Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 104330

The representation of functions (or objects which are in some generalize the notion of function) as constant linear combinations of sines and cosines at integer multiples of a given frequency, as Fourier transforms or as Fourier integrals.

1 vote

Sobolev inequalities and Wiener algebra

It is indeed false, (2) does not follow from (0) and $\nabla f\in L^2$. The idea is that since you consider the critical scaling, the potential inequality (on the Fourier side) fails both at $0$ and a …
Aleksei Kulikov's user avatar
2 votes
Accepted

A question on finite Fourier series

No problem, this is doable already for $N=2$. The idea is as follows: at the point of the maximum first derivative must vanish, and yet we want to separate the maximum into two by adding something sma …
Aleksei Kulikov's user avatar
4 votes
Accepted

Infinite direct sum decomposition of the heat semigroup on $\mathbb{R}^n$

I might be misunderstanding the definitions, but isn't it kinda trivial to do with the Fourier transform? On the Fourier side $Q_t$ is a multiplication by $e^{-t |x|^2}$, and $L^2(A)$ for any $A\subse …
Aleksei Kulikov's user avatar
10 votes
Accepted

Reference or proof of a theorem of L. Fejér on summability of Fourier series

I looked at that paper of Fejér, and although he does discuss some related things, I did not find that exact statement in his paper (granted, my German is probably not much better than yours). So, I w …
Aleksei Kulikov's user avatar
3 votes
Accepted

An asymptotic integral with complex phase

No, of course not. Essentially, you want to bound a $2$-dimensional object (the complex Fourier transform. I talk about the complex dimension here) from the at most $1$-dimensional bound (that is, eve …
Aleksei Kulikov's user avatar
3 votes
Accepted

On weighted Fourier transforms

The answer is yes. I will prove it using the assumption only for $\xi > 0$ (almost the same proof works if we only assume it for $\xi < 0$). For $z\in U = \{z: Im (z) \ge 0, |z| \ge 1\}$, consider th …
Aleksei Kulikov's user avatar
4 votes
Accepted

To find a $2\pi$-periodic function with a property

It is enough to consider the equation for $x = y$ only, so the rest of the equation will be useless for us. First, plugging $x = y = 0$ we get $g'(0)^2 + g(0)^2 = g'(0)$. Since $g$ is odd, $g(0) = 0$ …
Aleksei Kulikov's user avatar
5 votes
Accepted

$L^1$ norm for a product of cosines

Since I was asked to in the comments, let me record here some very simple observations. They boil down to the following inequality: for all $k \ge N$ we have $$I_{k - N} \min_{x\in \mathbb{R}} h_N(x) …
Aleksei Kulikov's user avatar
3 votes

Rate of decrease of the Fourier transform of standard mollifiers

So, morally, the only real way to bound the decay of $\hat{f}$ is to obtain some bounds on the derivatives of $f$. Let me do a little dilation and consider $f(x) = \exp(-1/(1-x)^p - 1/(1+x)^p)$, it d …
Aleksei Kulikov's user avatar
2 votes
Accepted

Example of a bounded function whose mean-zero mollification diverges at a point

$\newcommand{\ph}{\varphi}$ $\newcommand{\eps}{\varepsilon}$ Let $a_k$ be a very fast-growing sequence of integers (I think $a_k = 2^{1000k}$ should be enough). Consider the function $f$ defined as $$ …
Aleksei Kulikov's user avatar
16 votes
Accepted

Bounding a Fourier coefficient of a non-negative periodic function in terms of its $L^2$-norm

Your conjecture is indeed correct. The proof is based on the following simple yet powerful trick I learned many years ago: $|z| = \sup_{|v| = 1} \Re (zv)$. Therefore it is enough to only bound from ab …
Aleksei Kulikov's user avatar