Some Open Questions and Recent Results on Computable Banach Spaces

Abstract

We discuss some open questions and results in the geometry of computable Banach spaces.

Dedicated to the memory of Barry Cooper. Research supported by the Marsden Fund of New Zealand, and based on Downey’s Cooper Prize Lecture.

A Appendix-Some Proofs

1.1 A.1 Proof of Theorem 8

To prove Theorem 8, we will first need the following classical lemma. This proof is taken from [Qia21].

Lemma 5

(Mazur). Let X be an infinite dimensional Banach space, $B \subset X$ be a finite-dimensional subspace, and $\epsilon > 0$. Then there is an $x \in X$ with $\left\Vert x\right\Vert = 1$ so that

$$\begin{aligned} \left\Vert y\right\Vert \le (1 + \epsilon )\left\Vert y + \lambda x\right\Vert \end{aligned}$$

for all $y \in B, \lambda \in {\mathbb R}$. In fact, x can be chosen so that this inequality is strict whenever $\left\Vert y\right\Vert , \lambda \ne 0$.

When working with separable Banach spaces, this lemma can be slightly strengthened so that we only have to deal with the dense elements.

Lemma 6

In Lemma 5, further suppose that X is a separable Banach space and that $(q_i)_{i \in {\mathbb N}}$ is dense in the unit sphere of X. We can require the desired $x \in X$ to be some element from $(q_i)$.

Proof

Let X be some separable Banach space, and let $(q_i)_{i \in {\mathbb N}}$ be dense in the unit sphere of X. Let $B \subset X$ be some finite-dimensional subspace and $\epsilon > 0$ be some pre-determined constant. Further denote $x \in X$ to be some element that satisfies the requirements as given by Lemma 5 with $\left\Vert x\right\Vert = 1$. Note that by homogeneity ($y \in B \iff \frac{y}{\lambda } \in B$) it is sufficient to find some $z \in (q_i)$ which satisfies

$$\begin{aligned} \left\Vert y\right\Vert \le (1 + \epsilon )\left\Vert y + z\right\Vert \end{aligned}$$

for all $y \in B$. As $x \notin B$, we have that $\delta _x = \min _{y \in B}\left\Vert x + y\right\Vert$ is both well-defined and positive. Let $z \in X$ be any element where $\left\Vert z\right\Vert = 1$, since $\left\Vert y + x\right\Vert \le \left\Vert y + z\right\Vert + \left\Vert x - z\right\Vert$, we have

$$\begin{aligned} \delta _x = \min _{y \in B}\left\Vert y + x\right\Vert \le \delta _z + \left\Vert x - z\right\Vert \end{aligned}$$

From the inequality above, we can choose some z sufficiently close to x with $\left\Vert z\right\Vert = 1$ so that $\left\Vert x - z\right\Vert \le \epsilon (1 + \epsilon )^{-1}\delta _z$, we show that this choice works

$$\begin{aligned} \left\Vert y\right\Vert \le (1 + \epsilon )\left\Vert y + x\right\Vert = (1 + \epsilon )\left\Vert y + x - z + z\right\Vert \end{aligned}$$

$$\begin{aligned} \le (1 + \epsilon )\left( \left\Vert y + z\right\Vert + \left\Vert x - z\right\Vert \right) \le (1 + \epsilon )\left( \left\Vert y + z\right\Vert + \epsilon (1 + \epsilon )^{-1}\delta _z\right) \end{aligned}$$

And by definition of $\delta _z$, we get that

$$\begin{aligned} (1 + \epsilon )\left( \left\Vert y + z\right\Vert + \epsilon (1 + \epsilon )^{-1}\delta _z\right) \le (1 + \epsilon )\left\Vert y + z\right\Vert + \epsilon \left\Vert y + z\right\Vert \end{aligned}$$

$$\begin{aligned} = (1 + 2\epsilon )\left\Vert y + z\right\Vert \end{aligned}$$

Since Lemma 5 works for all values of $\epsilon$, the conclusion follows. In fact, the exact same argument shows that we can always choose the desired $x \in X$ to be some computable point when X is a computable Banach space.    $\square$

We are now ready to prove Theorem 8.

