6 Spaces of functions

Nontrivialities arise when we generalize from finite-dimensional to infinite-dimensional vector spaces. The theory of normed vector spaces provide the foundation of dual spaces, weak topology, and convex theory;
Banach spaces include \(L^p\) spaces of integrable functions we will later be working with, so we take time
to develop some of their properties. This section, works with \(\mathbb F\) which can be either real or complex.

  1. Normed vector spaces, Hahn-Banach: existence of a basis, normed space of linear operators, characterization of continuous linear functionals and operators, graphs, the Hahn-Banach theorem, variational formula of the norm, weak topology and convergence.
  2. Banach theorems: Baire’s (category) theorem, open mapping theorem, inverse mapping theorem, principle of uniform boundedness, bounded inverse theorem, closed graph theorem.

Key takeaways:

  1. A basis in infinite-dimensional vector spaces spans by finite linear combinations, and the dual space (definition 6.4) consists of the set of bounded or, equivalently, continuous linear functionals under the operator norm (definition 6.2); \(\mathcal B(V, W)\) is Banach if \(W\) is Banach 6.5.
  2. The existence of a basis for infinite-dimensional vector space is is nontrivial and nonconstructive (theorem 6.1).
  3. Continuity of linear maps cannot be taken for granted in infinite-dimensional vector spaces. A linear map is continuous iff bounded (theorem 6.2). Every infinite-dimensional normed vector space has a discontinuous linear functional (proposition 6.1).
  4. Universal property of the weak topology 6.5.
  5. The Hahn-Banach theorem 6.3 can be viewed as an existence result of a minimization problem: what is the minimum norm of the extension of a bounded linear functional to the whole space?
  6. Baire’s theorem 6.4: every closed cover of a Banach space contains an element which has nonempty interior.
  7. Baire’s theorem leads to the principle of uniform boundedness (corollary 6.1): every family of pointwise bounded linear operators from a Banach space to a normed vector space is uniformly bounded. Completeness (for continuous / bounded linear maps) plays a role similar to compactness (for continuous maps).
  8. The closed graph theorem 6.6 is nontrivial since demonstrating continuity needs to show that (1) the limit exists, and (2) the limits are equal; but CGT reduces this to checking the closure of the graph.

Normed spaces, Hahn-Banach

Definition 6.1 ((Hamel) basis) A subset \(B\subset V\) of a vector space \(V\) is a (Hamel) basis if it is linearly independent and every \(v\in V\) can be written as a finite linear combination of elements in \(V\).

Zorn’s lemma is in fact equivalent to the axiom of choice; it is responsible for the nonconstructive existence result for the basis of infinite-dimensional vector spaces. We will take it for granted. Fun quote from Wikipedia:

If you are building a mathematical object in stages and find that (i) you have not finished even after infinitely many stages, and (ii) there seems to be nothing to stop you continuing to build, then Zorn’s lemma may well be able to help you.
— William Timothy Gowers

Lemma 6.1 (Zorn's lemma) Given a partially ordered set \(P\) in which every chain (subset with pairwise comparable elements) has an upper bound in \(P\), then \(P\) contains at least one maximal element.

Theorem 6.1 (existence of basis) Every infinite-dimensional vector space \(V\) has a basis.

Proof Consider the space \(A\subset 2^V\) of all linearly independent subsets of \(V\), partially-ordered by set inclusion. Every chain \(B_1\subset B_2\cdots \in A\) is upper-bounded \(B_j\nearrow B\) (which is still linearly independent!), then the maximal element is precisely a basis, else the vector not in the finite linear span could be added to extend the basis.

Definition 6.2 ((normed) space of (bounded) linear operators) Given two vector spaces \(U, V\), the set of linear operators from \(U\) to \(V\) is also a vector space, which we denote \(\mathcal L(U, V)\). If \(U, V\) are normed, then \(\mathcal L(U, V)\) is equipped with the operator norm \[ \|T\| = \sup_{\|f\in U\|\leq 1} \|Tf\| \] The set of bounded linear operators is denoted \(\mathcal B(U, V)\).

