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 $$\lim_{n\to\infty}F_n = F, \ \text{in } \mathcal X_{t_0}.$$

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  • $\begingroup$ Strong = uniform $\endgroup$ – Daniel Camarena Perez Nov 9 '18 at 17:31
  • $\begingroup$ @DanielCamarenaPerez Can you explain more? $\endgroup$ – Q. Huang Nov 9 '18 at 17:33
  • $\begingroup$ @DanielCamarenaPerez: The strong operator topology does not equal the uniform (i.e. operator norm) topology when $X$ is infinite dimensional. $\endgroup$ – Nate Eldredge Nov 9 '18 at 17:55
  • $\begingroup$ @Nate Eldredge how define the strong topology? $\endgroup$ – Daniel Camarena Perez Nov 9 '18 at 18:00
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ Then can you explain the last sentence in the picture above? $\endgroup$ – Q. Huang Nov 9 '18 at 17:56
  • $\begingroup$ I added a source link for the book by Engel & Nagel. $\endgroup$ – Q. Huang Nov 9 '18 at 18:11
  • $\begingroup$ @Q.Huang: See edits. $\endgroup$ – Nate Eldredge Nov 9 '18 at 18:33
  • $\begingroup$ 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... $\endgroup$ – Q. Huang Nov 9 '18 at 18:54
  • $\begingroup$ $\|F(t)-F(0)\|=\|F(t)\|\ge {\|F(t)h\|}_{L^1}={\|h\|}_{L^1}= t$ if $h=1_{[0,t]}$? $\endgroup$ – Daniel Camarena Perez Nov 9 '18 at 18:55

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