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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 …
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 …
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 …
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 …
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 …
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 …
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, $g(0) = 0$ …
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
$$I_{k - N} \min_{x\in \mathbb{R}} h_N(x) …
3
votes
Rate of decrease of the Fourier transform of standard mollifiers
So, morally, the only real way to bound the decay of $\hat{f}$ is to obtain some bounds on the derivatives of $f$. Let me do a little dilation and consider $f(x) = \exp(-1/(1-x)^p - 1/(1+x)^p)$, it d …
2
votes
Accepted
Example of a bounded function whose mean-zero mollification diverges at a point
$\newcommand{\ph}{\varphi}$
$\newcommand{\eps}{\varepsilon}$
Let $a_k$ be a very fast-growing sequence of integers (I think $a_k = 2^{1000k}$ should be enough). Consider the function $f$ defined as
$$ …
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 …