Proof

(Proof of Theorem 8). In light of Lemmas 6 and 2, we can simply carry out the classical construction. Fix some sequence of computable reals $(\epsilon _i)_{i \in {\mathbb N}}$ such that $\prod _{i = 0}^\infty (1 + \epsilon _i) < \infty$. We will construct a basic sequence $(u_i)_{i \in {\mathbb N}}$ inductively. Having constructed $u_0, \ldots , u_n$, find some x in the effective dense sequence for X such that $\text {bc}(u_0, \ldots , u_n, x) \le \prod _{i = 0}^{n + 1}(1 + \epsilon _i)$. The existence of such an element is guaranteed by Lemma 6. Furthermore, this process is computable as the basis constants are computable.    $\square$

1.2 A.2 Complexity of Computable Basis

Whilst we don’t have space to prove all of the claims in the paper, we will give a brief skectch of how to prove $\varSigma _3^0$ completeness of the index sets of computable Banach spaces with computable bases. In doing so, we also sketch the ideas used by Bosserhof [Bos08] as per [Qia21]. Below, let $\text {BASIS}_C$ denote the index-set of computable Banach spaces that have a computable Schauder basis.

Theorem 11

$\text {BASIS}_C$ is $\varSigma ^0_3$ complete.

We first introduce the construction used in [Bos08]. Let Z denote the Banach space constructed in [Dav73] that lacks the approximation property. It was proven in [Bos08] that this space is computable and also exhibits the local basis property.

Theorem 12

([Bos08]). There exists a computable Banach space without $\text {AP}$ but has $\text {LBS}$.

In particular, this implies that Z can be approximated by a sequence of “nice” subspaces.

Theorem 13

([Bos08]). There is a computable linearly independent sequence $(x_i)_{i \in {\mathbb N}} \subseteq Z$, a computable increasing function $\sigma : {\mathbb N} \rightarrow {\mathbb N}$ and an universal constant C such that $[x_0, \ldots ] = Z$ and

$$\begin{aligned} (\forall n \in {\mathbb N})(\text {bc}([x_0, \ldots , x_{\sigma (n)}]) < C) \end{aligned}$$

We first need the following definitions.

Definition 7

([Bos08]). For any $n \in {\mathbb N}$, $Z_n$ is defined as:

$$\begin{aligned} Z_n = [x_0, \ldots , x_{\sigma (n)}] \end{aligned}$$

where $(x_i)_{i \in {\mathbb N}}$ is given by Theorem 13. For any $\tau :{\mathbb N} \rightarrow {\mathbb N}$, the Banach space $Y_\tau$ is defined as:

$$\begin{aligned} Y_\tau = \left( \oplus _{i}Z_{\tau (i)}\right) _{c_{0}} \end{aligned}$$

which is the sequence space where norms of elements within each sequence tends to 0, and the norm on the sequence is the supremum norm on the elements.

An important feature of this space is that it has a basis. Intuitively, as the columns have universally bounded basis constants, we can simply “join up” the bases of the columns in the larger space, and the resulting sequence will be a basis.

Lemma 7

([Bos08]). The space $Y_{\tau }$ as defined in Definition 7 has a basis for any $\tau : {\mathbb N} \rightarrow {\mathbb N}$.

The key idea is that $Y_\tau$ is a Banach space with basis, however each of its components can be made arbitrarily “large” such that no computable sequence can span it. For the sake of simplicity, also denote $Y = \left( \oplus _i Z\right) _{c_0}$. The following lemma is crucial.

Lemma 8

([Bos08]). For any basic sequence $(y_i)_{i \in {\mathbb N}} \in Y^{\mathbb N}$ and $n \in {\mathbb N}$, we have

$$\begin{aligned} \text {emb}^n(Z) \nsubseteq [y_0, y_1, \ldots ] \end{aligned}$$

Where $\text {emb}^n : Z \rightarrow Y$ is the map defined by

$$\begin{aligned} \text {emb}^n(x) = (0, \ldots ,0, x, 0, \ldots ) \in Y \end{aligned}$$

mapping $x \in Z$ to n-th position of a sequence that is otherwise entirely zero.

