A nonlinear Lazarev-Lieb theorem: $L^2$-orthogonality via motion planning

TL;DR

This paper generalizes a result by Lazarev and Lieb on orthogonality of integrable functions using smooth circle-valued functions, employing topological methods and a relaxed motion planning approach.

Contribution

It provides a lower bound on the equivariant topology of smooth circle-valued functions, extending previous orthogonality results with novel topological techniques.

Findings

Established a lower bound for the equivariant topology of function spaces.

Generalized Lazarev and Lieb's orthogonality theorem.

Introduced a relaxed motion planning algorithm with topological implications.

Abstract

Lazarev and Lieb showed that finitely many integrable functions from the unit interval to $\mathbb{C}$ can be simultaneously annihilated in the $L^2$ inner product by a smooth function to the unit circle. Here we answer a question of Lazarev and Lieb proving a generalization of their result by lower bounding the equivariant topology of the space of smooth circle-valued functions with a certain $W^{1,1}$-norm bound. Our proof uses a relaxed notion of motion planning algorithm that instead of contractibility yields a lower bound for the $\mathbb{Z}/2$-coindex of a space.

Equations

$${\int }_0^1{f}_j(x)h(x)dx=0.$$

$$\Vert h{\Vert }_{W^{1,1}}={\int }_0^1|h(x)|dx+{\int }_0^1|h^{\prime }(x)|dx$$

$${\int }_0^1{f}_j(x)h(x)dx=0.$$

$$\Vert h_2-h_1{\Vert }_q$$

$$\le {\left({\int }_0^1|h_2(x)-h_1(x){|}^p\cdot (\text{diam}(S^1){)}^{q-p}dx\right)}^{1/q}$$

$$\le (\text{diam}(S^1){)}^{(q-p)/q}\cdot \Vert h_2-h_1{\Vert }_p^{p/q}$$

$$d_{0,\infty }(h_1,h_2)$$

$$\text{ for some }S\subseteq [0,1]\text{ with }\mu (S)<\delta \}.$$

$${\int }_0^1|h_2(x)-h_1(x)|dx\le {\int }_S\text{diam}(S^1)dx+{\int }_{[0,1]\setminus S}\delta dx\le \delta (\text{diam}(S^1)+1).$$

$${\int }_{[0,1]}|h_2(x)-h_1(x)|d\mu$$

$$\le \text{diam}(S^1)\mu (S)+\delta \mu ([0,1])$$

$${\psi }_j:h\mapsto {\int }_0^1{f}_j(x)h(x)dx.$$

$${\mu }_f(S)={\int }_0^1|f_j(x)|dx.$$

$$|{\psi }_j(h_2)-{\psi }_j(h_1)|$$

$$\le {\int }_{[0,1]}|h_2-h_1|d{\mu }_f$$

$$\le \Vert h_2-h_1{\Vert }_1.$$

$$y\in Y$$

$$U$$

$$\text{if}\varphi (y_0),\varphi (y_1)\in U,w<\delta ,$$

$$\text{then}\varphi (s_w(y_0,y_1)(t))\in V\text{ for all }t\in [0,1].$$

$${\tilde{\alpha }}_k(x)=\{\begin{array}{ll}{\alpha }_k(x)& x\in E\\ s_{w(x)}({\alpha }_k(u(x)),{\alpha }_k(l(x)))(t(x))& x\notin E\end{array}$$

$$\text{where }w(x)=\min (d(x,E),t(x),1-t(x))$$

$$t(x)=\frac{d(u(x),x)}{d(u(x),l(x))}$$

$$({\alpha }_{k+1}{)}_{\ge 0}:S_{\ge 0}^{k+1}\to Y$$

$$({\alpha }_{k+1}{)}_{\le 0}:S_{\le 0}^{k+1}\to Y$$

$$\varphi \circ ({\alpha }_{k+1}{)}_{\le 0}=\varphi \circ \rho \circ {\tilde{\alpha }}_k\circ \pi \circ (-)=\sigma \circ (\varphi \circ {\tilde{\alpha }}_k)\circ \pi \circ (-)$$

$${\tilde{\alpha }}_k(x)⪯\rho ({\tilde{\alpha }}_k(-x))$$

$${\tilde{\alpha }}_k(x)⪯{\alpha }_k(l(x))={\alpha }_k(-u(-x))=\rho ({\alpha }_k(u(-x)))⪯\rho ({\tilde{\alpha }}_k(-x)).$$

$${\alpha }_{k+1}(x)={\tilde{\alpha }}_k(\pi (x))⪯\rho ({\tilde{\alpha }}_k(-\pi (x)))=\rho ({\tilde{\alpha }}_k(\pi (-x)))={\alpha }_{k+1}(m(x))$$

$${\beta }_n(-x)=\varphi ({\alpha }_n(-x))=\varphi (\rho ({\alpha }_n(x)))=\sigma (\varphi ({\alpha }_n(x)))=\sigma ({\beta }_n(x))$$

$$\varphi (g)(x)=\{\begin{array}{ll}1& g(x)\text{ even}\\ -1& g(x)\text{ odd}\end{array}$$

$$s_w(g_0,g_1)(t)(x)=\{\begin{array}{ll}{g}_0(x)& x<1-t\\ g_1(x)& x\ge 1-t\end{array}$$

$$\Vert \varphi (g_2)-\varphi (g_1){\Vert }_1$$

$$\le {\int }_0^1|g_2(x)-g_1(x)|dx$$

$$\le \Vert g_2-g_1{\Vert }_1.$$

$$s_w(g_0,g_1)(t)(x)=\left(1-\tau \left(\frac{x-(1-t)}{w}\right)\right)g_0(x)+\tau \left(\frac{x-(1-t)}{w}\right)g_1(x).$$

Full text

1. Introduction

In 1965 Hobby and Rice established the following result:

Theorem 1.1** (Hobby and Rice [4]).**

Let $f_{1},\ldots,f_{n}\in L^{1}([0,1];\mathbb{R})$. Then there exists ${h\colon[0,1]\rightarrow\{\pm 1\}}$ with at most $n$ sign changes, such that for all $j$,

[TABLE]

If we restrict the $f_{j}$ to lie in $L^{2}([0,1];\mathbb{R})$, we can view this as an orthogonality result in the $L^{2}$ inner product. The Hobby–Rice theorem and its generalizations have found a multitude of applications, ranging from mathematical physics [6] and combinatorics [1] to the geometry of spatial curves [2].

