746 reputation
624
bio website
location
age
visits member for 4 years, 5 months
seen 3 hours ago

Tyler Hilton


Dec
15
comment Show that the Open Mapping Theorem requires both spaces to be complete
Ahh I think I get it. If $X = \ell^1$ was complete, then the norms would be equivalent. But we defined $||\cdot||_f$ as a sum of continuous norm + a discontinuous linear functional. So does this imply that $||\cdot||_f$ is discontinuous?
Dec
15
comment Show that the Open Mapping Theorem requires both spaces to be complete
I am not sure about your last time, I don't understand it. What do you mean by the "identity ..." and how can you say its not open.
Dec
15
comment Show that the Open Mapping Theorem requires both spaces to be complete
I've been thinking about it. Could you give me a hint?
Dec
14
comment Difference between an eigenvalue and a spectral value
Ah, i see. I'll have to go through this again as its a bit difficult for me (and I have very little experience with measure theory - but I don't think that's entirely relevant here)
Dec
14
comment Difference between an eigenvalue and a spectral value
Yes, I understand it now. Thanks for explaining it.
Dec
14
comment Difference between an eigenvalue and a spectral value
Ignore comments, the first paragraph on wiki link cleared up the confusion for me. But my question is that when an operator $A$ has eigenvalues, is it true that $\ker(A- \lambda I) \neq {0}$ or $\ker(A) \neq {0}$
Dec
14
comment Difference between an eigenvalue and a spectral value
Actually, my bad. We are talking about a specific $\lambda = 0$. $A$ could potentially have other eigenvalues. But $0$ is not one of them and this implies $\ker(A) = {0}$ NOT $\ker(A-\lambda I)$. Infact because $\ker(A - \lambda I) \neq {0}$ it follows there are spectral values. Now I am trying to think why 0 is a spectral value.
Dec
14
comment Difference between an eigenvalue and a spectral value
So if I have this correctly, $S$ is injective so $\ker(S - \lambda I) = {0}$ and thus $0$ is not an eigenvalue. But since $S$ is not surjective, $\ker(S - \lambda I) \neq {0}$. How can this be? How can the kernel be both empty and non empty
Dec
14
asked Difference between an eigenvalue and a spectral value
Dec
12
awarded  Popular Question
Dec
10
comment Show that the Open Mapping Theorem requires both spaces to be complete
Actually from your example, I would think the norms are equivalent, no? Norms are equivalent if there exists some $c >0$ such that $||\cdot||_1 \leq c||\cdot||_\infty$. Take $c = n$ and the norms would become equivalent. Also, is there a quick way to see that the limit does not exist in $\ell^1$ for your second part.
Dec
9
asked Show that the Open Mapping Theorem requires both spaces to be complete
Dec
8
awarded  Caucus
Dec
3
comment Why is the operator norm of a diagonal matrix it's largest value?
How can we be sure that $\lambda_i$ is an eigenvalue
Nov
28
comment Domain of a bounded linear operator on a Hilbert Space
I am also looking for some proof that a closed linear, bounded operator has a closed domain and if so, is the domain the entire space?
Nov
20
comment Fredholm operator norm
Hi, could you explain two parts to me? First why normalize h ? And secondly how did you come about your inequality. Ie why can you write $T_k \bar{h}(s_0) \geq |T_k g(s_0)| - \epsilon \sup{k}$. Why substract the episolon*sup k?
Nov
19
awarded  Popular Question
Nov
11
comment Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
Actually, if I can bother you with one more. I don't understand the last inequality. It dosnt seem to match the triangle inequality (or the reverse triangle inequality). In your answer you say you have to replace $\int f$ with $\int |f|$ but are we allowed to do that? edit: OMG. ofcourse, f is integrable <-> |f| is integrable lol.
Nov
11
comment Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$
Nope, Im okay. Thanks. I've moved on to understanding $(**)$ but that seems easy enough. I've marked your answer the the correct one even thuough mookid's was good, I just needed more explanation.
Nov
11
accepted Proving that f is measurable if g is measurable where $g(x, y) = |f(x) - f(y)|$