t.b.

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On extended leave from the site. I visit occasionally, but irregularly, mostly without logging in. For the time being it is impossible to predict if and when I will find the time to contribute again.

	43,139 reputation	bio	website		visits	member for	2 years, 4 months
3112195 badges		location			seen	May 18 at 10:40

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Oct 29	awarded	examples-counterexamples
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Sep 16	comment	differentiable square root of nonnegative smooth function Here's a related thread and a related MO thread. There are quite a few articles linked in those threads, maybe you can find something in one of them.
Sep 16	comment	Balanced but not convex? That's not balanced: If $x$ is near one of the corners then $-x$ is not in the pentagram. Similar stars work if you take an even number of corners.
Sep 16	comment	About an integral over measurable sets +1 now. Just a small suggestion for your more rigorous argument: I would define $g = f \chi_{F}$ and $g_n = f\chi_{F_n}$ then $\|g_n\|,\|g\| \leq \|f\|$ and $g_n \to g$ pointwise a.e. Thus, dominated convergence gives $$\int_{F} f = \int g = \lim_{n} \int g_n = \lim_{n} \int_{F_n} f$$ which is the desired conclusion.
Sep 16	reviewed	Approve suggested edit on what is the solution of this ODE
Sep 16	comment	About an integral over measurable sets Yes, that's absolutely right, Lebesgue dominated convergence leads to the result, as you intend. In my example I tried to point out that you could get $0 = \infty$ in the last displayed equation if you're not careful. The exercise is however much more elementary to solve than by using dominated convergence. Among the first facts one proves in measure theory is: if $F_n$ is a decreasing sequence of sets, $F = \bigcap F_n$ and for some $k$ we have $\mu(F_k) \lt \infty$ then $\mu(F) = \lim\limits_{n\to\infty} \mu(F_n)$. (this follows from $\sigma$-subadditivity of $\mu$)
Sep 16	comment	About an integral over measurable sets Hint: assume $f \geq 0$ and consider the finite measure $d\nu = f d\mu$.
Sep 16	comment	Specific homotopy between complex conjugation and the identity. Maybe I'm being silly, but: wouldn't this yield a homotopy $h$ between the identity and $z \mapsto z^{-1} = \bar z$ on $S^1$ by setting $h(t,z) = \frac{H(t,z)}{\lvert H(t,z)\rvert}$? This couldn't be because $z \mapsto z$ and $z \mapsto z^{-1}$ are distinguished by the degree. Probably such a degree argument can be done directly on $\mathbb{C}^\ast$.
Sep 16	revised	About continuous linear functional on the topology generated by linear functionals edited tags
Sep 16	comment	Entire functions representable in power series +1, although I think appealing to Baire is a bit of an overkill: if $f$ is not a polynomial, none of the $f^{(n)}$ is constant, hence each $F_n$ is countable, so $\bigcup_{n = 0}^\infty F_n \subsetneqq \mathbb{C}$.
Sep 16	comment	Entire functions representable in power series I see. This happens often :) So, does the other thread address your intended question?
Sep 16	comment	Entire functions representable in power series @abby: If $p$ is a polynomial of degree $n$ then $p^{(n+1)}(z) \equiv 0$, so $D_{n+1} = \mathbb{C}$.
Sep 16	revised	Taylor series of an entire function which is not a polynomial added 4 characters in body; edited title
Sep 16	comment	Entire functions representable in power series This question might be (the contrapositive of) what is intended -- ignore the part on Casorati-Weierstrass.