The theorem also holds for $f_{1},\ldots,f_{n}\in L^{1}([0,1];\mathbb{C})$, provided $h$ is allowed $2n$ sign changes, by splitting the $f_{j}$ into real and imaginary parts. Lazarev and Lieb showed that for complex-valued $f_{j}$, the function $h$ can be chosen in $C^{\infty}([0,1];S^{1})$, where $S^{1}$ denotes the unit circle in $\mathbb{C}$:

Theorem 1.2** (Lazarev and Lieb [5]).**

Let $f_{1},\ldots,f_{n}\in L^{1}([0,1];\mathbb{C})$. Then there exists $h\in C^{\infty}([0,1];S^{1})$ such that for all $j$,

[TABLE]

If $h$ is obtained by smoothing the function $h_{0}$ guaranteed by Theorem 1.1, then we would expect its $W^{1,1}$-norm, given by

[TABLE]

to be approximately $1+2\pi n$, since $|h(x)|=1$, and each sign change of $h_{0}$ contributes approximately $\pi$ to $\int_{0}^{1}|h^{\prime}(x)|dx$. However, Lazarev and Lieb did not establish any bound on the $W^{1,1}$-norm of $h$ and left this as an open problem; this was accomplished by Rutherfoord [9], who established a bound of $1+5\pi n$. Here we improve this bound to $1+2\pi n$; see Corollary 1.4.

The Hobby–Rice theorem has a simple proof due to Pinkus [8] via the Borsuk–Ulam theorem, which states that any map $f\colon S^{n}\rightarrow\mathbb{R}^{n}$ with $f(-x)=-f(x)$ for all $x\in S^{n}$ has a zero. Lazarev and Lieb asked whether there is a similar proof of their result and write: “There seems to be no way to adapt the proof of the Hobby–Rice Theorem (which involves a fixed-point argument).” Rutherfoord [9] offered a simplified proof of Theorem 1.2 based on Brouwer’s fixed point theorem. Here we give a proof using the Borsuk–Ulam theorem directly, which adapts Pinkus’ proof of the Hobby–Rice theorem. The advantage of this approach is that our main result gives a nonlinear extension of the result of Lazarev and Lieb; see Section 4 for the proof:

Theorem 1.3**.**

Let $\psi\colon C^{\infty}([0,1];S^{1})\rightarrow\mathbb{R}^{n}$ be continuous with respect to the $L^{1}$-norm such that $\psi(-h)=-\psi(h)$ for all $h\in C^{\infty}([0,1];S^{1})$. Then there exists $h\in C^{\infty}([0,1];S^{1})$ with $\psi(h)=0$ and $\|h\|_{W^{1,1}}\leq 1+\pi n$.

This is a non-linear extension of Theorem 1.2 since for given $f_{1},\ldots,f_{n}\in L^{1}([0,1];\mathbb{C})$ the map $\psi(h)=(\int_{0}^{1}f_{j}(x)h(x)dx)_{j}$ is continuous (see Section 2) and linear, so in particular, $\psi$ satisfies $\psi(-h)=-\psi(h)$. Using the $L^{1}$-norm is no restriction; as we show in the next section, the $L^{p}$ norms on $C^{\infty}([0,1];S^{1})$ for $1\leq p<\infty$ are all equivalent, so we could replace $L^{1}$ with any such $L^{p}$. In fact, the only relevant feature of the $L^{1}$-norm is that functions $h_{1},h_{2}$ are close in the $L^{1}$-norm if $h_{1},h_{2}$ are uniformly close outside of a set of small measure. As a consequence, we recover the result of Lazarev and Lieb, with a $W^{1,1}$-norm bound of $1+2\pi n$ since $\psi$ takes values in $\mathbb{C}^{n}\cong\mathbb{R}^{2n}$; see Section 2 for the proof:

Corollary 1.4**.**

Let $f_{1},\ldots,f_{n}\in L^{1}([0,1];\mathbb{C})$. Then there exists $h\in C^{\infty}([0,1];S^{1})$ with $\|h\|_{W^{1,1}}\leq 1+2\pi n$ such that for all $j$,

[TABLE]

Given a space $Z$ with a $\mathbb{Z}/2$-action $\sigma\colon Z\to Z$, the largest integer $n$ such that the $n$-sphere $S^{n}$ with the antipodal $\mathbb{Z}/2$-action (i.e. $x\mapsto-x$) admits a continuous map $f\colon S^{n}\to Z$ with $f(-x)=\sigma(f(x))$ for all $x\in S^{n}$ is called the $\mathbb{Z}/2$-coindex of $Z$, denoted $\operatorname{\mathrm{coind}}Z$. We show that the coindex of the space of smooth $S^{1}$-valued functions in the $L^{1}$-norm with $W^{1,1}$-norm at most $1+\pi n$ is between $n$ and $2n-1$; see Theorem 6.2. Determining the coindex exactly remains an interesting open problem. Our proof proceeds by constructing $\mathbb{Z}/2$-maps from $S^{n}$, i.e., commuting with the antipodal $\mathbb{Z}/2$-actions, via elementary obstruction theory, that is, inductively dimension by dimension.

We find it illuminating to phrase our proof using the language of motion planning algorithms. A motion planning algorithm (mpa) for a space $Z$ is a continuous choice of connecting path for any two endpoints in $Z$; see Section 3 for details and Farber [3] for an introduction. An mpa for $Z$ exists if and only if $Z$ is contractible. Here we introduce the notion of (full) lifted mpa, which does not imply contractibility but is sufficiently strong to establish lower bounds for the coindex of $Z$; see Theorem 3.5. We refer to Section 3 for details.

2. Relationship between topologies on $C^{\infty}([0,1];S^{1})$

We now make precise our introductory comments about the topologies on $C^{\infty}([0,1];S^{1})$ induced by the various $L^{p}$-norms and the $d_{0,\infty}$ metric.

Proposition 2.1**.**

The $L^{p}$-norms for $1\leq p<\infty$ induce equivalent topologies on $C^{\infty}([0,1];S^{1})$.

Proof.

For $1\leq p<\infty$, let $Z_{p}$ be $C^{\infty}([0,1];S^{1})$, equipped with the topology induced by the $L^{p}$-norm. Note that $\|h\|_{p}<\infty$ for all $h\in C^{\infty}([0,1];S^{1})$, so the identity maps $1_{p,q}\colon Z_{p}\rightarrow Z_{q}$ are well-defined as functions. It suffices to show that $1_{p,q}$ is continuous for all $p,q\in[1,\infty)$.

It is a standard fact that $1_{p,q}$ is continuous for $p\geq q$ when the domain has finite measure, as is the case here for $[0,1]$. For $p<q$, we have

[TABLE]

Since $S^{1}$ is bounded, $1_{p,q}$ is continuous. Hence the $Z_{p}$ are all homeomorphic. ∎

In the introduction, we claimed that “the only relevant feature of the $L^{1}$-norm is that functions $h_{1},h_{2}$ are close in the $L^{1}$-norm if $h_{1},h_{2}$ are uniformly close outside of a set of small measure.” To give content to this statement, we define a metric $d_{0,\infty}$ on $C^{\infty}([0,1];S^{1})$ by

[TABLE]

Proposition 2.2**.**

The function $d_{0,\infty}$ is a metric.

Proof.

