# Pre-Schauder Bases in Topological Vector Spaces

^{*}

^{†}

## Abstract

**:**

## 1. Introduction

- expanding if $X=\mathrm{span}\{{e}_{i}:i\in I\}$;
- fundamental if $X=\overline{\mathrm{span}}\{{e}_{i}:i\in I\}$; and
- total if ${X}^{*}={\overline{\mathrm{span}}}^{{w}^{*}}\{{e}_{i}^{*}:i\in I\}$.

## 2. Impact of Pre-Schauder Bases on the Vector Topology

- $y\in \cap {\mathcal{N}}_{x}:={\bigcap}_{V\in {\mathcal{N}}_{x}}V$.
- If ${({y}_{i})}_{i\in I}$ is a net in X converging to y, then it also converges to x.

**Remark**

**1.**

**Theorem**

**1.**

**Proof.**

**Lemma**

**1.**

**Proof.**

**Remark**

**2.**

**Lemma**

**2.**

**Proof.**

- $x\in {W}_{X}$. In this case, the range of the above map is $\mathbb{K}x$, which is contained in ${W}_{X}$ and thus endowed with the trivial topology. Since $\mathbb{K}$ is Hausdorff, the above map is not an isomorphism.
- $x\notin {W}_{X}$. Observe that, in particular, $x\ne 0$ and thus the above map is a linear isomorphism. Let us check that its inverse is continuous. Let $\epsilon >0$ and consider the open ball ${\mathsf{U}}_{\mathbb{K}}(0,\epsilon )$ of center 0 and radius $\epsilon $. In accordance with Remark 2, let V be a balanced neighborhood of 0 in X such that $x\notin V$. Observe that since $\mathbb{K}$ is a division ring, in accordance with Lemma 1, ${\mathsf{U}}_{\mathbb{K}}(0,\epsilon )(V\cap \mathbb{K}x)$ is a neighborhood of 0 in $\mathbb{K}x$. We prove next that the image of ${\mathsf{U}}_{\mathbb{K}}(0,\epsilon )(V\cap \mathbb{K}x)$ under the inverse of the above map is contained in ${\mathsf{U}}_{\mathbb{K}}(0,\epsilon )$. Indeed, an element of ${\mathsf{U}}_{\mathbb{K}}(0,\epsilon )(V\cap \mathbb{K}x)$ has the form $\lambda (\gamma x)$ where $\lambda \in {\mathsf{U}}_{\mathbb{K}}(0,\epsilon )$ and $\gamma \in \mathbb{K}$. If $\gamma =0$, then the image of $\lambda (\gamma x)$ under the inverse map is 0, which trivially is in ${\mathsf{U}}_{\mathbb{K}}(0,\epsilon )$. Thus, assume that $\gamma \ne 0$. In this case, observe that it must be $\left|\gamma \right|<1$, since otherwise we would have that $\left|\gamma \right|\ge 1$ so $\frac{1}{\left|\gamma \right|}\le 1$ and then the balancedness of V brings up the contradiction that$$x=\frac{1}{\gamma}(\gamma x)\in \frac{1}{\gamma}V\subseteq V.$$Therefore, $\left|\gamma \right|<1$ so $\left|\lambda \gamma \right|<\epsilon $ and hence the image of $\lambda (\gamma x)$ under the inverse map verifies that $\lambda \gamma \in {\mathsf{U}}_{K}(0,\epsilon )$.

**Remark**

**3.**

**Corollary**

**1.**

**Proof.**

**Theorem**

**2.**

**Proof.**

**Example**

**1.**

## 3. Renormings Concerning Schauder Bases

- normalized if $\parallel {e}_{n}\parallel =1$ for all $n\in \mathbb{N}$;
- binormalized if $\parallel {e}_{n}\parallel =\parallel {e}_{n}^{*}\parallel =1$ for all $n\in \mathbb{N}$;
- monotone if $\parallel {p}_{n}\parallel =1$ for all $n\in \mathbb{N}$;
- strictly monotone if $\parallel {p}_{n}(x)\parallel <\parallel {p}_{n+1}(x)\parallel $ for all $n\in \mathbb{N}$ and all $x\in X\backslash ker({e}_{n+1}^{*})$; and
- bimonotone if $\parallel {p}_{m}-{p}_{n}\parallel =1$ for all $m\in \mathbb{N}$ and all $n\in \mathbb{N}\cup \{0\}$ with $m>n$.