Definition 6.3 (bounded linear functional) A linear functional on a vector space \(V\) is a linear map of signature \(T:V\to \mathbb F\). It is bounded if \(\|T\|<\infty\) using the operator norm.

Theorem 6.2 (characterization of continuous linear maps) A linear map \(T:U\to V\) is continuous iff it is bounded.

Proof Given \(T\) bounded, for every \(f_j\to f\in U\), we have continuity from \[ \|Tf_k - Tf\| \leq \|T\| \, \|f_k - f\| \] Conversely, assume \(T\) unbounded, then there exists a sequence \(f_j\in V\) with \(\|f_j\|\leq 1\) and \(\|T f_{k\to \infty}\| \to \infty\). Then \[ \lim_{k\to \infty} \dfrac{f_k}{\|T f_k\|} = 0, \quad T\, \dfrac{f_k}{\|T f_k\|} \not \to 0 \]

Theorem 6.2 (characterization of continuous linear functionals) A linear functional \(\varphi:U\to \mathbb F\) which is not identally \(0\) is continuous iff it is bounded iff:

  1. \(\ker \varphi\) is a closed subspace of \(V\).
  2. \(\overline{\ker \varphi} \neq V\).
Proof Continuity implies (1) since \(\ker \varphi = \varphi^{-1}(\{0\})\) and continuous maps send closed singletons to closed sets. Conversely, \(\varphi\) unbounded implies \(\exists f_j\in V\) such that \(\|f_j\|\leq 1\) and \(|\varphi\, f_j|>j\) for each \(j\), note that \[ \forall j, \quad \dfrac{f_1}{\varphi\, f_1} - \dfrac{f_k}{\varphi\, f_k} \in \ker \varphi \] On the other hand, we have \[ \lim_{k\to \infty} \dfrac{f_1}{\varphi\, f_1} - \dfrac{f_k}{\varphi\, f_k} = \dfrac{f_1}{\varphi\, f_1} \notin \ker \varphi \] thus \(\overline{\ker \varphi}\neq \ker \varphi\) if \(\varphi\) unbounded. Finally prove equivalence with (2), we have (1) implies (2) since \(V\) is closed. Conversely, if \(\overline{\ker \varphi} \supsetneq \ker \varphi\), we will show that \(\overline{\ker \varphi} = V\). To do so, let \(f\in \overline{\ker \varphi} - \ker \varphi\), then \(\forall g\in V\): \[ g = \left(g - \dfrac{\varphi\, g}{\varphi\, f} f\right) + \dfrac{\varphi\, g}{\varphi\, f} f \] Recall that \(\overline{\ker \varphi}\) is a vector space. The term in the big paranthesis is in \(\ker \varphi\), and the trailing term is in \(\overline{\ker \varphi}\), thus \(g\in \overline{\ker \varphi}\) for every \(g\in V\).

Proposition 6.1 Every infinite-dimensional normed vector space has a discontinuous linear functional.

Proof: Consider a basis \(\{e_j\in V\}_{j\in \Gamma \supset \mathbb N}\). The linear functional which maps \(e_j\) to \(j\) is discontinuous.

Definition 6.4 (dual space) Given a normed vector space \(V\), its dual space \(V^*\) consists of the vector space of bounded linear operators under the operator norm (definition 6.2).

Definition 6.5 (weak topology, convergence) The weak topology on a normed vector space \(V\) is the coarsest topology in which all continuous linear functionals on the space remain continuous. A sequence \(f_j\) in a normed vector space \(V\) converges weakly to \(f\in V\) if \(\alpha\, x_n\to \alpha\, x\) for all \(\alpha \in V^*\).

We next proceed to prove the Hahn-Banach theorem.

Lemma 6.2 (extension lemma) Given a real normed vector space \(V\), a subspace \(U\subset V\), and a bounded linear functional \(\psi:U\to \mathbb R\). For every \(h\in V-U\), there exists a linear functional \(\varphi:U\oplus \mathbb Rh\to \mathbb R\) such that \(\varphi_{|U} = \psi\) and \(\|\varphi\| = \|\psi\|\).