By the continuity of maps in $C^{\infty}([0,1];S^{1})$, we have $d_{0,\infty}(h_{1},h_{2})=0$ iff $h_{1}=h_{2}$. For the triangle inequality, suppose:

• •

$|h_{2}(x)-h_{1}(x)|<\delta_{1}$ for all $x\in[0,1]\setminus S_{1}$, where $\mu(S_{1})<\delta_{1}$.

• •

$|h_{3}(x)-h_{2}(x)|<\delta_{2}$ for all $x\in[0,1]\setminus S_{2}$, where $\mu(S_{2})<\delta_{2}$

Then $|h_{3}(x)-h_{1}(x)|<\delta_{1}+\delta_{2}$ for all $x\in[0,1]\setminus(S_{1}\cup S_{2})$, and $\mu(S_{1}\cup S_{2})<\delta_{1}+\delta_{2}$. Hence $d_{0,\infty}(h_{1},h_{3})\leq\delta_{1}+\delta_{2}$. Taking the infimum over $\delta_{1},\delta_{2}$, we obtain $d_{0,\infty}(h_{1},h_{3})\leq d_{0,\infty}(h_{1},h_{2})+d_{0,\infty}(h_{2},h_{3})$. ∎

Proposition 2.3**.**

The metric $d_{0,\infty}$ and the norm $\|\cdot\|_{1}$ induce equivalent topologies on $C^{\infty}([0,1];S^{1})$.

Proof.

Let $Z_{0,\infty}$ be $C^{\infty}([0,1];S^{1})$, equipped with the topology induced by $d_{0,\infty}$; it suffices to show that the identity maps between $Z_{0,\infty},Z_{1}$ are continuous.

For the identity map $1\colon Z_{0,\infty}\rightarrow Z_{1}$, suppose $d_{0,\infty}(h_{1},h_{2})<\delta$, so that there exists $S\subseteq[0,1]$ with $\mu(S)<\delta$ such that $|h_{2}(x)-h_{1}(x)|<\delta$ on $[0,1]\setminus S$. Then

[TABLE]

This shows that $1\colon Z_{0,\infty}\rightarrow Z_{1}$ is continuous.

For the identity map $1\colon Z_{1}\rightarrow Z_{0,\infty}$, let $\varepsilon>0$ and suppose $\|h_{2}-h_{1}\|_{1}<\delta$ for $\delta=\varepsilon^{2}$. If $d_{0,\infty}(h_{1},h_{2})\geq\varepsilon$, then $|h_{2}(x)-h_{1}(x)|\geq\varepsilon$ on a set $S$ with $\mu(S)\geq\varepsilon$, implying $\|h_{2}-h_{1}\|_{1}\geq\varepsilon^{2}$, a contradiction. Hence $d_{0,\infty}(h_{1},h_{2})<\varepsilon$, and $1\colon Z_{1}\rightarrow Z_{0,\infty}$ is continuous. ∎

Now we expand our view to consider $L^{p}$ spaces under other measures $\mu$. We show that finite, absolutely continuous measures can only produce coarser topologies than Lebesgue measure:

Proposition 2.4**.**

Let $\mu$ be a finite measure on $[0,1]$ that is absolutely continuous with respect to Lebesgue measure. Let $Z_{1}$ be $C^{\infty}([0,1];S^{1})$, equipped with the topology induced by the $L_{1}$-norm with respect to Lebesgue measure, and let $Z_{1,\mu}$ be $C^{\infty}([0,1];S^{1})$, equipped with the topology induced by the $L_{1}$-norm with respect to $\mu$. Then the identity function $1\colon Z_{1}\rightarrow Z_{1,\mu}$ is continuous.

Proof.

By Proposition 2.3, it suffices to show that $1\colon Z_{0,\infty}\rightarrow Z_{1,\mu}$ is continuous. The argument is similar to the argument that $1\colon Z_{0,\infty}\rightarrow Z_{1}$ is continuous. Using $\lambda$ to denote Lebesgue measure, suppose $d_{0,\infty}(h_{1},h_{2})<\delta$, so that there exists $S\subseteq[0,1]$ with $\lambda(S)<\delta$ such that $|h_{2}(x)-h_{1}(x)|<\delta$ on $[0,1]\setminus S$. Then

[TABLE]

Note that since $\mu$ is finite, we have $\mu([0,1])<\infty$. As $\delta\rightarrow 0$, we have $\lambda(S)\rightarrow 0$, so $\mu(S)\rightarrow 0$ by absolute continuity, hence the right side approaches 0. This shows the desired continuity. ∎

The relationships between the topologies on $C^{\infty}([0,1];S^{1})$ can be summarized as follows, where $1<p_{1}<p_{2}<\infty$ and $\mu$ is a finite measure on $[0,1]$ which is absolutely continuous with respect to Lebesgue measure:

{Z_{\infty}}$${Z_{p_{2}}}$${Z_{p_{1}}}$${Z_{1}}$${Z_{0,\infty}}$${Z_{p_{2},\mu}}$${Z_{p_{1},\mu}}$${Z_{1,\mu}}$$\scriptstyle{\not\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}

Therefore, when establishing the continuity of $\psi$ for the sake of applying Theorem 1.3, we may use any $L^{p}$ norm on $C^{\infty}([0,1];S^{1})$, with respect to any finite measure $\mu$ on $[0,1]$ which is absolutely continuous with respect to Lebesgue measure. (If we use a measure $\mu$ other than Lebesgue measure, we can precompose $\psi$ with $1\colon Z_{1}\rightarrow Z_{1,\mu}$ before applying Theorem 1.3.)

With these results in hand, we can now deduce Corollary 1.4 from Theorem 1.3:

Proof of Corollary 1.4.

Let $\psi\colon C^{\infty}([0,1];S^{1})\rightarrow\mathbb{C}^{n}$ be given by component maps

[TABLE]

We claim $\psi_{j}$ is continuous. Since $f_{j}\in L^{1}([0,1];\mathbb{C})$, $f_{j}$ induces a finite measure $\mu_{f}$ which is absolutely continuous with respect to Lebesgue measure, given by

[TABLE]

By the above, we may view $C^{\infty}([0,1];S^{1})$ as having the topology induced by the $L^{1}$-norm $\|\cdot\|_{1}$ with respect to $\mu_{f}$. Then

[TABLE]

Therefore, $\psi_{j}$ is continuous, so $\psi$ is continuous. Viewing the codomain $\mathbb{C}^{n}$ of $\psi$ as $\mathbb{R}^{2n}$, we may apply Theorem 1.3 and get $\|h\|_{W^{1,1}}\leq 1+2\pi n$. ∎

3. Lifts of motion planning algorithms and the coindex

Our proof of Theorem 1.3 makes use of motion planning algorithms; see Farber [3]. We use $Y,Z$ in the following definitions to match our notation later:

Definition 3.1**.**

Let $Z$ be a topological space, and let $PZ$ be the space of continuous paths $\gamma\colon[0,1]\rightarrow Z$, equipped with the compact-open topology. Then a motion planning algorithm (or mpa) is a continuous map $s\colon Z\times Z\rightarrow PZ$, such that $s(z_{0},z_{1})(0)=z_{0}$ and $s(z_{0},z_{1})(1)=z_{1}$.