There is also a natural computability structure on the space $Y_\tau$ for certain classes of $\tau$.

Definition 8

A function $\tau : {\mathbb N} \rightarrow {\mathbb N}$ is lower semicomputable if there is a c.e set $A \subseteq {\mathbb N}$ such that

$$\begin{aligned} \tau (n) = \sup \{k \in {\mathbb N} : \langle n, k \rangle \in A\} \end{aligned}$$

for all $n \in {\mathbb N}$.

Lemma 9

([Bos08]). For any $\tau : {\mathbb N} \rightarrow {\mathbb N}$ that is lower semicomputable, the constructed space $Y_\tau$ equipped with the dense set $\{\text {emb}^j(x_i)\}_{i \le \sigma (\tau (j)), j \in {\mathbb N}}$ is a computable Banach space.

Finally, to construct a computable Banach space without any computable basis, it is sufficient to construct some lower semicomputable $\tau$ such that $Y_\tau$ does not contain any computable basis. Furthmore, by Lemma 8 and Theorem 13, we can construct $\tau$ by directly diagonalising against all computable basic sequences. The following is due to [Bos08], although presented in a slightly different fashion.

Lemma 10

([Bos08]). There is a lower semicomputable function $\psi : {\mathbb N}^3 \rightarrow {\mathbb N}$ such that for all $n, k, i \in {\mathbb N}$, if $\phi _n$ computes a basic sequence $(y_i)_{i \in {\mathbb N}} \in Y^{\mathbb N}$ with basis constant smaller than k, we have

$$\begin{aligned} \text {emb}^i(Z_{\psi (n, k, i)}) \nsubseteq [y_0, \ldots ] \end{aligned}$$

Corollary 1

([Bos08]). There exists a computable Banach space without computable basis.

Proof

By Lemmas 9 and 10, define $\tau : {\mathbb N} \rightarrow {\mathbb N}$ by

$$\begin{aligned} \tau (\langle n, k \rangle ) = \psi (n, k, \langle n, k \rangle ) \end{aligned}$$

The resulting space $Y_\tau$ is a computable Banach space where $\tau (\langle n, k \rangle )$ is large enough so that $\text {emb}^{\langle n, k \rangle }(Z_{\tau (\langle n, k \rangle )})$ is not spanned by $\phi _n$ (if it is a basic sequence with basis constant smaller than k). This implies that the space $Y_\tau$ cannot be spanned by any computable basic sequence¹, and therefore lacks basis.    $\square$

It is worth noting that although the space constructed in Corollary 1 has no computable basis, it is unclear how uncomputable the bases are.

Question 10

Let $Y_\tau$ be the space used in the proof of Corollary 1 that was constructed by [Bos08]. What are the corresponding Turing degrees for the bases in this space?

Using Lemma 1, it is easy to see that having a computable basis is $\varSigma _3^0$, and hence we need following lemma to show completeness.

Lemma 11

Recall the construction carried out in Lemma 7. If $\tau$ is a computable function, then $Y_\tau$ contains a computable basis.

Proof

As the basis constant of $Z_{\tau (i)}$ is uniformly bounded by some constant C, there is some basis $(a_{i, j})_{j \le \sigma (\tau (i))}$ with basis constant smaller than C for each $Z_{\tau (i)}$. It was proved in [Bos08] that the natural embedding of these bases into $Y_{\tau }$ (i.e. $\{\text {emb}^i(a_{i, j}) \vert i \in {\mathbb N}, j \le \sigma (\tau (i))\}$) forms a basis for $Y_{\tau }$. We will show that this is actually computable when $\tau$ is computable. If $\tau$ is computable, the sequence

$$\begin{aligned} x_0, x_1, \ldots , x_{\sigma (\tau (i))} \end{aligned}$$

will be computable as well since $(x_i)_{i \in {\mathbb N}}$ and $\sigma$ are both computable. Therefore, the rational span of the sequence will be computable as well. By continuity, we can therefore effectively find some basis that lies in the rational span of $(x_i)_{i \le \sigma (\tau (i))}$ with basis constant smaller than C. As this procedure is uniform, it gives a computable basis in $Y_\tau$.    $\square$

We are now ready to prove Theorem 11.

