2,209 reputation
1420
bio website
location
age 30
visits member for 3 years, 1 month
seen Oct 22 at 21:32

I am a PhD student at University of Alberta.


Sep
30
awarded  Explainer
Sep
30
awarded  Yearling
Sep
24
awarded  Autobiographer
Sep
19
revised Integration over subsets of the complex plane.
$t$ and $\theta$ were being used for the same variable
Sep
19
comment Integration over subsets of the complex plane.
I'm a bit embarassed I didn't make this connection. We do this all the time for continuous functions $f:\mathbb{R}^{2}\to\mathbb{R}$. Thanks for the help!
Sep
19
accepted Integration over subsets of the complex plane.
Sep
19
asked Integration over subsets of the complex plane.
Aug
9
comment Neighborhood Base (the definition)
While I agree with you in other contexts, this is not really what the question was about. It does not follow at all from the definition I provided that the neighbourhoods should contain $x$. Arturo Magidin's comments above have answered the question.
Jul
13
accepted Convolution of $L^1(G)$ functions with elements of $M(G)$.
Jul
13
comment Convolution of $L^1(G)$ functions with elements of $M(G)$.
Thanks very much!
Jul
13
comment Convolution of $L^1(G)$ functions with elements of $M(G)$.
Thanks Norbert. Can you refer me to where I might find the proof that $e_{\alpha}\to \delta_{e}$?
Jul
2
awarded  Socratic
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
19
comment Relationship between $T$ and $T^{*}$ for $T\in B(H)$
My previous comment suggested I was confused. What I should have said was that $\left(\text{im}(T^{*})^{\perp}\right)^{\perp} = \overline{\text{im}(T^{*})}$.
Jun
19
asked Relationship between $T$ and $T^{*}$ for $T\in B(H)$
Jun
10
comment Concluding that a linear operator on a Hilbert space is invertible
Thanks for the direct argument. :)
Jun
10
accepted Concluding that a linear operator on a Hilbert space is invertible
Jun
9
comment Concluding that a linear operator on a Hilbert space is invertible
Thanks for this!
Jun
9
asked Concluding that a linear operator on a Hilbert space is invertible