For $Z$ a locally compact Hausdorff space, using the compact-open topology for $PZ$ ensures that a function $s\colon Z\times Z\rightarrow PZ$ is continuous if and only if its uncurried form $\widetilde{s}\colon Z\times Z\times[0,1]\rightarrow Z$ given by $(z_{0},z_{1},t)\mapsto s(z_{0},z_{1})(t)$ is continuous; see Munkres [7, Thm. 46.11]. One basic fact is that an mpa for $Z$ exists if and only if $Z$ is contractible [3].

We weaken the definition above for our purposes:

Definition 3.2**.**

Let $Y,Z$ be topological spaces, and let $\phi\colon Y\rightarrow Z$ be continuous. Let $(\preceq)$ be a preorder on $Y$, and let $Y^{2}_{\preceq}=\{(y_{0},y_{1})\in Y^{2}:y_{0}\preceq y_{1}\}$, giving $Y^{2}$ the product topology and $Y^{2}_{\preceq}$ the resulting subspace topology.

A lifted motion planning algorithm (or lifted mpa) for $(Y,Z,\phi,\preceq)$ is a family of maps $s_{w}\colon Y^{2}_{\preceq}\rightarrow PY$ for $w\in(0,1]$ with $s_{w}(y_{0},y_{1})(0)=y_{0}$ and $s_{w}(y_{0},y_{1})(1)=y_{1}$, assembling into a continuous map $s\colon(0,1]\times Y^{2}_{\preceq}\rightarrow PY$, with the following continuity property:

[TABLE]

Definition 3.3**.**

A lifted mpa $s\colon(0,1]\times Y^{2}_{\preceq}\rightarrow PY$ for $(Y,Z,\phi,\preceq)$ is full if $y_{0}\preceq y_{1}$ for all $y_{0},y_{1}\in Y$. In this case we say $s$ is a full lifted mpa for $(Y,Z,\phi)$, omitting $(\preceq)$.

The continuity property essentially says that if two points $y_{1},y_{2}\in Y$ have images in $Z$ close to $\phi(y)\in Z$, then $s_{w}$ carries $(y_{0},y_{1})$ to a path whose image under $\phi$ is a path that stays close to $\phi(y)$, provided $w$ is small.

Note that an mpa $s\colon Z\times Z\rightarrow PZ$ satisfying $s(z,z)=c_{z}$ for all $z\in Z$ extends to a full lifted mpa for $(Z,Z,1_{Z})$ by taking $s_{w}=s$ for all $w$; the continuity property just restates the continuity of $s$ at diagonal points $(z,z)\in Z\times Z$.

This relaxed notion of mpa still provides lower bounds for the (equivariant) topology of $Z$ that are weaker than contractibility. Recall that for a topological space $Z$ with $\mathbb{Z}/2$-action generated by $\sigma\colon Z\to Z$ the $\mathbb{Z}/2$-coindex of $Z$ denoted by $\operatorname{\mathrm{coind}}Z$ is the largest integer $n$ such that there is a $\mathbb{Z}/2$-map $f\colon S^{n}\to Z$, that is, a map satisfying $f(-x)=\sigma(f(x))$.

Definition 3.4**.**

Let $x\in S^{k}$, and let $x=(x_{1},\ldots,x_{k+1})$. We say that $x$ is positive if its last nonzero coordinate is positive, and negative otherwise.

Our main tool in proving Theorem 1.3 will be the following theorem:

Theorem 3.5**.**

Let $Y,Z$ be topological spaces, equip $Y$ with a $\mathbb{Z}$-action generated by $\rho\colon Y\rightarrow Y$, and equip $Z$ with a $\mathbb{Z}/2$-action generated by $\sigma\colon Z\rightarrow Z$. Let $\phi\colon Y\rightarrow Z$ be continuous and equivariant, i.e., $\sigma\circ\phi=\phi\circ\rho$. Let $(\preceq)$ be a preorder on $Y$ and $s\colon(0,1]\times Y^{2}_{\preceq}\rightarrow PY$ a lifted mpa for $(Y,Z,\phi,\preceq)$ such that:

1. (1)

$y\preceq\rho(y)$. 2. (2)

$\rho(y_{0})\preceq\rho(y_{1})$* if and only if $y_{0}\preceq y_{1}$.* 3. (3)

$y_{0}\preceq y_{1}$* implies $y_{0}\preceq s_{w}(y_{0},y_{1})(t)\preceq y_{1}$, for all $w\in(0,1]$, $t\in[0,1]$.*

Then for each integer $n\geq 0$, there exists a $\mathbb{Z}/{2}$-map $\beta_{n}\colon S^{n}\rightarrow Z$. Moreover, for any choice of initial point $y^{*}\in Y$, the maps $\beta_{n}$ can be chosen such that $\beta_{n}$ maps each positive point of $S^{n}$ to a point in $Z$ of the form $\phi(y)$, with $y^{*}\preceq y\preceq\rho^{n}(y^{*})$, that is, the subspace of these points $\phi(y)$ and their antipodes $\sigma(\phi(y))$ in $Z$ has coindex at least $n$.

We will apply Theorem 3.5 by taking $Z$ to be $C^{\infty}([0,1];S^{1})$ with the topology induced by the $L^{1}$-norm, and $Y$ to be $C^{\infty}([0,1];\mathbb{R})$ with the $L^{1}$-norm, restricted to increasing functions. Using lifted mpa’s allows us to reason about paths in $Y$, which are simpler than paths in $Z$. The theorem encapsulates the inductive construction of a function $\alpha_{n}\colon S^{n}\rightarrow Y$, from which we produce ${\beta_{n}\colon S^{n}\rightarrow Z}$; the continuity property of a lifted mpa is needed for this construction to work. The last part of the theorem will give us the $W^{1,1}$-norm bound.

Proof of Theorem 3.5.

We will inductively construct a function $\alpha_{n}\colon S^{n}\rightarrow Y$ and then take $\beta_{n}=\phi\circ\alpha_{n}$. We will allow $\alpha_{n}$ to be discontinuous on the equator of $S^{n}$, but in such a way that $\phi\circ\alpha_{n}$ is continuous everywhere.

