Compositions with a compact operator are compact

If $E\subseteq W$ is bounded and nonempty, say $E\subseteq B(w_0,r)$, then $\mathrm{\parallel }Aw\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }(\mathrm{\parallel }{w}_0\mathrm{\parallel }+r)$ for every $w\in E$ by [A1], so $A(E)\subseteq B(0,\mathrm{\parallel }A\mathrm{\parallel }(\mathrm{\parallel }{w}_0\mathrm{\parallel }+r)+1)$ is bounded; and $A(\mathrm{\varnothing })=\mathrm{\varnothing }$ is bounded, so $A$ carries bounded sets to bounded sets.

If $E=\mathrm{\varnothing }$, take $R=1$, so $E\subseteq R{\bar{B}}_X$ holds immediately. If $E$ is a bounded subset of $X$ with $E\mathrm{\ne }\mathrm{\varnothing }$ and $E\subseteq B(x_0,R_0)$, then $\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }{x}_0\mathrm{\parallel }+R_0=:R$ for every $x\in E$ by [A2] and the triangle inequality, so $E\subseteq R{\bar{B}}_X$; the same holds in any normed space.

The set $B(\overline{T({\stackrel{‾}{B}}_X)})$ is compact: $\overline{T({\stackrel{‾}{B}}_X)}$ is compact by [A1], and $B$ is continuous by [A1], so the image under $B$ is compact by [A3].

For every bounded $E\subseteq W$ the image $A(E)$ is bounded by [step 1.1], so $\overline{T(A(E))}$ is compact by [A1]; hence $TA$ is compact.

For every bounded $E\subseteq X$, [step 1.2] gives $E\subseteq R{\bar{B}}_X$ for some real $R\ge 0$, so $T(E)\subseteq R\overline{T({\stackrel{‾}{B}}_X)}$ and hence $B(T(E))\subseteq RB(\overline{T({\stackrel{‾}{B}}_X)})$, which is compact by [step 1.3] and [A3] and therefore closed; thus $\overline{B(T(E))}\subseteq RB(\overline{T({\stackrel{‾}{B}}_X)})$ is a closed subset of a compact set, hence compact, and $BT$ is compact.

Both composites $TA$ and $BT$ are therefore compact.