I'm having trouble verifying that the Fourier coefficients of a convolution are the products of the coefficients of the convoluted functions.
Let $h(x)=(f\star g)(x)=\frac1{2\pi}\int_{-\pi}^\pi f(x-y)g(y)dy$ be the convolution. Then its Fourier coefficients are given by $${1\over2\pi}\int_{-\pi}^\pi (f\star g)(x)e^{-inx}dx={1\over4\pi^2}\int_{-\pi}^\pi\left(\int_{-\pi}^\pi f(x-y)g(y)dy\right)e^{-inx}dx$$
Changing the order of integration, we get $${1\over4\pi^2}\int_{-\pi}^\pi g(y) \left(\int_{-\pi}^\pi f(x-y)e^{-inx} dx\right)\,dy\ .$$
Now here I'd like to do the substitution $t=x-y$ in the inner integral, but this makes the limits of integration depend on $y$, which I do not want. I want to end up with something like $${1\over4\pi^2}\int_{-\pi}^\pi g(y) e^{-iny}\left(\int_{-\pi}^\pi f(t) e^{-int} dt\right)\,dy\ .$$ Any ideas how I can go about this issue?
IF it is a question, go back through the argument until you are able to confidently say that "Yes, these things are equal" or "No, those things are different."
If $X$ is locally compact Hausdorff, I know that $L^p(X,\mu)$ where $\mu$ is a Radon measure contains $C_c(X)$ (functions with compact support) as a dense subspace. The result seems to also be true when $G$ is a locally compact Hausdorff group and $m$ is the Haar measure (i.e., $\overline{C_c(G)}=L^p(G,m)$).
Does anyone have a reference for this fact?
I could of course have a look if the proof for Radon measure translates... I guess I'll do that
ohh
Haar measures are Radon!
I didn't know
that makes life so much easier
ah, it makes sense, since the Haar measure is constructed using Riesz representation
if B is a bilinear form that is invariant under the lie group representation $f: G \rightarrow GL(V) $ then it is invariant under $ df: g \rightarrow End(V)$
Partially because i dont feel comfrotable to deal with differentials
i know that differentiating a bilinear form results into : d(B(x,y)(v,w))= B(x,w)+B(v,y)
which looks very close to what the invariance under df should look like.
so when we say $ g $ a real sub algebra of $ M(n,C) $. Do we look at it first as a C-vectorspace, then, restrict scalar multiplication to R and call that A "real" lie sub algebra, or what do you mean?
@Jakobian In the step $i-1\rightarrow i<n$, they first replace the $i$-th entry with $z$, which should be represented by a unitriangular matrix and hence has determinant $1$, then divide the $i$-th entry by $\lVert z\Vert$ and multiply the $i+1$-th entry by $\lVert z\rVert$, which also has determinant $1$, no?
Note that we're not talking about the change from $(u_{i-1, j})_j$ to $(u_{i, j})_j$, but even if we were, that has determinant $\frac{1}{\|z_{i-1}\|}$ and not $1$
because if we try to write $u_{i, i}$ in terms of $u_{i-1, j}$ we don't get coefficient $\frac{1}{\|z_i\|}$
we have something in terms $v_i$ that then need to be written in terms of $u_{i-1, i}$ so we get $\frac{1}{\|z_{i-1}\|\cdot \|z_i\|}$
the determinant is then $\frac{1}{\|z_{i-1}\|}$ and not $1$
and it checks out, the total transformation from $(v_1, ..., v_n)$ to $(u_{i, 1}, ..., u_{i, n})$ is then of determinant $\frac{1}{\|z_1\|\cdot ...\cdot \|z_{i-1}\|}$
so if we want it to be $1$, I'm pretty sure we need to have $u_{i, i+1} = \prod_{k=1}^i \|z_k\| v_{i+1}$
we even get a less problematic formula - with empty product for $i = 0$
I'd be very grateful if someone could check my work. I'm trying to solve the following problem; find the Fourier series of $h(t)=e^{3it}f(t-4)$, when $f$ has period $2\pi$ and satisfies $f(t)=1$ for $|t|<2$, $f(t)=0$ for $2<|t|<\pi$.
Previously I worked an exercise where I showed that if $f$ has Fourier coefficients $(c_n)$, then the function $t\mapsto e^{iat}f(t)$ has Fourier coefficients $(c_{n-a})$ for $a\in\mathbb Z$. And similarly, the function $t\mapsto f(t-b)$ has Fourier coefficients $(e^{-inb}c_n)$ for $b\in\mathbb R$.
