Let $X$ be a normed space, $\mathcal{F}(X)$ the algebra of all operators on $X$ with finite fank, then $\mathcal{F}(X)$ is the unique minimal ideal of $\mathcal{K}(X)$ the algebra of all compact operators on $X$.
I want to show that any non-zero ideal $\cal{I}$ of $\mathcal{K}(X)$ necessarily contains all operators with rank one. But I can not show a rank one operator $P$ can be written as $AT$, where $T\in\cal{I}$, $A\in\mathcal{K}(X)$.
Thanks a lot.
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 $\mathcal{F}(X)=\operatorname{span}\{x\bigcirc f: x\in X, f\in X^*\}$.
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 $\mathcal{F}(X)\subset I$.
