2,020 reputation
617
bio website
location
age
visits member for 2 years, 1 month
seen yesterday

I guess I'm a graduate student now. That's a little weird.

This will remain the account that I use to anonymously ask stupid questions.


Jan
20
accepted What is $\{ f \in F_{p^m}[x_1, \ldots, x_n] : f(a) = 0, \forall a \in A^n\}$?
Jan
19
asked What is $\{ f \in F_{p^m}[x_1, \ldots, x_n] : f(a) = 0, \forall a \in A^n\}$?
Jan
17
comment Non-compact manifold with compact boundary
Yep, edited. \{ \}
Jan
17
revised Non-compact manifold with compact boundary
added 4 characters in body
Jan
17
answered Non-compact manifold with compact boundary
Jan
2
answered $L_X( \omega (Y_1,\ldots, Y_n)) = (L_X \omega) (Y_1, \ldots, Y_n) + \sum_{i = 1}^n \omega (Y_1, \ldots, Y_{i-1}, L_X Y_i, \ldots, Y_n)$
Jan
2
asked $L_X( \omega (Y_1,\ldots, Y_n)) = (L_X \omega) (Y_1, \ldots, Y_n) + \sum_{i = 1}^n \omega (Y_1, \ldots, Y_{i-1}, L_X Y_i, \ldots, Y_n)$
Dec
23
awarded  Yearling
Dec
15
awarded  Caucus
Dec
10
revised Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$.
added 207 characters in body
Dec
10
comment Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$.
@NateEldredge Haha, that's true too. This was just what immediately came to mind, so I wrote it down for posterity. :)
Dec
10
revised Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$.
deleted 3 characters in body
Dec
10
answered Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$.
Dec
9
revised Let $X$ and $Y$ be Banach, and $T : X \to Y$ a bounded linear map. When is $T$ closed?
added 46 characters in body
Dec
9
comment Let $X$ and $Y$ be Banach, and $T : X \to Y$ a bounded linear map. When is $T$ closed?
I am trying to argue that the Schwartz class S is not normable by displaying it as a dense subspace of $L^1$, and then trying to argue that (given normability) the inclusion can be upgraded to surjectivity (which is nonsense), by showing that the range is closed. Since I don't know what the norm looks like, I wanted to make use of some less analytical property of $S$ in order to show that the range is closed.
Dec
9
comment Let $X$ and $Y$ be Banach, and $T : X \to Y$ a bounded linear map. When is $T$ closed?
Thanks, good point. I think in my mind (and at least in the problem I was working on), $T$ was implicitely injective.
Dec
9
revised Let $X$ and $Y$ be Banach, and $T : X \to Y$ a bounded linear map. When is $T$ closed?
added 64 characters in body
Dec
9
comment Let $X$ and $Y$ be Banach, and $T : X \to Y$ a bounded linear map. When is $T$ closed?
Sorry, yes, I mean that $im T$ is closed, not that the graph of $T$ is closed.
Dec
9
asked Let $X$ and $Y$ be Banach, and $T : X \to Y$ a bounded linear map. When is $T$ closed?
Dec
9
revised Why should the open mapping theorem be expected?
added 30 characters in body