When is $\mathbb{B}(\mathbb{X})$ compact?

Question

Let $\mathbb{X}$ be a (real or complex) Banach Space and let $\mathbb{B}(\mathbb{X})$ be the space of all bounded linear operators of $\mathbb{X}$.

Let $\tau_u$ be the Uniform Topology on $\mathbb{B}(\mathbb{X})$ induced by the norm on $\mathbb{B}(\mathbb{X})$and let $\tau_{SOT}$ be the Strong Operator Topology on $\mathbb{B}(\mathbb{X})$.

When is $\mathbb{B}(\mathbb{X})$ compact with respect to $\tau_u$ ($\tau_{SOT}$) ?

Answer 1

It works if and only if $\mathbb{X}$ is the trivial vector space, $\left\{ 0\right\}$.

I. Suppose $\mathbb{X}=\left\{ 0\right\}$. Then, letting $L\in\mathbb{B}\left(\mathbb{X}\right)$, we know that the operator norm of $L$, denoted $\left\Vert L\right\Vert$, is given by:$$\left\Vert L\right\Vert =\sup_{x\in\mathbb{X};\left\Vert x\right\Vert _{\mathbb{X}}\leq1}\left\Vert L\left(x\right)\right\Vert _{\mathbb{X}}$$ where $\left\Vert \cdot\right\Vert _{\mathbb{X}}$ is the norm on $\mathbb{X}$. Since the only element of $\mathbb{X}$ is $0$, we have that:

$$\left\Vert L\right\Vert =\sup_{x\in\mathbb{X};\left\Vert x\right\Vert _{\mathbb{X}}\leq1}\left\Vert L\left(x\right)\right\Vert _{\mathbb{X}}=\left\Vert L\left(0\right)\right\Vert _{\mathbb{X}}=\left\Vert 0\right\Vert _{\mathbb{X}}=0$$ since, by definition, any normed vector space must give the zero vector a norm of zero. Since $\mathbb{B}\left(\mathbb{X}\right)$ is a normed vector space, $L$ having zero operator norm forces $L$ to be the zero operator (the zero vector in $\mathbb{B}\left(\mathbb{X}\right)$). Consequently, as Banach spaces, $\mathbb{B}\left(\mathbb{X}\right)$ is isomorphic to $\left\{ 0\right\}$ (and hence, to $\mathbb{X}$) whenever $\mathbb{X}=\left\{ 0\right\}$ , which shows that $\mathbb{B}\left(\left\{ 0\right\} \right)$ is indeed compact whenever $\mathbb{X}=\left\{ 0\right\}$.

II. For the other direction, suppose that $\mathbb{X}\neq\left\{ 0\right\}$. Then, the identity map $\textrm{Id}_{\mathbb{X}}$ (which is most definitely an element of $\mathbb{B}\left(\mathbb{X}\right)$) is not the same thing as the zero map on $\mathbb{X}$. Hence, $\mathbb{B}\left(\mathbb{X}\right)$ is a vector space with a non-zero element ($\textrm{Id}_{\mathbb{X}}$). Noting that $\left\Vert \textrm{Id}_{\mathbb{X}}\right\Vert =1$ the homogeneity property of vector space norms then guarantees that $n\textrm{Id}_{\mathbb{X}}$ (the sum of $\textrm{Id}_{\mathbb{X}}$ with itself $n$ times, where $n$ is an integer—necessarily an element of the vector space $\mathbb{B}\left(\mathbb{X}\right)$) is an element of $\mathbb{B}\left(\mathbb{X}\right)$ with norm:$$\left\Vert n\textrm{Id}_{\mathbb{X}}\right\Vert =n$$ Consequently, for all integers $m$ and $n$:$$\left\Vert n\textrm{Id}_{\mathbb{X}}-m\textrm{Id}_{\mathbb{X}}\right\Vert =n-m$$ As such, $\left\{ n\textrm{Id}_{\mathbb{X}}\right\} _{n\in\mathbb{Z}}$ is an infinite subset of $\mathbb{B}\left(\mathbb{X}\right)$ that contains no convergent sequence. Consequently, $\mathbb{B}\left(\mathbb{X}\right)$ cannot be compact. Thus, $\mathbb{B}\left(\mathbb{X}\right)$ is never compact whenever $\mathbb{X}\neq\left\{ 0\right\}$.

Q.E.D.

Answer 2

As written in the MCS's answer $\mathbb{B}(\mathbb{X})$ is compact in the norm topology iff $\mathbb{X}$ is the trivial vector space. The same statement holds true for the strong operator topology.

If $\mathbb{X}=\{0\}$, then $\mathbb{B}(\mathbb{X})=\{0\}$ and is thus compact in whatever topology you put.

On the other hand, if $\mathbb{X}\neq \{0\}$, pick $x_0\in \mathbb{X}\setminus \{0\}$ and look at the following open cover (both in norm topology and in SOT) $$ \mathbb{B}(\mathbb{X}) = \bigcup_{n\geq 1} \{ T\in \mathbb{B}(\mathbb{X}) \ : \ \Vert T x_0 \Vert < n \}.$$ Clearly this does not have a finite subcover (as for example some $\lambda \cdot Id_\mathbb{X}$ would be missing). Hence, it is not compact.

Note that we will not get away using sequences as $(\mathbb{B}(\mathbb{X}), \tau_{SOT})$ is not metrizable (see https://mathoverflow.net/questions/298775/is-the-strong-operator-topology-metrizable).