m$ . Since $B^0_m$ and $B^0_n$ are disjoint, it follows that $m=n$ . ▪

Proof Define $F'=\{0\}\cup \{\theta \in (0,2\pi ):e^{i\theta }\in F\}$ , and define

$$ \begin{align*} A^{\prime}_n = \{\theta\in\mathbb{R}:e^{i\theta}\in A_n\}, \ \ \ B^{\prime}_n = \{\theta\in\mathbb{R}:e^{i\theta}\in B_n\}. \end{align*} $$

The assumption on $\beta $ ensures that it extends to a $C^1$ function on $\mathbb {R}$ satisfying ${\sigma }\beta =\beta {\sigma }$ . The sets $A^{\prime }_n$ , $B^{\prime }_n$ are countable, dense, and invariant under translation by $t=2\pi $ . The $A^{\prime }_n$ are pairwise disjoint, as are the $B^{\prime }_n$ , and the conditions $F\cap A_n=\emptyset $ , $g(F)\cap B_n=\emptyset $ , $1\notin A_n,B_n$ imply the conditions $F'\cap A^{\prime }_n=\emptyset $ , $\beta (F')\cap B^{\prime }_n=\emptyset $ . Choose $\delta>0$ so that $D\beta (\theta )>\delta $ , $\theta \in \mathbb {R}$ , and $|\theta _1-\theta _2|<\delta $ implies $|e^{i\theta _1} - e^{i\theta _2}|<{\varepsilon }$ . Apply Theorem 3.6 to get an entire function $\alpha $ such that such that $\alpha (\mathbb {R})\subseteq \mathbb {R}$ , $\alpha {\sigma }={\sigma } \alpha $ , and

1. (i) $|\alpha (\theta )-\beta (\theta )|<\delta $ for $\theta \in \mathbb {R}$ and $\alpha (\theta )=\beta (\theta )$ for $\theta \in F'$ .

2. (ii) $|D\alpha (\theta )-D\beta (\theta )|<\delta $ for $\theta \in \mathbb {R}$ .

3. (iii) $\alpha (A^{\prime }_n)=B^{\prime }_n (n\in {\mathbb N})$ .

4. (iv) $D\alpha>0$ .

(Note that (iv) follows from (ii) and the choice of $\delta $ .) From (i) we get $\alpha (0)=\beta (0)=0$ . For the function $\gamma (z)=\alpha (z)-z$ we have $\gamma (z+2\pi )=\gamma (z)$ , $\alpha (z)=z+\gamma (z)$ . Consider the branches of $\log $

$$ \begin{align*} \begin{array}{r@{\ }c@{\ }lc} \operatorname{Log} z & = & \ln |z| + i \operatorname{Arg} z & (-\pi < \operatorname{Arg} z < \pi), \\ \log z & = & \ln |z| + i \arg z & (0 < \arg z < 2\pi). \end{array} \end{align*} $$

We have $\operatorname {Log} z = \log z$ when $\operatorname {Im} z> 0$ and $\log z = \operatorname {Log} z + 2\pi i$ when $\operatorname {Im} z < 0$ . By the periodicity of $\gamma $ we have

(4.1) $$ \begin{align} \gamma(-i\log z) = \gamma(-i\operatorname{Log} z)\quad (\operatorname{Im} z\not=0). \end{align} $$

Define

$$ \begin{align*} h(z)=e^{i\alpha(-i\log z)} \end{align*} $$

for $z\not =0$ not a positive real number. We have

$$ \begin{align*} h(z)=e^{i\alpha(-i\log z)}=e^{i(-i\log z)}e^{i\gamma(-i\log z)} = ze^{i\gamma(-i\log z)}. \end{align*} $$

Similarly, if we set $H(z)=e^{i\alpha (-i\operatorname {Log} z)}$ when $z\not =0$ is not a negative real number, we get $H(z) = ze^{i\gamma (-i\operatorname {Log} z)}$ , and by (4.1), $h(z)=H(z)$ for $\operatorname {Im} z\not =0$ . Thus, h extends to an analytic function defined on $\mathbb {C}\setminus \{0\}$ by setting $h(z)=H(z)$ when z is a positive real number.

When $z=e^{i\theta }$ , $0<\theta <2\pi $ , we have $-i\log z = \theta $ , so $h(z)=e^{i\alpha (\theta )}$ . As $\theta $ runs over $(0,2\pi )$ from $0$ to $2\pi $ , $\alpha (\theta )$ does the same since $\alpha (0)=0$ and $\alpha (2\pi ) = \alpha (0)+2\pi =2\pi $ , so $h(z)=h(e^{i\theta })=e^{i\alpha (\theta )}$ runs over the arc $T\setminus \{1\}$ counterclockwise from $1$ to $1$ . From (i) and the choice of $\delta $ we get

$$ \begin{align*} |h(z)-g(z)| = |h(e^{i\theta})-g(e^{i\theta})| = |e^{i\alpha(\theta)} - e^{i\beta(\theta)}|<{\varepsilon} \end{align*} $$

and this implies the first part of (a). If $\theta \in F'$ then $h(e^{i\theta }) = e^{i\alpha (\theta )} = e^{i\beta (\theta )} = g(e^{i\theta })$ and this implies the second part of (a).

Also, for $0<\theta <2\pi $ and $n\in {\mathbb N}$ ,

$$ \begin{align*} e^{i\theta}\in A_n \Leftrightarrow \theta\in A^{\prime}_n\Leftrightarrow \alpha(\theta)\in B^{\prime}_n \Leftrightarrow e^{i \alpha(\theta)}\in B_n, \end{align*} $$

so that for $z\in T\setminus \{1\}$ , we have $z\in A_n$ if and only if $h(z)\in B_n$ . Hence h restricts to an order-isomorphism of $T\setminus \{1\}$ mapping $A_n$ onto $B_n$ .

There remains to verify that $Dh(z)\not =0$ when $|z|=1$ . When $|z|=1$ , $z\not =1$ , we have

$$ \begin{align*} Dh(z)=\frac{d}{dz}e^{i\alpha(-i\log z)} = e^{i\alpha(-i\log z)} \cdot i\cdot D\alpha(-i\log z)\cdot \frac{-i}{z}, \end{align*} $$

which is nonzero since $-i\log z$ is real when $|z|=1$ and $D\alpha (\theta )\not =0$ for $\theta \in \mathbb {R}$ . Writing $\operatorname {Log}$ instead of $\log $ we have the analogous computation when $|z|=1$ , $z\not =-1$ . ▪