Proof

(Proof of Theorem 11). $\text {BASIS}_C \in \varSigma _3$ essentially follows from Lemma 1, it remains to show that $\text {BASIS}_C$ is $\varSigma _3$ hard. It is a well known fact that for any set $A \in \varSigma _3$, there is a computable function $g : {\mathbb N}^2 \rightarrow {\mathbb N}$ such that

$$\begin{aligned} x \in A \iff (\exists y)(W_{g(x, y)} \text { is infinite }) \end{aligned}$$

For all $x \in {\mathbb N}$, we construct a lower semicomputable function $h : {\mathbb N} \rightarrow {\mathbb N}$ in stages. Let $\{\psi _s\}$ be some computable enumeration of the function $\psi$ constructed in Lemma 10. We also define the function $C: {\mathbb N} \rightarrow {\mathbb N}$, initially $C_0(n) = n$ for all $n \in {\mathbb N}$. C(n) indicates the computable sequence that is diagonalised against at n. Initialise the construction by setting $h_s = 0$. At stage s, the following is carried out for each $n \le s$.

• If $C(n) = -1$, do nothing. Otherwise:

• Enumerate $W_{g(x, C(n)), s}$. If a new element is enumerated, set C(k) to $C(k - 1)$ for all $k > n + |W_{g(x, C(n)), s}|$ and $C(n + |W_{g(x, C(n)), s}|)$ to $-1$.

• View C(n) as a pair $\langle a, b\rangle$ and set $h_s(n)$ to $\max (h_{s - 1}(n), \psi _{s}(a, b, n))$.

Finally we define h as $h = \lim _{s \rightarrow \infty } h_s$. This is the end of the construction, we now verify its validity.

Lemma 12

The function h constructed is indeed a lower semicomputable function.

Proof

The constructed sequence $\{h_s\}$ is clearly a computable enumeration of h. So it remains to verify that $\{h_s\}$ converges. For any $n \in {\mathbb N}$, we have $C(n) \le n$. Therefore $h_s(n) \le \max _{\langle a, b\rangle \le n}\psi (a, b, n)$ for all s, and since $(h_s(n))_s$ is monotone, this implies convergence.    $\square$

We now show that the constructed h has the desired properties.

Lemma 13

In addition to h being lower semicomputable, it also exihibit the following properties

• If $x \in A$, h is computable (although this might be non-uniform).

• If $x \notin A$, $Y_h$ contains no computable basis.

Proof

Suppose $x \in A$, thus there is some y such that $W_{g(x, C(y))}$ is infinite. By the construction, this means that

$$\begin{aligned} -1 = C(y + 1) = C(y + 2) = C(y + 3) = \ldots \end{aligned}$$

Therefore, to compute h(k) for any $k > y$, we just have to run the computable construction for finitely many steps until $C(k) = -1$, in which case the current value of h(k) will be its final value. And since there are only finite many values h(k) for $k \le y$, this can be computed non-uniformly. Hence, h is a computable function.

Now suppose $x \notin A$, in which case $W_{g(x, y)}$ is finite for all $y \in {\mathbb N}$. We will show that for all $\langle a, b \rangle \in {\mathbb N}$, there is some $n \in {\mathbb N}$ where $C(n) = \langle a, b \rangle$, implying that $h(n) \ge \psi (a, b, n)$ and therefore $Y_h$ cannot contain any computable basis. At each stage s of the construction, there will be some index $i_s$ where $C_s(i_s) = \langle a, b \rangle$. So it suffices to show that $(i_s)_s$ eventually stabilises. But by the construction, $i_s$ can only increase when some new element has been enumerated in $W_{g(x, C(k))}$ for some $C(k) < \langle a, b \rangle$. And since $\{k : C(k) < \langle a, b \rangle \}$ is finite, and each set of the form $W_{g(x, y)}$ is finite as well, $i_s$ can only increase for a finite number of steps until it eventually converges, and the proof is complete.    $\square$

Therefore, as the construction of h is uniform in x, we have established a reduction from an arbitrary $\varSigma _3$ set to $\text {BASIS}_C$, proving that $\text {BASIS}_C$ is indeed $\varSigma _3$ hard.    $\square$