Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Fourier transform of an $L^1$ function is uniformly continuous

+2
−0

$\def\Rbb{\mathbf{R}}$$\def\Cbb{\mathbf{C}}$$\def\intw{\int_{\Rbb^n}}$If $f\in L^1(\Rbb^n)$, denote the Fourier transform of $f$ as $$ \hat{f}(x) = \int_{\Rbb^n}f(t)e^{-2\pi x\cdot t}\ dt $$

Problem. If $f\in L^1(\Rbb^n)$, show that $\hat{f}:\Rbb^n\to\Cbb$ is uniformly continuous.

Note. This is a well-known theorem in the basic $L^1$ theory of Fourier transforms. It is the very first theorem ("Theorem 1.1") in Stein-Weiss's Introduction to Fourier Analysis on Euclidean spaces. The authors give the theorem at the beginning without a proof and simply say that it is "obvious". As an exercise in real analysis, I will write my own answer below. There are different ways to approach it; the essential tool is the Dominated Convergence Theorem, which allows one to exchange the limit and integral.

History
Why does this post require attention from curators or moderators?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

1 answer

+3
−0

Following the definition of uniform continuity, one basically needs to estimate: $$ \begin{align} |\hat{f}(x)-\hat{f}(y)| &= |\int_{\Rbb^n}f(t)(e^{-2\pi i x\cdot t}-e^{-2\pi i y\cdot t})\ dt|\\ &\le \intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt, \end{align} $$ where $h=x-y$.

If one can show that the integral on the right converges to zero as $h\to0$, then the uniform continuity follows.

It suffices to show that one can take the limit under the integral sign: $$ \lim_{h\to 0}\intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt = \intw \lim_{h\to 0}|f(t)||e^{-2\pi ih\cdot t}-1|\ dt\tag{*} $$ But that follows from the dominated convergence theorem (DCT) since the integral is dominated by $2\|f\|_1$ (by an easy application of triangle inequality on the exponential term): $$ \intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt\le \intw 2|f(t)|\ dt $$

Note. The usual version of DCT is written in terms of sequences. One can get the version for limit in a continuous variable using Heine's sequential characterization for the limit of functions. In order to show (*), it suffices to show that for every sequence $h_n$ with $h_n\to 0$, one has

$$ \lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt $$
History
Why does this post require attention from curators or moderators?
You might want to add some details to your flag.

0 comment threads

Sign up to answer this question »