Specifically, let $\alpha_{k}\colon S^{k}\rightarrow Y$ be a function, not necessarily continuous. Let $m\colon S^{k}\rightarrow S^{k}$ be given by $(x_{1},\ldots,x_{k},x_{k+1})\mapsto(x_{1},\ldots,x_{k},-x_{k+1})$, so that $m$ mirrors points across the plane perpendicular to the last coordinate axis. Then we say that $\alpha_{k}$ is good if

1. ($\alpha$-1)

For $x$ positive, $y^{*}\preceq\alpha_{k}(x)\preceq\rho^{k}(y^{*})$, and $\alpha_{k}(-x)=\rho(\alpha_{k}(x))$. 2. ($\alpha$-2)

For $x$ in the open upper hemisphere, $\alpha_{k}(x)\preceq\alpha_{k}(m(x))$. 3. ($\alpha$-3)

$\alpha_{k}$ is continuous on the open upper hemisphere. 4. ($\alpha$-4)

$\phi\circ\alpha_{k}$ is continuous.

Let $u,l\colon B^{k+1}\rightarrow S^{k}$ be the projections to the closed upper and lower hemispheres, that is, $u(x)$ is the unique point in the closed upper hemisphere sharing its first $k$ coordinates with $x$, and similarly for $l(x)$ for the lower hemisphere. Then we have the following claim:

Claim**.**

If $\alpha_{k}\colon S^{k}\rightarrow Y$ is good, then $\alpha_{k}$ extends to $\widetilde{\alpha}_{k}\colon B^{k+1}\rightarrow Y$, such that:

1. ($\widetilde{\alpha}$-1)

For all $x\in B^{k+1}$, we have $y^{*}\preceq\widetilde{\alpha}_{k}(x)\preceq\rho^{k+1}(y^{*})$. 2. ($\widetilde{\alpha}$-2)

For all $x\in B^{k+1}$, we have $\alpha_{k}(u(x))\preceq\widetilde{\alpha}_{k}(x)\preceq\alpha_{k}(l(x))$. 3. ($\widetilde{\alpha}$-3)

$\widetilde{\alpha}_{k}$ is continuous in the interior of $B^{k+1}$. 4. ($\widetilde{\alpha}$-4)

$\phi\circ\widetilde{\alpha}_{k}$ is continuous.

Proof of Claim.

Let $E\subset S^{k}$ be the equator, the set of points neither in the open upper or lower hemisphere. The set $E$ is compact, so the distance $d(x,E)$ for $x\in B^{k+1}$ is well-defined and nonzero for $x\notin E$. Define $\widetilde{\alpha}_{k}\colon B^{k+1}\rightarrow X_{k+1}$ by

[TABLE]

Note that $l(x)=m(u(x))$, so ($\alpha$-2) implies $\alpha_{k}(u(x))\preceq\alpha_{k}(l(x))$, so $s_{w(x)}(\alpha_{k}(u(x)),\alpha_{k}(l(x)))$ is well-defined, and (3) gives $\alpha_{k}(u(x))\preceq\widetilde{\alpha}(x)\preceq\alpha_{k}(l(x))$, establishing ($\widetilde{\alpha}$-2).

By ($\alpha$-1), we have $\rho(y^{*})\preceq\rho(\alpha_{k}(x))\preceq\rho^{k+1}(y^{*})$ for $x$ negative, so $y^{*}\preceq\alpha_{k}(x)\preceq\rho^{k+1}(y^{*})$ for all $x\in S^{k}$. Along with the inequality above, this implies $y^{*}\preceq\widetilde{\alpha}_{k}(x)\preceq\rho^{k+1}(y^{*})$, establishing ($\widetilde{\alpha}$-1).

The function $\widetilde{\alpha}_{k}$ is continuous for $x\notin E$, since $u(-)$, $l(-)$, $d(-,-)$, $d(-,E)$ are all continuous, $u(x),l(x)\notin E$, and $\alpha_{k}$ is continuous on the open upper (and hence lower) hemisphere. In particular, $\widetilde{\alpha}_{k}$ is continuous in the interior of $B^{k+1}$, establishing ($\widetilde{\alpha}$-3).

It remains to show $\phi\circ\widetilde{\alpha}_{k}$ is continuous at $x\in E$. Let $V$ be a neighborhood of $\phi(\widetilde{\alpha}_{k}(x))=\phi(\alpha_{k}(x))\in Z$, and obtain $\delta>0$ and a neighborhood $U$ of $\phi(\alpha_{k}(x))\in Z$ as in the lifted mpa definition. Since $u(-),l(-),d(-,E)$ are continuous, there exists a neighborhood $W\subseteq B^{k+1}$ of $x$ such that for all $x^{\prime}\in W$ we have $d(x^{\prime},E)<\delta$ and $u(x^{\prime}),l(x^{\prime})\in(\phi\circ\alpha_{k})^{-1}(U)$, using the continuity of $\phi\circ\alpha_{k}$ given by ($\alpha$-4). Then $\phi(\alpha_{k}(u(x^{\prime}))),\phi(\alpha_{k}(l(x^{\prime})))\in U$, so the lifted mpa property implies $\phi(\widetilde{\alpha}_{k}(x))\in V$, which shows $\phi\circ\widetilde{\alpha}_{k}$ is continuous at $x$, establishing ($\widetilde{\alpha}$-4). ∎

We use the claim above to inductively construct $\alpha_{k}\colon S^{k}\rightarrow Y$, by extending each $\alpha_{k}$ to a map $\widetilde{\alpha}_{k}\colon B^{k+1}\rightarrow Y$, using $\widetilde{\alpha}_{k}$ for the upper hemisphere of $\alpha_{k+1}$, and extending to the negative hemisphere via $\alpha_{k+1}(-x)=\rho(\alpha_{k+1}(x))$. Specifically, we have the following claim:

Claim**.**

For all $k\geq 0$ there exists $\alpha_{k}\colon S^{k}\rightarrow Y$, not necessarily continuous, such that $\alpha_{k}$ is good.

Proof of Claim.

We use induction. For the base case, use $\pm 1$ to denote the points of $S^{0}$; then let $\alpha_{0}$ map $\pm 1$ to $y^{*},\rho(y^{*})$, respectively. Then $\alpha_{0}$ is good.

Given $\alpha_{k}$ good and $\widetilde{\alpha}_{k}$ obtained through the previous claim, we now construct $\alpha_{k+1}\colon S^{k+1}\rightarrow Y$. Let $\pi\colon S^{k+1}_{\geq 0}\rightarrow B^{k+1}$ be the projection of the closed upper hemisphere onto the first $k+1$ coordinates. We define maps on the two closed hemispheres as follows:

[TABLE]

Finally, we define $\alpha_{k+1}$ by $x\mapsto(\alpha_{k+1})_{\geq 0}(x)$ for $x$ positive and $x\mapsto(\alpha_{k+1})_{\leq 0}(x)$ for $x$ negative.

For $\alpha_{k+1}$, ($\alpha$-1) holds by construction, due to ($\widetilde{\alpha}$-1). Next, since $\widetilde{\alpha}_{k}$ is continuous in the interior of $B^{k+1}$, we have that $(\alpha_{k+1})_{\geq 0}$ is continuous on the open upper hemisphere, hence $\alpha_{k+1}$ is also, so ($\alpha$-3) holds also.

