2
$\begingroup$

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

enter image description here

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.