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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.

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  • $\begingroup$ Have you tried approximating $f$ by a compactly supported smooth (or continuous) function, that is, take $\varphi\in C_c^{\infty}(\mathrm{R}^n)$ with $||f-\varphi||_{L^1}<\frac{\epsilon}{2}$ or something alike? I think you could make the argument work this way... $\endgroup$ Sep 30 '11 at 1:31
  • $\begingroup$ Thanks for the hint. But could you please elaborate a little bit more. Why do we need to approximate it by a compactly supported continuous function? $\endgroup$ Sep 30 '11 at 2:05
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    $\begingroup$ I haven't done the calculation, but it's a very common trick, and it might work here. You'll have $$ | \hat{f}(a+h)-\hat{f}(a) | \leq 2 ||f-\varphi||_{L^1} + | \int \varphi(x)(e^{2i\pi x\cdot (a+h)}-e^{2i\pi x\cdot a})dx|$$ in other words, $$|\hat{f}(a+h)-\hat{f}(a)|\leq 2||f-\varphi||_{L^1}+ |\hat{\varphi}(a+h)-\hat{\varphi}(a)|$$ and you only have to show the property for smooth compactly supported functions. Finally, you can say something like $$|\hat{\varphi}(a+h)-\hat{\varphi}(a)|\leq\int_{|x|\leq R}||\varphi||_{L^{\infty}}|e^{2i\pi x\cdot h}-1|dx$$ and I think that should almost do it. $\endgroup$ Sep 30 '11 at 2:29
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    $\begingroup$ Alternatively, you can remember the Riemann-Lebesgue Lemma, that tells you that $\hat{f}$ is continuous and tends to $0$ as $|x|$ tends to $\infty$, and such functions are always uniformly continuous (easy exercise). $\endgroup$ Sep 30 '11 at 2:35
  • $\begingroup$ Thanks a lot for your kind explanations. $\endgroup$ Sep 30 '11 at 3:27
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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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  • $\begingroup$ Thanks a lot for your answer. $\endgroup$ Oct 11 '11 at 2:16
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    $\begingroup$ 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. $\endgroup$
    – Sam Wong
    Sep 17 '18 at 23:59
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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 $|\eta|<\delta$, $$ \begin{align} |\hat{f}(\xi+\eta)-\hat{f}(\xi)| &\le\int_{\mathbb{R}^n}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\ &=\int_{|x|<R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x +\int_{|x|\ge R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\ &\le\|f\|_{L^1}\frac{\epsilon}{2\|f\|_{L^1}}+\;2\frac{\epsilon}{4}\\ &=\epsilon \end{align} $$

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  • $\begingroup$ Thanks a lot for providing the details. $\endgroup$ Oct 11 '11 at 2:16
  • $\begingroup$ How do we justify that for all $|x| \leq R$, we have $\left| e^{-2\pi i x. \eta}-1 \right| \leq \frac{\epsilon}{2\|f\|_{L^1}}$? $\endgroup$ Feb 24 '16 at 15:43
  • $\begingroup$ @pikachuchameleon: by our choice of $\delta$ and the fact that $2\pi| x||\eta|\le\frac{\epsilon}{2\|f\|_{L^1}}$ $\endgroup$
    – robjohn
    Feb 24 '16 at 16:23
  • $\begingroup$ @robjohn, I'm not sure how you can conclude what pikachuchameleon said from that. Don't you need a smaller $\delta$ for that bound? $\endgroup$ May 14 '17 at 15:53
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    $\begingroup$ @GuillermoMosse Just use the Taylor expansion of $e^{ix}$ and the triangle inequality. $\endgroup$
    – Sam Wong
    Sep 18 '18 at 0:07
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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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    $\begingroup$ 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. $\endgroup$ Oct 19 '12 at 22:29
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Observe that for real $x,y \in \mathbb R$, $$|e^{ix}-e^{iy}|\leq |x-y| \wedge 2.$$ Then $$|\hat f (\xi +h )-\hat f (\xi)| \leq \int |f(x)| (|2\pi x h|\wedge 2) \mathrm dx, $$ so we have a bound not depending on $\xi$, which goes to zero as $|h| \to \infty$ by the dominated convergence theorem.

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