The compact open topology and the operator norm

Question

Let $T:[-1,1]\rightarrow B(H)$ be a continuous family of bounded operators where $B(H)$ is endowed with the compact open topology for an infinite dimesional Hilbert space $H$. Is it true that if $T_0=0$ then for every $\epsilon>0$ there is a $\delta>0$ such that $|T_x|<\epsilon$ whenever $|x|<\delta$?

Answer 1

No, the compact open topology on $B(H)$ is the topology of uniform convergence on compact sets, so it is stronger than the strong operator topology, i.e. the topology of pointwise convergence. But once we restrict to bounded subsets of $B(H)$ (e.g. the unit ball), the two topologies coincide. So it suffices to construct a continuous (in the strong operator topology) family of operators $T_t$ such that $\|T_t\|=1$ for $t>0$ and $T_t=0$. An example is as follows: take your Hilbert space to be $L^{2}[0,1]$ and let $M_x$ be the operator of multiplication by $x$, i.e. $(M_x f)(t) = tf(t)$. As our family we will take $T_t = M_x^{\frac{1}{t}}$ for $t>0$ and $T_t=0$ for $t\leq 0$. Note that $T_t$ is constructed by applying the functional calculus to a fixed operator $M_x$; the function, which we are using, is $x\mapsto x^{\frac{1}{t}}$. This is very concrete -- it is the operator of multiplication by the function $x^{\frac{1}{t}}$. Note that, as $t\to 0^{+}$, this function, defined on $[0,1]$, converges pointwise to the characteristic function of $\{1\}$. Borel functional calculus transfers pointwise convergence of bounded functions to strong convergence of operators, so $M_x^{\frac{1}{t}}$ tends strongly to the spectral projection of $M_x$ onto $\{1\}$. But $1$ is not an eigenvalue of $M_x$, so the spectral projection is trivial, i.e. $M_x^{\frac{1}{t}}$ tends to $0$.