Mathematics

#7: Post edited by $user avatar$ Snoopy‭ · 2024-01-26T13:23:52Z (10 months ago)

Copy Link

Raw

Markdown

----
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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

----
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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

#6: Post edited by $user avatar$ Snoopy‭ · 2024-01-26T13:22:37Z (10 months ago)

Copy Link

Raw

Markdown

---
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).~~
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

---
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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

#5: Post edited by $user avatar$ Snoopy‭ · 2024-01-26T13:21:25Z (10 months ago)

Copy Link

Raw

Markdown

---
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).
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

---
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).
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

#4: Post edited by $user avatar$ Snoopy‭ · 2024-01-26T13:21:00Z (10 months ago)

Copy Link

Raw

Markdown

---
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 $h\to 0$, 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).
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

---
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).
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

#3: Post edited by $user avatar$ Snoopy‭ · 2024-01-26T13:20:39Z (10 months ago)

Copy Link

Raw

Markdown

---
Following the definition of uniform continuity, one basically needs to estimate:
$$
|\hat{f}(x)-\hat{f}(y)|
~~= |\int_{\Rbb^n}f(t)(e^{-2\pi i x\cdot t}-e^{-2\pi i y\cdot t})\ dt|~~
~~\le \intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt,~~
$$
where $h=x-y$. If one can show that the integral on the right converges to zero as $h\to 0$, 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).
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

---
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 $h\to 0$, 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).
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

#2: Post edited by $user avatar$ Snoopy‭ · 2024-01-26T00:15:41Z (10 months ago)

Copy Link

Raw

Markdown

---
Following the definition of uniform continuity, one basically needs to estimate:
$$
|\hat{f}(x)-\hat{f}(y)|
= |\int_{\Rbb^n}f(t)(e^{-2\pi i x\cdot t}-e^{-2\pi i y\cdot t})\ dt|
\le \intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt,
$$
where $h=x-y$. If one can show that the integral on the right converges to zero as $h\to 0$, then the uniform continuity follows.
It suffices to show that one can take the limit under the integral sign:
$$
~~\lim_{h\to 0}\intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt = \intw \lim_{h\to 0}|f(t)||e^{-2\pi ih\cdot t}-1|\ dt~~
$$
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).
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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) for the limit of functions.

---
Following the definition of uniform continuity, one basically needs to estimate:
$$
|\hat{f}(x)-\hat{f}(y)|
= |\int_{\Rbb^n}f(t)(e^{-2\pi i x\cdot t}-e^{-2\pi i y\cdot t})\ dt|
\le \intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt,
$$
where $h=x-y$. If one can show that the integral on the right converges to zero as $h\to 0$, 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).
**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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) 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
$$
\lim_{n\to \infty}\intw |f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt = \intw \lim_{n\to \infty}|f(t)||e^{-2\pi ih_n\cdot t}-1|\ dt
$$

#1: Initial revision by $user avatar$ Snoopy‭ · 2024-01-26T00:09:20Z (10 months ago)

Copy Link

Raw

Markdown

---
Following the definition of uniform continuity, one basically needs to estimate:
$$
|\hat{f}(x)-\hat{f}(y)| 
= |\int_{\Rbb^n}f(t)(e^{-2\pi i x\cdot t}-e^{-2\pi i y\cdot t})\ dt|
\le \intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt,
$$
where $h=x-y$. If one can show that the integral on the right converges to zero as $h\to 0$, then the uniform continuity follows. 

It suffices to show that one can take the limit under the integral sign:
$$
\lim_{h\to 0}\intw |f(t)||e^{-2\pi ih\cdot t}-1|\ dt = \intw \lim_{h\to 0}|f(t)||e^{-2\pi ih\cdot t}-1|\ dt
$$
 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). 

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](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) for the limit of functions.

Communities

Post History