Since $\widetilde{\alpha}_{k}$ satisfies $\widetilde{\alpha}_{k}(-x)=\rho(\widetilde{\alpha}_{k}(x))$ for positive $x$ on the boundary sphere $S^{k}\subset B^{k+1}$, we have $(\alpha_{k+1})_{\leq 0}(x)=\rho^{2}((\alpha_{k+1})_{\geq 0}(x))$ for positive $x$ on the equator $S^{k}\subset S^{k+1}$, and $(\alpha_{k+1})_{\leq 0}(x)=(\alpha_{k+1})_{\geq 0}(x)$ for negative $x$ on the equator. Hence $\phi\circ(\alpha_{k+1})_{\geq 0},\phi\circ(\alpha_{k+1})_{\leq 0}$ agree on the equator, since $\phi\circ\rho^{2}=\sigma^{2}\circ\phi=\phi$. Moreover, both composites are continuous; for the second, we have

[TABLE]

and $\sigma,\phi\circ\widetilde{\alpha}_{k},\pi,(-)$ are continuous. Hence ($\alpha$-4) holds.

Before showing ($\alpha$-2), we show that ($\widetilde{\alpha}$-2) implies

[TABLE]

for all $x\in B^{k+1}$ not on the equator. For such $x$, $u(-x)$ is on the open upper hemisphere and hence is positive. By ($\widetilde{\alpha}$-2), we have

[TABLE]

This proves the inequality above.

Now we show ($\alpha$-2). For $x\in S^{k+1}$ in the open upper hemisphere, we have

[TABLE]

by the inequality above. Hence ($\alpha$-2) holds. ∎

Taking $\beta_{n}=\phi\circ\alpha_{n}$, Theorem 3.5 follows from the claims above. To see that $\beta_{n}$ is a $\mathbb{Z}/2$-map, note that for $x\in S^{n}$ positive, we have

[TABLE]

The other conclusions of the theorem are clear. ∎

For a full lifted mpa, the preorder conditions of Theorem 3.5 are trivially satisfied, so we get:

Corollary 3.6**.**

Let $Y,Z$ be topological spaces, equip $Y$ with a $\mathbb{Z}$-action generated by $\rho\colon Y\rightarrow Y$, and equip $Z$ with a $\mathbb{Z}/2$-action generated by $\sigma\colon Z\rightarrow Z$. Let $\phi\colon Y\rightarrow Z$ be continuous and equivariant, i.e., $\sigma\circ\phi=\phi\circ\rho$. If there is a full lifted mpa for $(Y,Z,\phi)$, then there exists a $\mathbb{Z}/2$-map $\beta_{n}\colon S^{n}\rightarrow Z$ for all integers $n\geq 0$.

4. Constructing a lifted mpa

The goal of this section is to prove our main result, Theorem 1.3, by constructing a lifted mpa satisfying the conditions of Theorem 3.5. As a warm-up, we use Theorem 3.5 to prove the Hobby-Rice theorem, Theorem 1.1:

Proof of Theorem 1.1.

The idea is to lift the space of functions with range in $\{\pm 1\}$ to nondecreasing functions with range in $\mathbb{Z}$. By describing a continuous map from pairs of such functions to paths between them, we will produce a lifted mpa, which will imply the result by Theorem 3.5.

Let $Y$ be the space of nondecreasing functions $g\colon[0,1]\rightarrow\mathbb{Z}$ with finite range, and let $Z$ be the space of functions $h\colon[0,1]\rightarrow\{\pm 1\}$. Equip $Y,Z$ with the $L^{1}$-norm, and define $\rho(g)=g+1$, $\sigma(h)=-h$, and

[TABLE]

Let $g_{0}\preceq g_{1}$ if $g_{0}(x)\leq g_{1}(x)$ for all $x\in[0,1]$. Finally, for $g_{0}\preceq g_{1}$ define $s_{w}(g_{0},g_{1})$ to be the path (in $t$) of functions following $g_{0}$ on $[0,1-t)$ and $g_{1}$ on $[1-t,1]$:

[TABLE]

Note that $s_{w}$ is independent of $w$. The conditions of Theorem 3.5 are straightforward to check, except perhaps the continuity property in the lifted mpa definition, which we check now.

We are given $g\in Y$, and we may assume $V$ is a basis set, so that $V$ consists of all $h\in Z$ with $\|h-\phi(g)\|<\varepsilon$ for some $\varepsilon>0$. By our choice of $U$ we may ensure that $g_{0},g_{1}\in Y$ have the same parity as $g$ except on a sets $S_{0},S_{1}$ with $\mu(S_{i})<\varepsilon/4$. Then functions $g^{\prime}$ along the path $s_{w}(g_{0},g_{1})$ have the same parity as $g$ except on $S_{0}\cup S_{1}$, where $\mu(S_{0}\cup S_{1})<\varepsilon/2$, which implies $\|\phi(g^{\prime})-\phi(g)\|<\varepsilon$.

Hence the conditions of Theorem 3.5 are satisfied, so we obtain a $\mathbb{Z}/2$-map $\beta_{n}\colon S^{n}\rightarrow Z$. Applying the Borsuk–Ulam theorem to $\psi\circ\beta_{n}\colon S^{n}\rightarrow\mathbb{R}^{n}$, where $\psi\colon h\mapsto(\int_{0}^{1}f_{j}(x)h(x)dx)_{j}$, we obtain $x\in S^{n}$ with $\psi(\beta_{n}(x))=0$. Hence also $\psi(\beta_{n}(-x))=0$, so we may assume $x$ is positive. Taking $y^{*}=0$ in the last part of Theorem 3.5, we may ensure that $\beta_{n}$ maps each positive point of $S^{n}$ to a point in $Z$ of the form $\phi(g)$ with $0\leq g\leq n$, so that $\phi(g)$ has at most $n$ sign changes. This completes the proof. ∎

Now we prove our main result, Theorem 1.3:

Proof of Theorem 1.3.

Consider the space $C^{\infty}([0,1];\mathbb{R})$ with the $L^{1}$-norm, and let $Y$ be the subspace of nondecreasing functions in $C^{\infty}([0,1];\mathbb{R})$, equipped with the action $\rho\colon g\mapsto g+\pi$. Let $Z$ be $C^{\infty}([0,1];S^{1})$ with the $L^{1}$-norm, equipped with the action $\sigma\colon h\mapsto-h$.

Define $\phi\colon Y\rightarrow Z$ by $\phi(g)(x)=e^{ig(x)}$; then $\phi$ is continuous since $x\mapsto e^{ix}$ is 1-Lipschitz:

[TABLE]

Define $(\preceq)$ on $Y$ as $(\leq)$ pointwise. Then properties (1) and (2) of Theorem 3.5 and the commutativity property $\phi\circ\rho=\sigma\circ\phi$ evidently hold.