Proof idea: write an inequality for the value of \(\psi(h)\). Prove that a valid value exists by triangle inequality and boundedness of \(\psi\).
Proof We only need to define \(\varphi(h) = c\in \mathbb R\) and figure out a value for \(c\) such that \(\varphi(f+\alpha h) = \psi(f) + \alpha c\) satisfies the norm condition. On one hand, \(\varphi_{|U} = \psi\implies \|\varphi\| \geq \|\psi\|\). On the other hand, \(\|\varphi\| \leq \|\psi\|\) requires \(c\) to satisfy \[ |\psi(f) + \alpha c| = \varphi(f) \leq \|\varphi\| \|f + \alpha h\| \leq \|\psi\| \|f + \alpha h\|, \quad \forall f\in U, \alpha \in \mathbb R. \] Without loss of generality we can set \(\alpha=1\) since \(f\) is rescalable, requiring \[ |\psi(f) + c| \leq \|\psi\| \|f + h\|, \quad \forall f\in U, \alpha \in \mathbb R. \] Expanding the absolute value \[ -\|\psi\|\|f+h\| - \psi(f) \leq c\leq \|\psi\|\|f+h\| - \psi(f), \quad \forall f\in U \] The equation will be implied by (in fact, about as strong as, since \(c\) must be uniform \[ \sup_f \left[ -\|\psi\|\|f+h\| - \psi(f) \right] \leq \inf_g \left[ \|\psi\|\|g+h\| - \psi(g) \right] \] To prove this, first note the triangle inequality \(-\|f+h\| \leq \|g+h\| - \|g-f\|\), then \[\begin{align} -\|\psi\|\|f+h\| - \psi(f) &\leq \|\psi\|(\|g+h\| - \|g-f\|) - \psi(f) \\ &\leq \|\psi\|(\|g+h\| - \|g-f\|) + \psi(g-f) - \psi(g)\\ &=\|\psi\|\|g+h\| - \psi(g) + \left[\psi(g-f) - \|\psi\|\|g-f\|\right]_{\leq 0} \\ &\leq \|\psi\| \|g+h\| - \psi(g) \end{align}\]

Definition 6.6 (graph of a function) Given a function \(T:V\to W\), the graph \(\mathrm{graph}(T)\) of \(T\) IS \[ \mathrm{graph}(T) = \{(f, T(f):f\in V\}\subset V\times W \]

Properties of functions can be read from the graph:

  1. \(T\) is a linear map iff \(\mathrm{graph}(T)\) is a subspace of \(V\times W\).
  2. \(T:V\to W\) is an extension of \(S:U\to W\) iff \(\mathrm{graph}(S)\subset \mathrm{graph}(T)\).
  3. Given a linear map \(T:V\to W\), then \(\|T\|\leq c\) iff \(\|g\| \leq c\|f\|\) for all \((f, g) \in \mathrm{graph}(T)\).

Theorem 6.3 (Hahn-Banach theorem) Given a normed vector space \(V\) with subspace \(U\) a bounded linear functional \(\psi\in U^*\), then there exists \(\tilde \psi\in V^*\) with \(\|\tilde \psi\|_V = \|\psi\|_U\) and \(\tilde \psi_{|U} = \psi\).

Proof

We prove the theorem for the real case. Let \(\mathcal A\) be the collection of subsets \(E\subset V\times \mathbb R\) satisfying:

  1. \(E=\mathrm{graph}(\varphi)\) for some linear functional \(\varphi\) on some subspace of \(V\).
  2. \(\mathrm{graph}(\psi)\subset E\iff \varphi\) is an extension of \(\psi\).
  3. \(\psi\) generating \(E\) has norm at most \(\|\psi\|\).
Then \(\mathcal A\) satisfies the hypothesis of Zorn’s lemma 6.1: for every chain of subsets, their union is also in \(\mathcal A\). The extension lemma 6.2 guarantees that the maximal element is the graph of a linear functional defined on all of \(V\).

The next proposition anticipates the variational formula of the \(p\)-norm (proposition 7.6.