**Example**

**2.**

- The Banach space loses all kind of smooth properties after applying the previous renorming. We refer the reader to [9] where it is shown that a uniformly Frechet smooth Banach space with a Schauder basis can be equivalently renormed to remain uniformly Frechet smooth and to make the basis monotone.
- The renorming ${\parallel \xb7\parallel}_{m}$ does not assure that the Schauder basis become binormalized even if ${({e}_{n})}_{n\in \mathbb{N}}$ is normalized with the original norm.

- It is an equivalent norm on X. In fact,$$\parallel \xb7\parallel \le {\parallel \xb7\parallel}_{m}\le {\parallel \xb7\parallel}_{M}\le 2\mathrm{bc}\left({({e}_{n})}_{n\in \mathbb{N}}\right)\parallel \xb7\parallel .$$
- $\parallel {e}_{n}{\parallel}_{M}={\parallel {e}_{n}^{*}\parallel}_{M}^{*}=1$ for all $n\in \mathbb{N}$.
- $\parallel {p}_{m}-{p}_{n}{\parallel}_{M}=1$ for all $m\in \mathbb{N}$ and all $n\in \mathbb{N}\cup \{0\}$ with $m>n$.

**Theorem**

**3.**

- 1.
- It is an equivalent norm on X closer to the original norm that ${\parallel \xb7\parallel}_{M}$. In fact,$$\parallel \xb7\parallel \le {\parallel \xb7\parallel}_{m}\le {\parallel \xb7\parallel}_{md}\le {\parallel \xb7\parallel}_{M}\le 2\mathrm{bc}\left({({e}_{n})}_{n\in \mathbb{N}}\right)\parallel \xb7\parallel .$$
- 2.
- $\parallel {e}_{n}{\parallel}_{md}={\parallel {e}_{n}^{*}\parallel}_{md}^{*}=1$ for all $n\in \mathbb{N}$.
- 3.
- $\parallel {p}_{n}{\parallel}_{md}=1$ for every $n\in \mathbb{N}$.

**Proof.**

- It is trivial that ${\parallel \xb7\parallel}_{md}$ defines a norm on X. Notice in first place that ${\parallel \xb7\parallel}_{m}\le {\parallel \xb7\parallel}_{M}$ by Equation (11). By Equation (15), ${\parallel \xb7\parallel}_{d}\le {\parallel \xb7\parallel}_{M}$, thus ${\parallel \xb7\parallel}_{md}={max\{\parallel \xb7\parallel}_{m}{,\parallel \xb7\parallel}_{d}{\}\le \parallel \xb7\parallel}_{M}$. Finally, by Equation (11), we obtain the desired Equation (19).
- Obviously, $\parallel {e}_{n}{\parallel}_{md}=\parallel {e}_{n}{\parallel}_{d}=\parallel {e}_{n}{\parallel}_{m}=\parallel {e}_{n}\parallel =1$ for all $n\in \mathbb{N}$. Thus, it suffices to show that $\parallel {e}_{n}^{*}{\parallel}_{md}^{*}\le 1$ for all $n\in \mathbb{N}$. Indeed, if $x\in X$, then $|{e}_{n}^{*}{(x)|\le \parallel x\parallel}_{d}\le {\parallel x\parallel}_{md}$ for all $n\in \mathbb{N}$. This shows that $\parallel {e}_{n}^{*}{\parallel}_{md}^{*}\le 1$ for all $n\in \mathbb{N}$.
- Again, since $\parallel {e}_{n}{\parallel}_{md}=1$ for all $n\in \mathbb{N}$, it suffices to show that $\parallel {p}_{n}{\parallel}_{md}\le 1$ for every $n\in \mathbb{N}$. Indeed, fix arbitrary elements $n\in \mathbb{N}$ and $x\in X$ with ${\parallel x\parallel}_{md}\le 1$. We will follow two steps:
- Step 1
- Since ${\parallel x\parallel}_{m}\le 1$, we have that $\parallel {p}_{k}(x)\parallel \le 1$ for all $k\in \mathbb{N}$, therefore$$\parallel {p}_{k}({p}_{n}(x))\parallel =\parallel {p}_{min\{k,n\}}(x)\parallel \le 1$$
- Step 2
- Since ${\parallel x\parallel}_{d}\le 1$, we have that $|{e}_{k}^{*}(x)|\le 1$ for all $k\in \mathbb{N}$. Now,$${e}_{k}^{*}({p}_{n}(x))=\left\{\begin{array}{cc}\hfill 0& \phantom{\rule{4.pt}{0ex}}\mathrm{if}\phantom{\rule{4.pt}{0ex}}k>m\hfill \\ \hfill {e}_{k}^{*}(x)& \phantom{\rule{4.pt}{0ex}}\mathrm{if}\phantom{\rule{4.pt}{0ex}}k\le m\hfill \end{array}\right.$$

As a consequence, $\parallel {p}_{n}{(x)\parallel}_{md}=max\{\parallel {p}_{n}{(x)\parallel}_{m},\parallel {p}_{n}(x){\parallel}_{d}\}\le 1$.