It remains to construct the lifted mpa $s$. Let $\tau\colon\mathbb{R}\rightarrow[0,1]$ be a smooth, nondecreasing function with $\tau(x)=0$ for $x\leq-1$, and $\tau(x)=1$ for $x\geq 1$. (For example, take an integral of a mollifier.) Then define $s_{w}\colon Y_{\preceq}^{2}\rightarrow PY$ by

[TABLE]

Since $\tau$ is smooth, and since $x\mapsto(x-(1-t))/w$ is smooth for $w\neq 0$, the function $s_{w}(g_{0},g_{1})(t)\colon[0,1]\rightarrow\mathbb{R}$ is smooth. Also, $s_{w}(g_{0},g_{1})(t)$ is nondecreasing:

[TABLE]

Therefore, $s_{w}(g_{0},g_{1})$ takes values in $PY$. Since $g_{0}\leq g_{1}$, we have $g_{0}\leq s_{w}(g_{0},g_{1})(t)\leq g_{1}$, so property (3) of Theorem 3.5 holds.

Next we show $s_{w}(g_{0},g_{1})(t)$ is continuous in $w,g_{0},g_{1},t$. First we establish a helpful result. Let $B$ be the subspace of $L^{\infty}([0,1];\mathbb{R})$ consisting of smooth functions, and let $\widetilde{Y}$ be the space $L^{1}([0,1];\mathbb{R})$, of which $Y$ is a subspace; then pointwise multiplication $(b,g)\mapsto b\cdot g$ defines a continuous map $B\times\widetilde{Y}\rightarrow\widetilde{Y}$, via the following inequality, using Hölder’s inequality:

[TABLE]

Since $(w,g_{0},g_{1},t)\mapsto g_{0}$, $(w,g_{0},g_{1},t)\mapsto g_{1}$ are continuous maps $(0,1]\times Y\times Y\times[0,1]\rightarrow Y$, by the result above it suffices to show that

[TABLE]

is a continuous map to $B$; the subtraction from 1 in the first term is handled by virtue of the fact that $B$ is a normed linear space, so that pointwise addition and scalar multiplication by $-1$ each define a continuous map.

Since $\tau$ is constant outside of the compact set $[-1,1]$, $\tau$ is uniformly continuous, hence it suffices to prove that

[TABLE]

is a continuous map to $B$. Note that

[TABLE]

Since $w\mapsto 1/w$ is a continuous map $\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$, the map $(w,g_{0},g_{1},t)\mapsto(x\mapsto x/w)$ is a continuous map to $B$, as is $(w,g_{0},g_{1},t)\mapsto(x\mapsto-(1-t)/w)$, so the map above is indeed a continuous map to $B$. Hence $s_{w}(g_{0},g_{1})(t)$ is continuous in $w,g_{0},g_{1},t$.

It remains to show the continuity property for a lifted mpa. Let $g\in Y$, then for $g_{0},g_{1}\in Y$ we have

[TABLE]

where we use the fact that $S^{1}$ has diameter 2 in the last step. This inequality implies the continuity property for a lifted mpa.

Therefore, we may apply Theorem 3.5 to obtain a $\mathbb{Z}/2$-map $\beta_{n}\colon S^{n}\rightarrow Z$. Then $\psi\circ\beta_{n}\colon S^{n}\rightarrow\mathbb{R}^{n}$ is a $\mathbb{Z}/2$-map, so by the Borsuk–Ulam theorem, we have $\psi(\beta_{n}(x))=0$ for some $x\in S^{n}$, and we may assume $x$ is positive. Taking $y^{*}=c_{0}$ in the last part of Theorem 3.5, we have $\rho^{n}(y^{*})=c_{n}$, so we may ensure that $h=\beta_{n}(x)$ is of the form $\phi(g)$ for $g\in Y$, where $g$ is an increasing function with range in $[0,\pi n]$. This gives the desired $W^{1,1}$-norm bound:

[TABLE]

which implies $\|h\|_{W^{1,1}}\leq 1+\pi n$. ∎

5. Improving the bound further

In the introduction we argued that a $W^{1,1}$-norm bound of $1+2\pi n$ in Theorem 1.2 might be expected from smoothing the Hobby–Rice theorem. In this section, we show an improved bound for Theorem 1.2 in the case where the $f_{j}$ are real-valued. The idea is to modify the $S^{1}$ step of our construction so that some functions in the image of $\alpha_{k}$ have smaller range within $[0,\pi k]$, and to modify the later steps so that functions $h$ in the image of $\alpha_{k}$ with large range have $\psi(\phi(h))\neq 0$.

Theorem 5.1**.**

Let $f_{1},\ldots,f_{n}\in L^{1}([0,1];\mathbb{R})$. Then there exists $h\in C^{\infty}([0,1];S^{1})$ such that for all $j$,

[TABLE]

Moreover, for any $\varepsilon>0$, $h$ can be chosen such that

[TABLE]

Proof.

Define $Y,Z,\rho,\sigma,\phi,s$ as in the proof of Theorem 1.3, let $y^{*}=c_{0}$, and let $(\preceq)$ be $(\leq)$. We will produce $\alpha_{n}\colon S^{n}\rightarrow Y$ and $\beta_{n}\colon S^{n}\rightarrow Z$ by the inductive construction in the proof of Theorem 3.5, but we modify the first step by defining $\alpha_{1}\colon S^{1}\rightarrow Y$ by $e^{ix}\mapsto c_{x}$ for $x\in[0,2\pi)$. This $\alpha_{1}$ differs from the $\alpha_{1}$ obtained in the proof of Theorem 3.5, which only gives constant functions at $\pm 1\in S^{1}$, but is still good in the sense introduced in the proof of Theorem 3.5. Using this $\alpha_{1}$ as our base case, we inductively construct $\alpha_{k}$ as before with the following additional condition:

[TABLE]

Here $\mu_{f}$ is as in the proof of Corollary 1.4, that is,

[TABLE]

and $\pi_{1}\colon S^{k}\rightarrow[-1,1]$ is the projection to the first coordinate.

The condition $(P_{\alpha_{k},\delta})$ holds for $k=1$ and all $\delta>0$ by our definition of $\alpha_{1}$. To show that the condition carries through the inductive step, it suffices to show that given $\delta>0$, there exists $\delta^{\prime}>0$ such that given $\alpha_{k}$ such that $(P_{\alpha_{k},\delta^{\prime}})$ holds, we can extend $\alpha_{k}$ to $\widetilde{\alpha}_{k}$ as in the first claim in the proof of Theorem 3.5 such that $(P_{\widetilde{\alpha}_{k},\delta})$ holds.