Proposition 6.2 (variational formula of the norm) Given a normed vector space \(V\) and \(0\neq f\in V\), \[ \|f\| = \sup_{\|\varphi\|_{V^*} = 1} |\varphi(f)| \] In other words, there exists \(\varphi\in V'\) with \(\|\varphi\|=1\) such that \(\|f\| = \varphi(f)\).

Proof Consider the \(1\)-dimensional subspace spanned by \(f\), and define \(\psi(\alpha f) = \alpha \|f\|\), which on this subspace satisfies \(\|\psi\|=1\) and \(\psi(f)=\|f\|\). Apply the Hahn-Banach extension.

Banach theorems

Definition 6.7 (completeness, Banach space) A metric space is complete if every Cauchy sequence in the space converges. A Banach space is a complete normed metric space.

Proposition 6.3 (connection between complete and closed metric spaces) A complete subset of a metric space is closed, and a closed subset of a complete metric space is complete.

Proof Given a complete subset \(V\) of a metric space, every convergent sequence is also a convergent Cauchy sequence whose limit stays inside \(V\). Given a closed subset, the convergent limit of every Cauchy sequence must say within \(V\) by closure property, so \(V\) is complete.

Recall that to argue for Cauchy convergence we can drop to subsequences (proposition 1.7).

Proposition 6.4 (characterization of Banach space) A normed vector space \(V\) is complete (i.e. a Banach space) iff \(\sum_{k=1}^\infty g_k\) converges for every sequence \(g_j\in V\) satisfying \(\sum_{k=1}^\infty \|g_k\|<\infty\).

Proof Given \(V\) complete, the partial sum of \(g\) is a Cauchy sequence thus converges. Conversely, given a Cauchy sequence, drop to a subsequence \(f_j\) and set \(f_0=0\) so that
\[ \sum_{k=1}^\infty \|f_k - f_{k-1}\| < \infty \] then by assumption \(\sum_{k=1}^\infty (f_k - f_{k-1})=\lim_{n\to \infty} f_n\) exists.

Proposition 6.5 (Banach dual space) \(\mathcal B(V, W)\) is a Banach space (under the operator norm) if \(W\) is a Banach space. In particular, the dual space of a Banach space is a Banach space.

Proof Given a Cauchy sequence \(T_j\in \mathcal B(V, W)\), for every function \(f\in V\), the sequence \(T_j f\in W\) is a Cauchy sequence; define \(T\, f = T_{j\to \infty}\, f\). It is linear and bounded, so it remains to show that \(\lim_{k\to \infty} \|T_k - T\| = 0\). This follows from \[ \|(T_j - T)\, f\| = \lim_{k\to \infty} \|T_j f - T_k f\| \]

Theorem 6.4 (Baire's theorem) Given a Banach space:

  1. the intersection of countably many dense open sets is nonempty.
  2. if \(X\) is the countable union of closed subsets \(G_1, \cdots\), then some \(G_j\) contains an open ball.
Proof idea: for (1), by denseness construct a Cauchy sequence; argue (2) by contradiction.
Proof

We prove first prove (1): given a countable sequence of dense open sets \(U_j\), we want to show that \(\bigcap_{k=1}^\infty U_k \neq \infty\).

  1. Choose \(f_1\in U_1\) and \(r_1\in (0, 1)\) with \(\bar B_{r_1}(f_1)\subset U_1\) (denoting closure).
  2. For each \(j\), choose \(0<r_j<1/j\), \(\bar B_{r_j}(f_j) \subset U_j\) and \(\bar B_{r_j}(f_j)\subset \bar B_{r_{j-1}}(f_{j-1})\). We can do this because, fixing \(r_{j-1}, f_{j-1}\), the ball \(B_{r_{j-1}}(f_{j-1})\cap U_j\) is a nontrivial (by \(U_j\) dense) open set, from which we can choose \(f_j\) and the radius such that \(\bar B_{r_j}(f_j)\subset U_{j-1}\).
  3. Then \(f_j\) is naturally a Cauchy sequence because \(|f_{j>N} - f_{k>n}| < 1/N\).
By completeness, \(f_j\to f\in X\), completing the proof of (1). To prove (2) suppose for contradiction that \(X=\bigcup G_j\), each of which has empty interior, then \(U_j = G_j^c\) is open and dense; by (1) there exists \(f\in \bigcap (G_j^c) = (\bigcup G_j)^c\), contradicting \(\bigcup G_j=X\).

