# The compact open topology and the operator norm

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 $$|x|<\delta$$?

• by $|T_x|$ do you meant the norm of the operator? – erz Feb 22 '19 at 4:53
• @erz Yes, by $|T_x|$ I mean the operator norm...I've heard that this statement is true as a consequence of the Banach-Steinhaus theorem, but I don't know how... – ernest Feb 22 '19 at 5:06

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 $$0$$.
• I see...I guess that does not make any difference even if I just consider continuous families of unitary operators $T:[0,1]\rightarrow U(H)$ in the compact open topology, that's true? – ernest Feb 22 '19 at 18:12
• What would be the statement that you are after in that case? All unitaries have norm $1$. – Mateusz Wasilewski Feb 22 '19 at 19:07
• I would like like something like this: Let $U:[0,1]\rightarrow U(H)$ be a continuous family of unitary operators in the compact open topology, then the for every $\epsilon>0$ there is a $\delta>0$ such that $|U_t-U_0|<\epsilon$ wehenever $t<\delta$...is something like this possible, or this has the same problem? – ernest Feb 22 '19 at 19:22
• It's the same issue. If $U_t$ is the operator of multplication by $\exp(itx)$ on $L^2(\mathbb{R})$ then there is no continuity. – Mateusz Wasilewski Feb 22 '19 at 20:16