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14h
comment Convergence problem in different norms
Incidentally, this fact also holds if $H$ is replaced by a Banach space, but @Norbert's proof won't generalize, because not all Banach spaces have the approximation property.
15h
comment Question about adjoint map and strong operator topology (SOT)
There seems to be a typo in Lemma 2: in the statement, the conclusion is only strong convergence, but the proof actually gives convergence in norm (which is what we need).
15h
comment Convergence problem in different norms
If you're familiar with the proof of the Arzela-Ascoli theorem, the same device is used there. The key is that a set of operators bounded in norm is equicontinuous. And the relevant step in Arzela-Ascoli shows that an equicontinuous sequence converging pointwise also converges uniformly on compact sets.
15h
comment Convergence problem in different norms
A hint: show that strong convergence implies uniform convergence on compact sets. (It will help to know that strongly convergent sequences are bounded in norm, which comes from the uniform boundedness principle.)
15h
comment Do most nowhere dense sets have measure $0$?
It's certainly possible to define a measure on any set of sets; for instance, the zero measure. The more difficult question is whether there is a natural measure to use.
16h
comment What is a co-dimension?
As you can see, the word "codimension" is used in many contexts in somewhat different (though related) ways, so to clarify your question you should give some more context.
17h
answered Existence of finite Borel measure $\mu$ with prescribed $\mu(X)$
1d
comment Unit ball on $B(H)$ and the weak -topology
Do you mean the weak operator topology? Technically B(H) is a Banach space and as such has a weak topology induced by its dual, but I assume that is not what you mean.
1d
comment Compactly supported distribution is finite linear combination of …
See the comments on this question.
1d
comment When does open and connected imply path-connected?
Note that local path connectedness is not necessary. Consider the rationals, which are not locally path connected, but every connected open set is path connected (because every such set is empty). Your property 2 also fails here: the path components are singletons and are not open.
1d
comment relatively compact family of analytic function
But you don't have a subset of $\mathbb{C}$, which would be a set of complex numbers. You have a set of functions. They aren't elements of $\mathbb{C}$.
1d
comment relatively compact family of analytic function
Relatively compact in what topology? What have you tried? What facts do you know that might be relevant?
2d
comment Let $D$ be a subset of $L^2[0,1]$ defined in the following way
@MarkFd: Unfortunately, the closed graph theorem applies to closed operators that are defined everywhere on a Hilbert space. Our operator is only defined on the subset $D$, and we don't know yet whether it is closed (that's what we have to prove).
Apr
22
comment Is my conjecture correct? Any advice on how to solve this conjecture?
Also, you talk about GCDmany being given a number that is less than $n$ bits; but I thought it was supposed to take as input several numbers?
Apr
22
comment Is my conjecture correct? Any advice on how to solve this conjecture?
I don't quite understand. Can you state concisely exactly what property it is that you think the sequence 1,3,4,6,7,... has?
Apr
21
comment Markov process which is not martingale
For one thing, the state space of a martingale has to be $\mathbb{R}$ or at least a vector space...
Apr
21
comment Completion of metric spaces
Now that you have edited out the metric, your question no longer makes sense. And $\RR$ doesn't parse. Please edit your question to make your space and your metric explicit.
Apr
21
comment Is $\sum_{k=1}^{n} k^k / \sum_{k=1}^{n} k \in \mathbb{N}$ for some $n > 1$?
The reverse implication in your second sentence is false when a,b are not relatively prime. 6 and 9 each divide 18, but 6*9=54 does not.
Apr
21
comment Restarting a Markov chain
This isn't a research level question suitable for this site (a good hint is "I am new to this subject") but would be fine at math.se. Voting to migrate.
Apr
20
comment What does “twice as likely” mean?
@ErickWong: Oh, I see. Sorry for the confusion.