Corollary 6.1 (principle of uniform boundedness) Given \(\mathcal A\subset \mathcal B(V, W)\) with \(V\) Banach and \(W\) normed satisfying pointwise boundedness (below), then \(\sup_{T\in \mathcal A} \|T\|<\infty\). \[ \sup_{T\in \mathcal A} \|Tf\| < \infty, \quad \text{ for every } f\in V \]

Proof Given our hypothesis, \(V_n\nearrow V\) with \[ V_n = \{f\in V: \|T\, f\|\leq n, \quad \forall T\in \mathcal A\} \] Moreover, \(V_n\) is closed since each \(T\) is continuous. By Baire’s theorem there exists \(n, h\in V, r>0\) with \(B_r(h)\subset V_n\). For any \(g\in V\) with \(\|g\|<1\), we have \(rg+h\in B_r(h)\subset V_n\implies \|T(rg+h)\|\leq n\). \[ \|Tg\| = \left\| \dfrac{T(rg + h)}{r} - \dfrac{Th}{r} \right\| \leq \dfrac{\|T(rg+h)\|}{r} + \dfrac{\|Th\|}{r} \leq \dfrac{n+\|Th\|}{r}, \quad \forall g\in V \] Taking \(\sup\) over all \(T\) completes the proof.

Theorem 6.5 (open mapping theorem) Given Banach spaces \(V, W\), a surjective bounded linear map \(T:V\to W\) is open (i.e.  sends open sets to open sets); note that this does not require \(T\) injective, so \(T\) is not guaranteed to be a homeomorphism.

Proof idea: Show that \(0\in \mathrm{int}\, T(B(0, 1))\). Do this by constructing, for each \(g\in B(0, 1)\subset W\), a bounded \(f\in V\) (which is the limit of a Cauchy sequence) s.t. \(g=T\, f\). Next use linearity to extend to all open balls.
Proof Given open \(U\subset V\), we need to show that \(\forall f\in U\) such that \(B(f, a)\subset U\), there exists \(b<a\) such that, by linearity \[ T\, B(f, b) = T\, f + bT(B) \subset T\, U \] It thus suffices to show that \(\exists r: B(0, r)\subset T(B)\) (i.e. \(0\) is an interior point of \(T(B)\)). Surjectivity and linearity of \(T\) implies that \[ W = \bigcup_{k=1}^\infty T(kB) = \bigcup_{k=1}^\infty k T\, B \] Apply Baire’s theorem to \(k \overline{T\, B}\) and use linearity (which respects closure) shows that \(\overline{T\, B}\) has a nonempty interior; this is where we use the Banach condition on \(W\).
We need to be especially careful when working with closure and interiors of infinite-dimensional spaces. Given \(T\, g_{\in B}\in \mathrm{int}\, {\overline{T\, B}}\), \[ 0 \in \mathrm{int}\, \overline{T\, (B - g)} \subset \mathrm{int} \, 2\, \overline{T\, B} \implies \overline{B(0, \exists r)} \subset \overline{T\, B} \] We proceed to show that this implies \(0\in \mathrm{int}(T\, B)\); again, this is not a trivial argument. By definition of closure, \(\forall h\in W\) with \(\|h\|\leq r\) we have \[ \forall \epsilon>0, \exists f\in B: \|h - T\, f\| < \epsilon \] Substitute \(h\mapsto rh/\|h\|\) yields, \(\forall h\in W-\{0\}, \epsilon>0\) \[ \exists f\in \dfrac{\|h\|}{r}B: \|h - T\, f\|<\epsilon \tag{6.1} \] Now pick arbitrary \(g\in B\subset W\). Apply equation (6.1) with \((h, \epsilon)\mapsto (g, 1/2)\) to obtain \[ \exists f_1\in B/r: \|g - T\, f_1\|<1/2 \] Apply equation (6.1) again with \(h\mapsto g - T\, f_1\in B/2, \epsilon = 1/4\) to obtain \[ \exists f_2\in B/2r: \|g - T\, f_1 - T\, f_2\| < 1/4 \] Repeat this construction inductively to obtain \(f_1, f_2, \cdots \in V\) and let \[ f = \sum_{k=1}^\infty f_k, \quad \|f\|\leq \sum_{k=1}^\infty \|f_k\| \leq \sum_{k=1}^\infty \dfrac 1 {2^{k-1}r} = \dfrac 2 r \] Here we used the Banach condition on \(V\). By construction, \(\|g-T\, f\| < 2^{-n}\) for every \(n\), then by continuity of \(T\) we have \(g=T\, f\). Thus \(T\, B(0, 2 / r)\subset B(0, 1)\implies B(0, r / 2)\subset T\, B\).

