Post History
#5: Post edited
by
Snoopy
·
2024-01-26T13:49:02Z (11 months ago)
#4: Post edited
by
Snoopy
·
2024-01-26T13:23:27Z (11 months ago)
$\def\Rbb{\mathbf{R}}$ $\def\Cbb{\mathbf{C}}$ $\def\intw{\int_{\Rbb^n}}$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_](https://mathscinet.ams.org/mathscinet/relay-station?mr=304972). 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](https://en.wikipedia.org/wiki/Dominated_convergence_theorem), which allows one to exchange the limit and integral.
- $\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_](https://mathscinet.ams.org/mathscinet/relay-station?mr=304972). 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](https://en.wikipedia.org/wiki/Dominated_convergence_theorem), which allows one to exchange the limit and integral.
#3: Post edited
by
Peter Taylor
·
2024-01-26T13:05:58Z (11 months ago)
Fix typo
- $\def\Rbb{\mathbf{R}}$ $\def\Cbb{\mathbf{C}}$ $\def\intw{\int_{\Rbb^n}}$
- 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 continous.- **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_](https://mathscinet.ams.org/mathscinet/relay-station?mr=304972). 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](https://en.wikipedia.org/wiki/Dominated_convergence_theorem), which allows one to exchange the limit and integral.
- $\def\Rbb{\mathbf{R}}$ $\def\Cbb{\mathbf{C}}$ $\def\intw{\int_{\Rbb^n}}$
- 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_](https://mathscinet.ams.org/mathscinet/relay-station?mr=304972). 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](https://en.wikipedia.org/wiki/Dominated_convergence_theorem), which allows one to exchange the limit and integral.
#2: Post edited
by
Snoopy
·
2024-01-26T00:09:36Z (11 months ago)
Fourier transform of an $L^1$ function is uniformly continuous in its domain
- Fourier transform of an $L^1$ function is uniformly continuous
#1: Initial revision
by
Snoopy
·
2024-01-26T00:06:28Z (11 months ago)
Fourier transform of an $L^1$ function is uniformly continuous in its domain
$\def\Rbb{\mathbf{R}}$ $\def\Cbb{\mathbf{C}}$ $\def\intw{\int_{\Rbb^n}}$
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 continous.
**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_](https://mathscinet.ams.org/mathscinet/relay-station?mr=304972). 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](https://en.wikipedia.org/wiki/Dominated_convergence_theorem), which allows one to exchange the limit and integral.