We accomplish this by modifying the definition of $\widetilde{\alpha}_{k}$ in the first claim in the proof of Theorem 3.5 to impose a universal upper bound on $w(x)$. Since $\mu_{f}$ is absolutely continuous with respect to Lebesgue measure $\lambda$, for $\delta^{\prime\prime}>0$ there exists $\delta^{\prime\prime\prime}>0$ such that $\lambda(S)\leq 2\delta^{\prime\prime\prime}$ implies $\mu_{f}(S)<\delta^{\prime\prime}$. Then we use $\delta^{\prime\prime\prime}$ as our upper bound on $w(x)$:

[TABLE]

This ensures that functions in the image of $\widetilde{\alpha}_{k}$ are equal to one of the functions $\alpha_{k}(u(x)),\alpha_{k}(l(x))$ except on a set $S$ with $\mu_{f}(S)<\delta^{\prime\prime}$. Hence we may take $\delta^{\prime}=\delta^{\prime\prime}=\delta/2$; then $(P_{\widetilde{\alpha_{k}}},\delta)$ holds as desired. This shows that for any $\delta>0$, $\alpha_{k}$ may be chosen such that $(P_{\alpha_{k},\delta})$ holds.

Now we apply the Borsuk–Ulam theorem as before. We have the following diagram:

[TABLE]

The composition $\psi\circ\phi\circ\alpha_{2n}$ is a $\mathbb{Z}/{2}$-map, so the Borsuk–Ulam theorem implies that it has a zero; that is, there exists $x\in S^{2n}$ such that for all $j$, we have

[TABLE]

Moreover, we may assume $x\in S^{2n}$ is positive.

But by the above, we have for the real parts, for all $j$,

[TABLE]

We can bound the last term as follows:

[TABLE]

Now if all $\int_{0}^{1}f_{j}(t)dt$ are [math], then we may take $h$ to be an arbitrary constant, which gives ${\|h\|_{W^{1,1}}=1}$. Hence we may assume that some $\int_{0}^{1}f_{j}(t)dt$ is nonzero. In this case, we may ensure that for the $x$ with $(\psi\circ\phi\circ\alpha_{2n})(x)=0$ guaranteed by the Borsuk–Ulam theorem, $\pi_{1}(x)$ is smaller than any constant we like, by taking $\delta$ small in $(P_{\alpha_{2n},\delta})$. In particular, choose $\delta$ sufficiently small such that $|\mathrm{Re}[e^{i\theta}]|<\delta$ implies $|\theta-\pi/2|<\varepsilon^{\prime}$ for $\theta\in[0,\pi]$.

Now we analyze the ranges of functions $\alpha_{k}(x)\colon[0,1]\rightarrow\mathbb{R}$ with $x$ positive and $|\pi_{1}(x)|<\delta$, using the fact that functions $\alpha_{k+1}(x)$ are produced as transition functions between two functions $\alpha_{k}(x^{\prime}),\alpha_{k}(x^{\prime\prime})$ with $\pi_{1}(x^{\prime})=\pi_{1}(x^{\prime\prime})=\pi_{1}(x)$. For $k=1$, $\alpha_{k}(x)$ has range in $[\pi/2-\varepsilon^{\prime},\pi/2+\varepsilon^{\prime}]$, and each increment of $k$ extends the right end of this interval by $\pi$. Hence $\alpha_{2n}(x)$ has range in

[TABLE]

Hence taking $h=\phi(\alpha_{2n}(x))$ gives $\|h\|_{W^{1,1}}\leq 1+\pi(2n-1)+2\varepsilon^{\prime}$. Choosing $\varepsilon^{\prime}<\varepsilon/2$ gives the desired result. ∎

6. A lower bound

We ask whether $\|h\|_{W^{1,1}}\leq 1+2n\pi$ is the best possible bound in Theorem 1.2. We prove a lower bound of $1+n\pi$ in the case that the $f_{j}$ are real-valued, which implies the same lower bound in the case that the $f_{j}$ are complex-valued.

Theorem 6.1**.**

There exist $f_{1},\ldots,f_{n}\in L^{1}([0,1];\mathbb{R})$, such that for any $h\in C^{1}([0,1];S^{1})$ with

[TABLE]

we have $\|h\|_{W^{1,1}}>\pi n+1$.

Proof.

Consider the case $n=1$, and take $f_{1}$ constant and nonzero. Suppose for contradiction that $\|h\|_{W^{1,1}}\leq\pi+1$, and write $h(x)$ as $e^{ig(x)}$ for $g\in C^{1}([0,1];\mathbb{R})$, so that $\int_{0}^{1}|g^{\prime}(x)|dx\leq\pi$. Since $g$ is continuous, $g$ attains its minimum $m$ and maximum $M$ on $[0,1]$. By adding a constant to $g$, we may assume $m=0$; then we have $M\leq\pi$.

Since $f_{1}$ is constant, we have $\int_{0}^{1}h(x)dx=0$, so $\int_{0}^{1}\text{Im}(h(x))dx=0$. But $\text{Im}(h(x))$ is continuous in $x$ and nonnegative, so $\text{Im}(h(x))=0$ for all $x$. Hence $h$ is constant at either $1$ or $-1$, but this contradicts $\int_{0}^{1}h(x)dx=0$. Therefore, $\|h\|_{W^{1,1}}>\pi+1$ for $n=1$.

Now allow $n$ arbitrary, and take each $f_{j}$ to be the indicator function on a disjoint interval $I_{j}$. If $\|h\|_{W^{1,1}}\leq\pi n+1$, then $\int_{I_{j}}|g^{\prime}(x)|dx\leq\pi$ for some $j$, and we obtain a contradiction as above. Therefore, $\|h\|_{W^{1,1}}>\pi n+1$. ∎

This $W^{1,1}$-norm bound establishes an upper bound for the coindex of the space of smooth circle-valued functions with norm at most ${1+\pi n}$:

Theorem 6.2**.**

For integer $n\geq 1$ let $Y_{n}$ denote the space of $C^{\infty}$-functions $f\colon[0,1]\to S^{1}$ with $\|f\|_{W^{1,1}}\leq 1+\pi n$. Then

[TABLE]

Proof.

In the proof of Theorem 1.3 we constructed a $\mathbb{Z}/2$-map $\beta_{n}\colon S^{n}\to Y_{n}$, which shows that ${\operatorname{\mathrm{coind}}Y_{n}\geq n}$. Let $f_{1},\dots,f_{n}$ be chosen as in Theorem 6.1. Then the map $\psi\colon Y_{n}\to\mathbb{R}^{2n}$ given by $\psi(h)=(\int_{0}^{1}f_{j}(x)h(x)dx)_{j}$ has no zero and is a $\mathbb{Z}/2$-map. Thus $\psi$ radially projects to a $\mathbb{Z}/2$-map $Y_{n}\to S^{2n-1}$. A $\mathbb{Z}/2$-map $S^{2n}\to Y_{n}$ would compose with $\psi$ to a $\mathbb{Z}/2$-map $S^{2n}\to S^{2n-1}$, contradicting the Borsuk–Ulam theorem. This implies $\operatorname{\mathrm{coind}}Y_{n}\leq 2n-1$. ∎

Problem 6.3**.**

Determine the homotopy type of $Y_{n}$.

Acknowledgements

The first author would like to thank Marius Lemm for bringing [5] to his attention.