**Theorem**

**4.**

- 1.
- ${\parallel \xb7\parallel}_{{d}_{1}}={\parallel \xb7\parallel}_{d}$.
- 2.
- $\parallel {p}_{j}{(x)\parallel \le \parallel x\parallel}_{{d}_{k}}$ for all $j\le k$ and all $x\in X$.
- 3.
- ${\parallel \xb7\parallel}_{{d}_{k}}\le 2k\mathrm{bc}\left({({e}_{n})}_{n\in \mathbb{N}}\right)\parallel \xb7\parallel $ for all $k\in \mathbb{N}$.
- 4.
- ${\parallel \xb7\parallel}_{{d}_{{k}_{1}}}\le {\parallel \xb7\parallel}_{{d}_{{k}_{2}}}$ if ${k}_{1}\le {k}_{2}$.
- 5.
- $\parallel {e}_{n}{\parallel}_{{d}_{k}}={\parallel {e}_{n}^{*}\parallel}_{{d}_{k}}=1$ for all $n\in \mathbb{N}$.
- 6.
- ${\u2225{\sum}_{i\in A}{p}_{i}-{p}_{i-1}\u2225}_{{d}_{k}}=1$ for every $A\in {\mathcal{P}}_{k}^{\times}(\mathbb{N})$.

**Proof.**

**Theorem**

**5.**

**Corollary**

**2.**

- 1.
- $\parallel \xb7\parallel \le {\parallel \xb7\parallel}_{m}\le {\parallel \xb7\parallel}_{{d}_{\infty}}$.
- 2.
- $\parallel {e}_{n}{\parallel}_{{d}_{\infty}}={\parallel {e}_{n}^{*}\parallel}_{{d}_{\infty}}=1$ for all $n\in \mathbb{N}$.
- 3.
- ${\u2225{\sum}_{i\in A}{p}_{i}-{p}_{i-1}\u2225}_{{d}_{\infty}}=1$ for every $A\in {\mathcal{P}}_{\mathtt{f}}^{\times}(\mathbb{N})$.

**Proof.**

## 4. Conclusions

- It is a trivial fact that in real or complex normed spaces the coefficient functionals of a pre-Schauder basis are continuous if and only if the canonical projections are also continuous. This is not trivial at all in real or complex topological vector spaces. We have accomplished this equivalence not only in real or complex topological vector spaces but also on topological vector spaces over an absolutely valued division ring (Corollary 1).
- We have placed on the spotlight the strong impact of a pre-Schauder basis on the vector topology in the sense that the existence of a pre-Schauder basis in a topological vector space over an absolutely valued division ring forces the existence of a dense Hausdorff subspace, which is precisely the linear span of the basis (Theorem 1).
- We have demonstrated the existence of Schauder bases in the non-Hausdorff setting (Theorem 2). Therefore, it makes sense to keep studying Schauder bases in the non-Hausdorff setting.
- We construct (Theorem 3) an equivalent renorming that turns a normalized Schauder basis into a binormalized and monotone Schauder basis and this renorming is not far from the original norm, at least it is closer to the original norm than the typical bimonotone renorming given in Equation (10). This way, our renorming has a lower chance of losing geometrical properties than the renorming given in Equation (10).

## Author Contributions

## Funding

## Conflicts of Interest

## Abbreviations

MDPI | Multidisciplinary Digital Publishing Institute |

DOAJ | Directory of open access journals |

TLA | Three letter acronym |

LD | linear dichroism |

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**MDPI and ACS Style**

García-Pacheco, F.J.; Pérez-Fernández, F.J.
Pre-Schauder Bases in Topological Vector Spaces. *Symmetry* **2019**, *11*, 1026.
https://doi.org/10.3390/sym11081026

**AMA Style**

García-Pacheco FJ, Pérez-Fernández FJ.
Pre-Schauder Bases in Topological Vector Spaces. *Symmetry*. 2019; 11(8):1026.
https://doi.org/10.3390/sym11081026

**Chicago/Turabian Style**

García-Pacheco, Francisco Javier, and Francisco Javier Pérez-Fernández.
2019. "Pre-Schauder Bases in Topological Vector Spaces" *Symmetry* 11, no. 8: 1026.
https://doi.org/10.3390/sym11081026