Basic Theorems Regarding Ideals in a Linear Space 2

# Basic Theorems Regarding Ideals in a Linear Space 2

Theorem 1: Let $X$ be a Banach algebra and let $J \subseteq X$ be a linear subalgebra. Then:a) If $J$ is proper modular left ideal of $X$ and $u$ is a right modular unit of $J$ then $\| u - j \| \geq 1$ for all $j \in J$.b) If $J$ is a proper modular right ideal of $X$ and $u$ is a left modular unit of $J$ then $\| u - j \| \geq 1$ for all $j \in J$. |

*The above proposition tells us that for proper modular left ideals of $X$, if $u$ is a corresponding right modular unit then the norm distance from $u$ to $J$ is at least $1$.*

**Proof of a):**Let $J$ be a proper modular left ideal of $X$ and let $u$ be a right modular unit of $J$. Suppose instead that for some $j^* \in J$ we have that $\| u - j^* \| < 1$. We look at the spectral radius of $u - j^*$

\begin{align} \quad r(u - j^*) = \inf \{ \| (u - j^*)^n \|^{1/n} : n \in \mathbb{N} \} \leq \| (u - j^*)^1 \|^1 = \| u - j^* \| < 1 \end{align}

- By the theorem on the Quasi-Invertibility of x When r(x) < 1 in a Banach Algebra page we have that since $X$ is a Banach algebra and $r(u - j^*) < 1$ that $u - j^*$ is quasi-invertible. Let $y$ be the quasi-inverse of $u - j^*$. So in particular, $y \circ (u - j^*) = 0$ and we have that is:

\begin{align} \quad y + (u - j^*) - y(u - j^*) = 0 \end{align}

- Or equivalently:

\begin{align} \quad u &= j^* - y + y(u - j^*) \\ &= j^* - y + yu - yj^* \\ &= j^* -y(1 - u) - yj^* \end{align}

- Now observe that $j^* \in J$ by definition. Also, since $J$ is a left modular ideal and $u$ is a right modular unit of $J$ we have that $X(1 - u) \subseteq J$. Since $y \in X$ (and hence $-y \in X$, we see that $-y(1 - u) \in J$ too. Also since $J$ is a left ideal we have that $XJ \subseteq J$ and so $-yj^* \in J$. Since $J$ is a linear subspace we have from the above equation that $u \in J$.

- But this is a contradiction since $J$ is assumed to be a proper ideal of $X$. So the assumption that such an element $j^*$ exists is false.

- So $\| u - j \| \geq 1$ for every $j \in J$. $\blacksquare$

**Proof of b)**Similarly to above, let $J$ be a proper modular right ideal of $X$ and let $u$ be a left modular unit of $J$. Suppose there exists a $\hat{j} \in J$ such that $\| u - \hat{j} \| < 1$. Then $r(u - \hat{j}) < 1$ and so since $X$ is a Banach algebra, from the aforementioned theorem, $u - \hat{j}$ is quasi-invertible, so let $y$ be the quasi-inverse of $u - \hat{j}$. Then in particular $(u - \hat{j}) \cdot y = 0$, that is:

\begin{align} \quad (u - \hat{j}) + y - (u - \hat{j})y = 0 \end{align}

- Or equivalently:

\begin{align} \quad u &= \hat{j} - y + (u - \hat{j})y \\ &= \hat{j} - y + uy - \hat{j}y \\ &= \hat{j} - (1 - u)y - \hat{j}y \end{align}

- We note that $\hat{j} \in J$, $(1 - u)y \in (1 - u)X \subseteq J$ (since $u$ is a left modular unit of $J$), and $-hat{j}y \in JX \subseteq J$ (since $J$ is a right ideal of $X$), and thus, $u \in J$. But again, $J$ is assumed to be a proper ideal and so $u$ cannot be in $J$. So the assumption that such a $\hat{j}$ exists is false.

- Thus $\| u - j \| \geq 1$ for all $j \in J$. $\blacksquare$

Theorem 2: Let $X$ be a Banach algebra and let $J \subseteq X$ be a linear subalgebra. If $J$ is a proper modular left (right) ideal of $X$ then the closure $\overline{J}$ is a proper modular left (right) ideal of $X$. |

**Proof:**Let $J$ be a proper modular left (right) ideal of $X$ and let $u$ be a right (left) modular unit of $J$.

- Since $J$ is a left (right) modular ideal of $X$ it is easy to verify that $\hat{J}$ is also a left (right) modular ideal of $X$, that is, $X \bar{J} \subseteq \bar{J}$ ($\bar{J}X \subseteq \bar{J}$), and also, since $J \subseteq \bar{J}$ and $u$ is a left (right) modular unit of $J$ we have that $u$ is a left (right) modular unit of $\bar{J}$ too by one of the propositions on the Basic Theorems Regarding Ideals in a Linear Space 1 page.

- All that remains to show is that $\bar{J}$ is proper.

- Suppose instead that $\bar{J}$ is not proper. Then $\bar{J} = X$, and so $J$ is dense in $X$. In particular, for the point $u \in J$ there must exist a point $y^* \in J$ such that:

\begin{align} \quad \| u - j^* \| < 1 \end{align}

- But this contradicts proposition 1. So the assumption that $\bar{J}$ is not proper is false. Thus $\bar{J}$ is a proper modular left (right) ideal of $X$. $\blacksquare$

Corollary 3: Let $X$ be a Banach algebra. If $M$ is a maximal modular left (right) ideal of $X$ then $M$ is closed. |

**Proof:**Let $M$ be a maximal modular left (right) ideal of $X$. Then $M$ is a proper modular left (right) ideal of $X$. By theorem 2 we have that $\overline{M}$ is a proper modular left (right) ideal of $X $}]. But [[$ \overline{M}$ contains $M$ and by the maximality of $M$ we have that $M = \overline{M}$, i.e., $M$ is closed. $\blacksquare$