Corollary 6.2 (bounded inverse theorem) Given \(T\in \mathcal B(V, W)\) injective (so that \(T^{-1}\) exists) and surjective, \(T^{-1}\) is a bounded linear map \(W\to V\).

Given two Banach spaces \(V, W\) their vector space product \(V\oplus W\) is a Banach space when equipped with the norm \(\|(f, g)\| = \max(\|f\|, \|g\|)\). This induces the product topology: a product sequence converges iff each component converges.

Theorem 6.6 (closed graph theorem) Given Banach spaces \(V, W\) and \(T:V\to W\), then \(T\) is a bounded linear map iff \(\mathrm{graph}(T)\subset V\times W\) is closed.

Proof Given \(T\) continuous, given \((f_j, T\, f_j)\in \mathrm{graph}(V\times W)\) converging to \((f, g)\in V\times W\), by the product topology this implies \((f, T\, f)\in \mathrm{graph}(T)\), which establishes closure of the grpah. Conversely, given the graph a closed subspace, it is a Banach space (proposition 6.3) with the inherited norm. Consider the bijective projection map \(\pi:\mathrm{graph}(T)\to V\) with \(\pi(f, T\, f) = f\), then \(\|\pi(f, T\, f)\|\leq \|(f, T\, f)\|\) thus is bounded. Apply the bounded inverse theorem to obtain that \(\pi^{-1}(f) = (f, T\, f)\) is bounded, implying \[ \|T\, f\| \leq \|(f, T\, f)\| = \|\pi^{-1}(f)\| \leq \|\pi^{-1}\|\|f\| \] Thus \(\|T\| \leq \|\pi\|^{-1}\), establishing the boundedness of \(T\).

Kolmorogorov-Arnold theorem

Let \(\mathbb I\) denote the compact interval \([0, 1]\) and \(C(\mathbb I)\) the Banach space of continuous functions \(\mathbb I\to \mathbb R\) equipped with the supremum norm \[ \|f\| = \sup_{x\in [0, 1]} |f(x)| \] Completeness follows from the completeness of \(\mathbb R\) and uniform continuity on a compact interval. We also equip \(C(I)^n\) with the following norm \[ \|f\in C(I)^n\| = \max_{j\in [n]} \|f_j\| \tag{6.2} \]

Lemma 6.3 (rational-distinguishing coefficients) For each \(n\in \mathbb N\) there exist real numbers \(\lambda_1, \cdots, \lambda_n\in \mathbb R\) such that for all \(x_1, \cdots, x_n, y_1, \cdots, y_n\in \mathbb Q\), \[ \sum_j \lambda_j x_j = \sum_j \lambda_j y_j \implies x_{\forall j} = y_j \]

Proof: Choose \(\lambda_j\) to be linearly independent over \(\mathbb Q\).

To reduce clutter, given a choice of coefficients \(\lambda\in \mathbb R^n\) and embeddings \(\phi\in C(\mathbb I)^n\), define the separately-continuous embedding functions \(\Phi_j: \mathbb I^n\to \mathbb R\) by \[ \Phi_j(x_1, \cdots, x_n) = \sum_{j=1}^n \lambda_j \phi_j(x_j) \tag{6.3} \]

Lemma 6.4 (approximate embeddings are dense open subset) Fixing \(\lambda\) satisfying lemma 6.3 and \(f\in C(\mathbb I^2)\) with \(\|f\|=1\), let \(U_f\subset C(\mathbb I)^5\) consist of tuples \((\phi_1, \cdots, \phi_5)\in C(I)^5\) satisfying

  • \(\exists g\in C(\mathbb R) \text{ such that } \|g\|\leq \dfrac 1 7\), and
  • \(\displaystyle \forall x, y\in I^2: \left| f(x, y) - \sum_{j=1}^5 (g\circ \Phi_j)(x, y) \right| < \dfrac 7 8\), with \(\Phi_j\) defined as in equation (6.3).

