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I am trying to prove the following statement:

If $f \in L^1$, then $\hat f$ is uniformly continuous.

The argument given is as follows :

$|\hat f (\xi +h )-\hat f (\xi)| = \left| \int f(x) (e^{-2 \pi i x \cdot (\xi+h)}- e^{-2 \pi i x \cdot (\xi)})\mathrm dx \right| \leq 2 \|f\|_{L^1}$

Now I suppose we have to use the Dominated Convergence Theorem, but I am unable to see to what sequence of functions we apply the theorem to.

Any help is greatly appreciated.

  • 0
    See also http://math.stackexchange.com/questions/21710/uniform-continuity-question2012-05-29

3 Answers 3

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I) The proof for $L_1$ is simpler actually; here is an outline:

1) Prove that a linear map $f : E\to E'$ is continuous (even uniformly) iff it is continuous at zero (0); i.e.

$\begin{align} (\exists c\ \epsilon \ \mathbb{R}) \ |f(x)|\leq c|x| \end{align}$

2) Fourier transform is a linear functional defined on $L_1$. So, by (1) you only need to prove it is continuous at 0: We have:

$\begin{align} F(f) = \int_{\mathbb{R}} f(x)e^{-j\omega x}dx \end{align}$ Where F is the Fourier operator defined on L1. $\begin{align} |F(f)| = \left|\int_{\mathbb{R}} f(x)e^{-j\omega x}dx\right| \leq \int_{\mathbb{R}} |f(x)e^{-j\omega x}|dx \leq \int_{\mathbb{R}} |f(x)|dx = \left \| f \right \|_{L^1} < {\infty} \end{align}$ Thus $\begin{align} \left | F(f) \right | \leq 1 \left \| f \right \|_{L^1}, \end{align}$ This completes the proof (set c = 1)

II) Yes, you could use DCT, here's how: Take any sequence $x_n \to 0$ Set $ u_n(x) = f(x)e^{-i\omega (x\pm x_n)} \ \\ \ u(x) = f(x)e^{-i\omega x} $ Clearly, $ u_n \to u $

Now, $\ u\ is\ L^1,\ so\ is\ each\ u_n$

We also have $\left |u_n \right | \leq \left | f \right |\ and\ f\ is\ L^1$ Now use DCT to get: $ \int_{\mathbb{R}}|u_n-u| \to 0\ as \ n \to \infty $

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    You misunderstood the question. What you are showing in part I is that the Fourier transform is a continuous map from $L^1$ to $L^\infty$. The question was about the uniform continuity of the function $\hat{f}(\xi)$ as a function of $\xi$. The map $\xi \mapsto \hat{f}(\xi)$ is not linear in general. In part II you actually give a correct proof, but this is a duplicate of previous answers.2012-10-19
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I like Olivier's comment suggesting the use of the Riemann-Lebesgue Lemma, but here is a different approach. $ \begin{align} \hat{f}(\xi+\eta)-\hat{f}(\xi) &=\int_{\mathbb{R}^n}f(x)\left(e^{-2\pi ix\cdot(\xi+\eta)}-e^{-2\pi ix\cdot\xi}\right)\mathrm{d}x\\ &=\int_{\mathbb{R}^n}f(x)\left(e^{-2\pi ix\cdot\eta}-1\right)e^{-2\pi ix\cdot\xi}\;\mathrm{d}x\tag{1} \end{align} $ For any $f\in L^1$ and $\epsilon>0$, by Dominated Convergence, we can find an $R>0$ so that $ \int_{|x|>R}|f(x)|\mathrm{d}x<\frac{\epsilon}{4}\tag{2} $ Let $\delta=\frac{\epsilon}{4\pi R\|f\|_{L^1}}$. For $|x|\le R$ and $|\eta|<\delta$, $ \left|e^{-2\pi ix\cdot\eta}-1\right|\le\frac{\epsilon}{2\|f\|_{L^1}}\tag{3} $ whereas for all $x$, $ \left|e^{-2\pi ix\cdot\eta}-1\right|\le2\tag{4} $ Then, for $|\eta|<\delta$, $ \begin{align} |\hat{f}(\xi+\eta)-\hat{f}(\xi)| &\le\int_{\mathbb{R}^n}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\ &=\int_{|x|

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    @GuillermoMosse Just use the Taylor expansion of $e^{ix}$ and the triangle inequality.2018-09-18
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I don't know if your questions has been answered in full. For completeness, we apply DCT for the reasons you mentioned in your post. The punchline of the story is:

$\begin{align} \left|\widehat{f}(\xi + h) - \widehat{f}(\xi)\right| &= \left| \int f(x) \left(e^{-2 \pi i x \cdot (\xi + h)} - e^{-2 \pi i \xi \cdot x} \right)dx \right| \\ &\leq \int |f(x)| \left|e^{2 \pi i x \cdot h} - 1 \right| dx \end{align}$

which tends to zero as $h \to 0$, and this is enough to show uniform continuity.

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    Should we separate the integral domain by two parts like what @robjohn did? One is a neighborhood of the origin, and the other one is the rest of whole space.2018-09-17