So here is my attempt. From the previous exercise, the coefficients of $h(t)$ should be $(e^{-i(n-3)4}c_{n-3})$, so I need to find $(c_n)$. They are given by, assuming $n\neq0$, \begin{align}\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{-int}dt&=\frac1{2\pi}\int_{-2}^2 e^{-int}dt \\ &=\frac1{2\pi}\left[-\frac{e^{-int}}{in}\right]_{-2}^2 \\ &=\frac1{2\pi}\left(\frac{e^{i2n}}{in}-\frac{e^{-i2n}}{in}\right) \\ &=\frac{\sin(2n)}{\pi n}.\end{align} For $n=0$, we get simply $\frac2{\pi}$.
Recall the coefficient of $h(t)$ should be $(e^{-i(n-3)4}c_{n-3})$, so they are $$e^{-i(n-3)4}\frac{\sin (2(n-3))}{\pi(n-3)}\text{ for }n\neq 3,\quad \frac{2}{\pi} \text{ for }n=3 .$$ Therefor the (complex) Fourier series of $h(t)$ must be $$h(t)\sim\frac{2}{\pi}e^{i3t}+\sum_{\substack{k\in\mathbb Z \\ k\neq 3}}e^{-i(n-3)4}\frac{\sin (2(n-3))}{\pi(n-3)}e^{int}.$$
Unfortunately my book does not provide any answer to this exercise. Is this going in the right direction?
@Jakobian I prefer just doing orthogonalization instead of orthonormalization. Inductively set $u_i=v_i-\sum_{j<i}\frac{\langle v_i,u_j\rangle}{\langle u_j,u_j\rangle}u_j$. This has change of basis that is unitriangular. If you want an orthonormal basis, you can just rescale by a positive diagonal matrix after.
This suffices to get a deformation retraction from $\mathrm{SL}(n)$ to $\mathrm{SO}(n)$ and then you can argue the latter is path-connected.
@Mad If you are going to post a picture instead of typing it so we can read it clearly, at least make sure the picture is oriented so that it can be read.
and then if we divide $x^n+1$ by $x^m-1$ it will be $x^{n\mod{m}}+1$, but this won't be zero unless it happens that $n\mod{m} = 0$ and we are in char $2$
so $x^p-x^q$ can't divide $x^r+x^s$ unless we are in char $2$, no matter the other conditions on $p, q, r, s$
uh, not sure what a "a matrix of change of basis" is, but if BA = I then B represents the linear transformation on F^n that sends the columns of A to the standard basis (in order), and an invertible matrix A represents the transformation taking the standard basis to the basis given by its columns
its one of those things where if you look over F^n its something like VW^{-1} or WV^{-1} where you write the v's and w's as coordinates and stick them in matrices V or W
oh, with a transpose in it
i could never keep any of that stuff straight and had to re derive it every time when teaching, even if it was worked out in my notes and the book already i felt like i was doing it for the first time
sha's question the other day about some group action being f(x^{-1} y) vs. f(xy) or following the expected left action law vs. not gave me the same kind of vibe
@Jakobian: I wonder if there is an abstract reason using group theory, somehow. The matrix $P^{-1}MP$ is conjugate to $M$ in the general linear group. Although perhaps this idea doesn't go anywhere, since we can still change the basis of a non-invertible matrix $M$.
Well if I know that independent rows iff determinant non-zero, then this is clear that if $A$ is a change of basis matrix then $\det(A) \neq 0$, and if $\det(A)\neq 0$ then $\det(W)\neq 0$ so that $W$ is a basis
assuming I also know that $\det(AB) = \det(A)\det(B)$
I think this is a good enough justification for my own purposes
I think "independent rows iff determinant non-zero" is proven before introducing change of basis matrices?
its up to you when or how or whether you prove it :) when i last taught, that fact did arrive just before change of basis matrices
it's pretty common for US textbooks to put determinants well before any explicit discussion of abstract bases or linear independence, although they vary in how much they use determinants to illustrate or explain those concepts once they arrive
im looking at my last notes, i seem to have done a lot in terms of counting pivots in row echelon form, with facts about both linear dependence and determinants as corollaries to that, presumably because the textbook did it that way
