$L(\ell_{p})$ contains only one proper closed ideal

Question

I am trying to solve the following problem: Show that if $1<p<\infty$ and $T:\ell_{p}\rightarrow\ell_{p}$ is not compact then there is a complemented infinite dimensional subspace $E$ of $\ell_{p}$ so that $T$ is an isomorphism of $E$ onto a complemented subspace $T(E)$. Deduce that the Banach algebra $L(\ell_{p})$ (of bounded linear operators on $\ell_{p}$) contains exactly one proper closed two-sided ideal (the ideal of compact operators).

I've done the first part of the problem but I can't figure out the 'deduce' part. I was trying to go along the lines of assuming an ideal contains a non-compact operator and showing that then the ideal must be the whole algebra but I ran into the problem of trying to factor any given bounded operator through the non-compact operator and I got stuck.

Answer 1

Lemma 1. Every proper two sided ideal of $\mathcal{B}(\ell_p)$ for $p\in[1,+\infty)$ is contained in $\mathcal{K}(\ell_p)$.

Proof. Let $I$ be a two sided ideal in $\mathcal{B}(\ell_p)$. Assume that there is a $T\in I\setminus \mathcal{K}(\ell_p)$. Then we have infinite dimensional complemeted subspace $E\subset \ell_p$ such that $T$ isomorphically maps $E$ to $F:=T(E)$. Since $E$ is complemented subspace of $\ell_p$, then $E$ is isomorphic to $\ell_p$, i.e. we have an isomorphism $S:\ell_p\to E$. Let $(e_n)_{n\in\mathbb{N}}$ be a standard basis of $\ell_p$, then since $S$ is an isomorphism, then $g_n=S(e_n)$ for $n\in\mathbb{N}$ is a basis of $E$. By construction $T$ is an isomorphism from $E$ to $F$, so $f_n=T(g_n)$ for $n\in\mathbb{N}$ is a basis of $F$. Since $F$ is complemented in $\ell_p$ it is isomorphic to $\ell_p$. Hence there exist an isomorphism $R:F\to\ell_p$ sending basis $(g_n)_{n\in\mathbb{N}}\subset F$ to the basis $(e_n)_{n\in\mathbb{N}}\subset\ell_p$. By construction $RTS(e_n)=e_n$, but $(e_n)_{n\in\mathbb{N}}$ is a basis of $\ell_p$, so $RTS=1_{\ell_p}$. Since $T$ is an element of two sided ideal $I$ we conclude that $1_{\ell_p}=RTS\in I$. In this case $I=\mathcal{B}(\ell_p)$. Contradiction, so $I\subset\mathcal{K}(\ell_p)$

Remark. Let $X$ be a normed space, by $\mathcal{F}(X)$ we denote the space of finite rank operators on $X$. For a given $x\in X$ and $f\in X^*$ we define rank one operator $x\bigcirc f:X\to X:z\mapsto f(z)x$. One can show that $\mathcal{F}(X)=\operatorname{span}\{x\bigcirc f: x\in X, f\in X^*\}$.

Lemma 2. Every non-zero two sided ideal of $\mathcal{B}(X)$ of a normed space $X$ contains $\mathcal{F}(X)$

Proof. Let $I$ be a non-zero two sided ideal of $\mathcal{B}(X)$. Consider $T\in I\setminus\{0\}$, then there is $x_0\in X$ such that $T(x_0)\neq 0$. By Hahn-Banach theorem there exist $f_0\in X^*$ such that $f(x_0)=1$. Now consider arbitrary rank one operator $x\bigcirc f$ where $x\in X$ and $f\in X^*$. Note that $x\bigcirc f=(x\bigcirc f_0)T(x_0\bigcirc f)$, but $T\in I$, so $x\bigcirc f\in I$. Now by previous remark $\mathcal{F}(X)\subset I$.

Proposition. Every proper closed two sided ideal of $\mathcal{B}(\ell_p)$ for $p\in[1,+\infty)$ is $\mathcal{K}(\ell_p)$.

Proof. Let $I$ be a proper closed two sided ideal of $\mathcal{B}(\ell_p)$. By lemma $1$ we know that $I\subset \mathcal{K}(\ell_p)$. By lemma $2$ we have $\mathcal{F}(\ell_p)\subset I$. Since $I$ and $\mathcal{K}(\ell_p)$ are closed, then $\operatorname{cl}_{\mathcal{B}(\ell_p)}\mathcal{F}(\ell_p)\subset I\subset \mathcal{K}(\ell_p)$. Since $\ell_p$ have a basis, then it have the approximation property, so $\operatorname{cl}_{\mathcal{B}(\ell_p)}\mathcal{F}(\ell_p)=\mathcal{K}(\ell_p)$. Therefore $I=\mathcal{K}(\ell_p)$.