If you have approximation by smooth, I recon that dividing by the sup of the absolute value on the function should give approximation by bounded-by-1 smooth functions
i know i can prove it if i can find a sequence of continuous functions that approximate $\mathbf{1}_{y_0 \geq 0} - \mathbf{1}_{y_0 < 0}$ in $L^1$ and are bounded by $1$ in absolute value
because then I can make $|F(s_n)|$ arbitrarily close to $||F||$ all while keeping $||s_n|| \leq 1$
@robjohn: Yes, of course, but he's got to build up his basic skills. Sorta scary. Of course, you can't draw those pictures in Mathematica unless you already know the answers :)
@TedShifrin I have nice tools to draw what I drew knowing just the vertices of the triangle and radii of the circles (and the start and end angles for the arcs)
Suppose I have a set of smooth functions $\{f_k\}$ on the real line. Let $A_k$ be the set of $(x,y)$ such that $y\geq f_k(x)$. I can then take the convex hull of these $A_k$. Then there should also exist a function $F$ such that $y\geq F(x)$ defines the desired hull. Is there a name for this $F$ in relation to the $f$'s?
Yes, of course, @robjohn. My point was only that to use Mathematica as a problem-solving tool for such things you have to already know the answer. Same thing with something like drawing the intersection of two surfaces ...
@Astyx If $X$ is a scheme, would you call maps $\text{Spec} A \to X$ as geometric points if $A$ is a 0-dimensional algebra, or is that just restricted to $A$ being a field?
Suppose I have a set of smooth functions $\{f_k\}$ on the real line. Let $A_k$ be the set of $(x,y)$ such that $y\geq f_k(x)$. I can then take the convex hull of these $A_k$. Then there should also exist a function $F$ such that $y\geq F(x)$ defines the desired hull. Is there a name for this $F$ in relation to the $f$'s?
Yeah, @Astyx, I realized after I knew what the problem actually was. Obviously, we can continuize $\text{sgn}\, y_0$ without a problem unless there are infinitely many sign changes.
the idea is that for any function on that interval which changes sign finitely many times, say $g$, there is a sequence of continuous functions $x_{n}^g$ associated with $g$, such that $|x_n| \leq 1$ and $\int x_ng \rightarrow \int |g|$ as $n \rightarrow \infty$
the interesting aspect is that, if you move the red curve up vertically, then you only need to take the lower convex envelope of the blue curve (which is mostly just a straight line)
then approximate $y_0$ with polynomials uniformly, say by $P_n$, and call its associated continuous functions $x_m^{P_n}$, now $\int x_m^{P_n}P_n \rightarrow \int |P_n|$ as $m \rightarrow \infty$ which can be as close as we like to $\int |y_0|$ given $n$ sufficiently large. But $x_m^{P_n}P_n \rightarrow x_m^{P_n}y_0$ uniformly in $n$. So I think these combined conclude the result?
@Astyx I think so, more or less, the idea is you start with a constant $+1$ on the interval its positive, minus a tiny bit at the end where you shoot to $-1$ where it starts going negative, and you decrease the length of these tiny parts so that this thing converges almost everywhere to $sgn$ of that function
@Astyx okay so $y_0$ doesn't change sign at all when its $> \eps$ in absolute value, so you are suggesting thinking about $y_0 \mathbf{1}_{|y_0| > \eps}$ and replacing $y_0$ with this?
by contrast, I can forgive centrifugal force: if you're in a non-inertial reference frame and so from your perspective you're not accelerating, then it's legit to say that there's a centrifugal force.
i still counsel students not to use it but it does produce sensible algebra.
Here's a really stupid probability question that I don't know how to prove.
Let $A_k$, $k \geq 1$, be a sequence of events for which $\mathbb{P}(A_k) = 1$ for each $k \geq 1$. How do I show that $\mathbb{P}\left(\bigcap_{k=1}^{\infty}A_k\right) = 1$?
and show that for finitely many of them $\mathbb{P}(A_1 \cap ... A_n) = 1$
@Clarinetist in general if $\mu$ is a measure and $\{B_n\}$ is a decreasing sequence of sets of finite measure, then $\mu(\cap_{n} B_n) = \lim_n \mu(B_n)$
So consider the sequence $A_n = \bigcap_{k=1}^{n}A_k$; one has that $A_n \downarrow \bigcap_{k=1}^{\infty}A_k$ and that $\mathbb{P}(A_1) = 1 < \infty$. Thus $\mathbb{P}(\bigcap_{k=1}^{n}A_k) \downarrow \mathbb{P}\left(\bigcap_{k=1}^{\infty}A_k \right)$ as $n \to \infty$.
This I get. It's showing that these quantities are equal to $1$ that I don't quite get.
@porridge: Offhand, it seems to me that the sup bound on the polynomials may well go to infinity. You're going to introduce huge oscillations in the polynomials.
Maybe "geometric point" is a more specific notion, but in general, if you have an A-scheme X, and an A-algebra B, you define a B point of X to be a morphisme B -> X over A
i guess my issue is I don't know how to really visualize $\mathbf{1}_{|y_0| > \epsilon} y_0$, why does it only have finitely many sign changes? And even if it does, how do we construct a companion function that approximates sgn of this?
I don't need a limiting function, I just need to be able to get as close as I want to $\int_{a}^b |y_0(t)|dt$ via computing $|\int_{a}^b y_0(t)x(t) dt|$ for some $||x|| \leq 1$
@BalarkaSen non-degeneracy of the trace form is equivalent to the trace map not being the zero map by conjugation invariance, the latter is true since trace is inseparability degree times sum over embeddings into an algebraic closure and those are linearly independent by Dedekind
"Help, I don't have enough information to solve!" "What information are you given?" "Oh, I know this stuff." Then why didn't you say that in the first place
@Thorgott It must be true that the trace map on $L$, basechanged to $\text{tr}_L \otimes 1 : L \otimes_K \overline{K} \to \overline{K}$, decomposes as sum of traces on the various $K$-algebra factors on $L \otimes_K K$.
But if there are nilpotent factors the trace map is zero
@Semiclassical I agree, but with all the people wanting us to do their exams/homework, it's nice that someone might be trying to think for himself a little bit more. Little.
If $I$ is an invertible fractional ideal of a domain $A$, I can just write the inverse as the denominator ideal $(A : I)$, right?
$A \subseteq I (A : I) \subseteq A$ yeah ok
Uniqueness of inverse
If $E/\Bbb Q$ is a number field the different ideal is $\delta_E = \{x \in \mathcal{O}_E | xy \in \mathcal{O}_E \forall y\in E : \text{tr}_E(y\mathcal{O}_E) \subseteq \Bbb Z\}$
$a_1, \cdots, a_n$ be an integral basis of $\mathcal{O}_E$, tr-dualize to a $\Bbb Q$-basis $b_1, \cdots, b_n$ of $E$ so everything in $E$ which satisfies $\text{tr}_E(-\mathcal{O}_E) \subseteq \Bbb Z$ is a $\Bbb Z$-linear combo of $b_1, \cdots, b_n$
Is it literally an equal to? Yeah, $\delta_E^{-1} = \bigoplus_{i = 1}^n \Bbb Zb_i$.
Ah ok this is just taking dual of a lattice wrt some form
One second
Covolume of the lattice is discriminant, so volume of the dual lattice should be discriminant, or something like this.
If you see the discriminant $d_E$ as the norm of the different $\delta_E$, then you have $1=N(\mathcal O_E)=N(\delta_E\delta_E^{-1}) = N(\delta_E)N(\delta_E^{-1})=(d_E)N(\delta_E^{-1})$
