Compact Operators on Hilbert Spaces

Question

Given $H$ an Hilbert space and $A:H\rightarrow H$ a compact operator with norm less than one. Consider

$$B: L^2([0,1];H)\rightarrow L^2([0,1];H) $$ defined as

$(Bf)(x):=A[f(x)]$.

Is $B$ still a compact operator?

Answer 1

Not necessarily. If $H=\mathbb R$, then $A$ is of the form $Ax=\lambda x$ for some $\lambda$. If $\lambda\neq0$, then $f\mapsto\lambda f$ is not compact on $L^2(0,1)$.