Impact

~72k
people reached

- 0 posts edited
- 0 helpful flags
- 127 votes cast

Jul
31 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
I'm embarrassed I didn't make that observation myself about the finite dimensionality of $V$. It's a nice little argument. It also highlights that the $2$ in the $L^2$ condition isn't key as you wind up in a finite dimensional setting. Showing that the inner product, $\langle f(x-k_0), f(x-k)\rangle$, tends to zero is equivalent to proving the Riemann-Lebesgue lemma, but if you just approximate $f$ with a $C_c^\infty$ function, I think all you need is the dominated convergence theorem and Cauchy-Schwarz, so trigonometric polynomials never enter the picture. Thanks! |

Jul
29 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
Sorry, maybe I'm just being dense, but why does linear dependence imply the space is finite dimensional? Certainly if you have one dependence relation, you will have infinitely many, but it doesn't seem obvious at first glance that I should get a finite dimensional space. |

Dec
20 |
comment |
When is a collection of exponentials dense in $L^2(K), |K|<\infty$
Alexei Poltoratski gave a very nice series of lectures with a number of comments in this direction this summer at Clemson. I don't think he really discussed the R^d case, but notes similar to the lectures he gave are available here: internetanalysisseminar.gatech.edu/lectures_uncp Some of the references may point you in the right direction. |

May
8 |
comment |
Square roots of the Laplace operator
Now that I think about it, the analytic continuation might need some additional work... there might be some poles at certain integers $\alpha$, so you would need to add an additional term to balance things out. This computation has to be written down someplace. Sorry that I don't know of a good place. |

May
8 |
comment |
Square roots of the Laplace operator
Yes, much of this stuff is fairly disjointed, as I recall, which is unfortunate. Many of the basic computations are not terribly difficult (just tricky), so I think they are not written down in many places. The issue you are having in the computation in your comment comes from having to use some analytic continuation technique to extend the range of possible $\alpha$s in the inverse transform of $|\xi|^\alpha$. $|x|^{-n-1}$ is not locally integrable, as Tom pointed out, so its a bit more delicate. |

Nov
23 |
comment |
Carleson's Theorem (on the Adeles and other exotic groups)
I don't think you can accept more than one answer, but that's ok. I just hope what I wrote helped in some way! |

Oct
19 |
comment |
Can the supremum of continuous functions be discontinuous on a set of positive measure?
Hm, I suppose you could also have just made it 1 on $K_n$ and 0 on $E$ and avoided that $\sup(-f_n)$ business... |

Sep
13 |
comment |
Approximating high-dimensional integrals by low-dimensional ones
If X is very nice, perhaps you can utilize Gaussian quadrature? For example, in a fixed interval of $\mathbb{R}$, by sampling n special points one can compute exactly the integrals of all polynomials of degree <2n. It seems like this might be generalizable to, say, rectangular regions in higher dimensions; it is computationally infeasible to do this in high dimensions, though, as it would require an absurd number of samplings. That is ultimately why Monte Carlo methods are used to do integrations in $\mathbb{R}^100$ |

Sep
12 |
comment |
Function space between uniform continuity and Hölder continuity
What do you mean by containment in this case? Contained as sets of functions? Must there be some kind of isometric vector space embedding-type relationship? If so, what are the topologies on each of the spaces? |

Sep
1 |
comment |
Applications of PDE in mathematical subjects other than geometry & topology
Solving PDE is rather tricky business; even some of the simplest ones are so rich! The Cauchy-Riemann equations are innocent looking yet produce an incredible theory. Since there is much interest in solving these problems from an applied standpoint, people work hard and come up with ingenious ways to solve them, and in doing so produce a host of great abstract problems which don't relate that closely to the original problem anymore. |