Then \(\forall f\in C(\mathbb I^2)\), the set \(U_f\) of embeddings which allow \(g\)-approximations of \(f\) is a dense open subset of \(C(\mathbb I)^5\).

Proof idea: Openness is easy to argue, to open density, fix \(\phi^0\in C(\mathbb I)^n\) and

  1. Create a atrous redundant tiling \(R_1, \cdots, R_{2n+1}\) of \(\mathbb I\), each covering \(1/(2n+1)\) of \(\mathbb I\) with a fine enough resolution.
  2. Choose \(\epsilon\)-approximation \(\phi_j\sim \phi_j^0\) that is constant rational on each tile \(R_j\), distinct on the disconnected pieces of \(R_j\), and distinct from any \(\phi_{k\neq j}\) everywhere on their designated intervals \(R_j, R_k\).

Fixing \(f, \lambda\), the set \(U_f\) is apparantly open since the same \(g\) which works for \((\phi_j)\) will also work for a neighborhood of \((\phi_j)\) by continuity. To prove \(U_f\) dense, we need to demonstrate:

  • For every \(\phi^0_1, \cdots, \phi^0_5\in C(\mathbb I)^5\) and \(\epsilon>0\), there exists \(\phi_1, \cdots, \phi_5\in U_f\) such that \(\|\phi - \phi^0\|<\epsilon\) as in equation (6.2).

Fixing a positive integer \(N\) (we’ll specify it later), for each \(j\in \{1, \cdots, 5\}\) we can consider the set of all intervals \[ R_j = \bigcup_{s=0}^{N-1} \left[\dfrac s N, \dfrac{s+1}{N}\right] 1_{s\neq j-1\mod 5}, \quad j\in \{1, \cdots, 5\} \] Let \(R_{jk}\) denote the almost-disjoint subintervals in the union, we call each \(R_{jk}\) the \(k\)-th red interval of rank \(j\). The following properties are satisfied:

  • Each \(R_j\) covers at most \(4/5\), and each \(R_{jk}\) covers \(4/N\) of \(\mathbb I\).
  • Almost every \(x\in \mathbb I\) is contained in at most \(3\) of the subintervals \(R_{jk}\) for distinct \(j\)’s.

Fixing \(\phi^0\in C(\mathbb I)^5\) and \(\epsilon>0\), there clearly exists \(N\) large enough such that there exists \(\phi \in C(\mathbb I)^5\) satisfying:

  1. \(\|\phi - \phi^0\| < \epsilon\).
  2. each \(\phi_j\) constant and equal to a rational number on each \(R_{jk}\).
    • Follows from the arbitrariness of \(N\), uniform continuity of \(\phi_j^0\), and the density of \(\mathbb Q\subset \mathbb R\).
  3. \(\phi_j(R_{jk})\neq \phi_j(R_{jl})\) for \(k\neq l\).
  4. \(\phi_j(x) \neq \phi_k(z)\) for all \(x\in R_j, z\in R_k\).

In spirit, each \(\phi_j\) is responsible for approximating \(\phi_j^0\) on \(R_j\) and is unconstrained elsewhere. Next define the red rectangles \(r_{jk}\) of rank \(j\) to be the disconnected components of \(R_j\times R_j\subset \mathbb I^2\).

  1. The distance between any two points in one \(R_{jk}\) is \(\leq \sqrt{2\cdot (4/N)^2} = \sqrt{32}/N\).
  2. By construction of \(\lambda\) satisfying lemma 6.3, the contant value \[ \Phi_j(R_{jr}) = \Phi_k(R_{ks}) \implies (j, r) = (k, s) \]