# $C([0,T],(L(X),\mathcal T_{\text{strong}})) = C([0,T],(L(X),\mathcal T_{\text{uniform}}))$?

Given a Banach space $$X$$, we have two topologies on the space of all bounded linear operators $$L(X)$$, one is uniform operator topology $$\mathcal T_{\text{strong}}$$, the other is strong operator topology $$\mathcal T_{\text{uniform}}$$. We know that the topology $$\mathcal T_{\text{uniform}}$$ is normable with the operator norm $$\|\cdot\|$$.

Now $$C([0,T],(L(X),\mathcal T_{\text{uniform}}))$$ is the space of all function from $$[0,T]$$ to $$L(X)$$ that are continuous for the uniform operator topology. Clearly, it is normable with the supremum norm $$\|F\|_\infty:=\sup_{t\in[0,T]}\|F(t)\|.$$

$$C([0,T],(L(X),\mathcal T_{\text{strong}}))$$ is the space of all function from $$[0,T]$$ to $$L(X)$$ that are continuous for the strong operator topology. That is, $$F\in C([0,T],(L(X),\mathcal T_{\text{strong}}))$$ if and only if for each $$x\in X$$, the function $$t\to F(t)x$$ is continuous.

How to compare these two spaces $$C([0,T],(L(X),\mathcal T_{\text{strong}}))$$ and $$C([0,T],(L(X),\mathcal T_{\text{uniform}}))$$? Are they equal as sets? And how about their topologies? Are these two topologies equivalent?

The question is actually motivated by an argument in the book One-parameter semigroups for linear evolution equations by Engel & Nagel. See the last sentence in the picture below. How can they derive from $$\lim_{n\to\infty}F_n(\cdot)x = F(\cdot)x, \ \text{in } C([0,t_0],X), \quad\forall x\in X,$$ to $$\lim_{n\to\infty}F_n = F, \ \text{in } \mathcal X_{t_0}.$$

• Strong = uniform – Daniel Camarena Perez Nov 9 '18 at 17:31
• @DanielCamarenaPerez Can you explain more? – Q. Huang Nov 9 '18 at 17:33
• @DanielCamarenaPerez: The strong operator topology does not equal the uniform (i.e. operator norm) topology when $X$ is infinite dimensional. – Nate Eldredge Nov 9 '18 at 17:55
• @Nate Eldredge how define the strong topology? – Daniel Camarena Perez Nov 9 '18 at 18:00
• @DanielCamarenaPerez: It's the strong operator topology: en.wikipedia.org/wiki/Strong_operator_topology. That is a different thing from the "strong topology". The notation $\mathcal{T}_{\text{strong}}$ is just being used for shorthand. – Nate Eldredge Nov 9 '18 at 18:01

They are generally not equal as sets.

Take $$X = L^1([0,1])$$, and for $$h \in X$$ let $$F(t)h = 1_{[0,t]} h$$. You can check, using the dominated convergence theorem, that for any fixed $$h$$, if $$t_n \to t$$ we have $$1_{[0,t_n]}h \to 1_{[0,t]}h$$ in $$L^1$$. Hence $$F : [0,1] \to L(X)$$ is continuous when $$L(X)$$ is equipped with the strong operator topology. However, for any $$t > 0$$, if we take $$h = 1_{[0,t]}$$ then $$F(t)h = h$$ and thus $$\|F(t)\| \ge 1$$, whereas $$F(0)$$ is the zero operator. So $$F$$ is not continuous when $$L(X)$$ is equipped with the uniform topology.

That is, this particular function $$F$$ is in $$C([0,T],(L(X),\mathcal T_{\text{strong}}))$$ but not in $$C([0,T],(L(X),\mathcal T_{\text{uniform}}))$$.

For the proof in the book, there are a few more steps that have not been written out.

At this point, what has been shown is the following:

For each fixed $$x \in X$$, the sequence of functions $$t \mapsto F_n(t)x$$ converges in sup norm to some function called $$t \mapsto F(t)x$$, which is continuous.

Now you have to verify the following:

• For each $$t$$, the map $$x \mapsto F(t) x$$ is linear. Hence we can view $$F(t)$$ as a linear operator on $$X$$.

• For each $$t$$, the linear operator $$x \mapsto F(t)x$$ is bounded, i.e. $$\sup_{\|x\|=1} \|F(t)x\|_X < \infty$$. (You can use the uniform boundedness principle.) Hence we can view $$F$$ as a function from $$[0,t_0]$$ into $$L(X)$$.

• Since we showed above that $$t \mapsto F(t) x$$ is continuous for each $$x$$, this shows that $$F : [0,t_0] \to L(X)$$ is continuous with respect to the strong operator topology. So $$F$$ really is an element of $$\mathcal{X}_{t_0}$$.

• Show that $$F_n$$ converges to $$F$$ in the above norm. That is, you must show $$\sup_{t \in [0,t_0]} \|F_n(t) - F(t)\|_{L(X)} := \sup_{t \in [0,t_0]} \sup_{\|x\| = 1} \|F_n(t)x - F(t)x\|_{X} \to 0$$ as $$n \to \infty$$. This will be a typical sort of triangle inequality "$$\epsilon/2$$" argument.

Specifically, fix $$\epsilon > 0$$. Since $$\{F_n\}$$ is Cauchy, choose $$N$$ so large that $$\|F_k - F_m\|_{\infty} < \epsilon/2$$ for all $$m, k > N$$. Choose an arbitrary $$x \in X$$ with $$\|x\|_X = 1$$. Since $$F_n(\cdot) x \to F(\cdot) x$$ uniformly (which is the statement in the box above), we can find an $$m > N$$ such that for every $$t \in [0,t_0]$$ we have $$\|F_m(t) x - F(t)x\|_X < \epsilon/2$$. We also have $$\|F_k(t)x - F_m(t)x\| \le \|F_k(t) -F_m(t)\|_{L(X)} \|x\| \le \|F_k - F_m\|_\infty \|x\| <\epsilon/2$$ as above, so by the triangle inequality, $$\|F_k(t) x - F(t) x\|_X \le \|F_k(t) x - F_m(t) x\|_X + \|F_m(t) x - F(t) x\|_X < \epsilon.$$ So we have $$\|F_k(t) x - F(t) x\|_X < \epsilon$$. But $$x$$ and $$t$$ were arbitrary, so we have $$\sup_{t \in [0,t_0]} \sup_{\|x\| = 1} \|F_n(t)x - F(t)x\|_{X} \le \epsilon.$$ This holds for all $$k > N$$, so we have shown the desired convergence.

• Then can you explain the last sentence in the picture above? – Q. Huang Nov 9 '18 at 17:56
• I added a source link for the book by Engel & Nagel. – Q. Huang Nov 9 '18 at 18:11
• @Q.Huang: See edits. – Nate Eldredge Nov 9 '18 at 18:33
• Could you explain more for the "$\epsilon/2$" argument in the last step? I know that the continuity of the function $t\to F(t)x$ should play a role in this step, but I don't know how to use this... – Q. Huang Nov 9 '18 at 18:54
• $\|F(t)-F(0)\|=\|F(t)\|\ge {\|F(t)h\|}_{L^1}={\|h\|}_{L^1}= t$ if $h=1_{[0,t]}$? – Daniel Camarena Perez Nov 9 '18 at 18:55