Sep
1 |
comment |
Applications of PDE in mathematical subjects other than geometry & topology
A professor here at Cornell gave a talk several years ago where he began by briefly estimating that the majority of mathematics (as measured by number of articles) of the last 50-60 years relates to differential equations (he includes most of applied math, numerical analysis, much of analysis itself, and so on). I think by this measure, differential equations represents the greatest source of interesting problems in all of mathematics. It is probably rather difficult to find an area of mathematics which does not relate in some (potentially very loose) way to or use ideas from PDE. |

Aug
29 |
comment |
What does the σ in σ-algebra stand for?
Couldn't the $\sigma$ have originally come from the word for sum (which, in latin, is summa)? In the end, $\sigma$-algebras are designed to produce sets which play well with summation (capital $\Sigma$). There is the additional connection that the sign for the integral also comes from the latin word summa (and is supposed to be an elongated s), and $\sigma$-algebras are used to generate the modern integral. |

Aug
26 |
comment |
Analytic functions with algebraic Taylor coefficients at some point.
Ah, of course you're right. A careless error on my part. It thankfully doesn't change the main points, though. |

Aug
10 |
comment |
High dimensional beta integral (a typo in Stein's book “singular integrals”)
You're welcome. I can totally understand the confusion about the domain of integration since Stein mentions the beta integral which only integrates on [0,1] and makes a typo that $x=1$! Then again, one often learns more from correcting mistakes in books than from simply reading. There is something to be said for getting one's hands dirty and doing some hard work. |

Jul
9 |
comment |
Long time behavior of the heat equation on R
If μ is absolutely continuous with respect to Lebesgue measure, then the density should be dominated by some polynomial. Certainly in that case, the integral will grow at most like some power of t: in particular I believe it should grow like $t^{d/2}$, where d is the degree of the polynomial bound for the density function. |

Jul
9 |
comment |
Long time behavior of the heat equation on R
I don't think it needs to decrease. For instance if $\mu$ is simply Lebesgue measure, then $u(x,t)\equiv 1$, no? |

Jun
16 |
comment |
What is convolution intuitively?
Ignoring the extraneous first $x$, this formula is only correct if the functions are integrable. Convolution makes sense if both functions are square-integrable but not integrable, in which case this equation is incorrect (since Fubini does not apply). But in any case, what intuition do you glean from knowing, for a fixed $f$, that $T_f(g)=f*g$ has the property you stated? The operator $S_f(g)=g(x)\int_{-\infty}^{\infty}f(t)dt$ has the same property but is just a constant times the identity... |

Jun
2 |
comment |
A Generalization of Hadamard?
Obviously Mariano was mostly joking, but I think the grammar in the title works fine; one frequently uses 'of' in the sense of 'owing to', e.g. the lemma of Schwarz (I think my professor used this form mostly because it was easier to say, frankly, but there is also a historical trend for that particular lemma). We often leave the originator's name in theorems even if they were later generalized, as was indicated above with the Cauchy-Schwarz inequality (sorry Bunyakovsky). Plus isn't it a bit poetic that Hadamard might exist through mathematics in some generalized way even after his death? |

May
31 |
comment |
A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
The Cantor-Lebesgue function has derivative 0 almost everywhere but is continuous and strictly increasing. The issue is that for a function on $R$ to have distributional derivatives it must be absolutely continuous; simply having a well-defined almost everywhere derivative is not enough. |

May
31 |
comment |
A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
Suppose you are on $R^1$ and $l$ is just the point $\{0\}$. Then the function $f$ which is 0 when $x<0$ and 1 when $x\ge 0$ has distributional derivative 0 away from 0. So $f'$ is defined almost everywhere and $f'$ is the distributional derivative for any function compactly supported in $R^1-\{0\}$. But if $\phi$ is any $C_c^\infty$ supported function which is 1 in a neighborhood of $0$, then $-\int f\phi'=\phi(0)=1$ rather than 0 if $f'$ were actually the distributional derivative on all of $R$. You can come up with clever ones so that $f$ is continuous: see the Cantor-Lebesgue function. |