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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a Schwartz function. Suppose that $\left|\hat{f}(\omega)\right|\leq1$, $\left|\hat{f}(\omega)\right|\leq\left|\omega\right|^{-4}$. Show that: $ \left|f(3)-f(1)\right|<1000$

One thing is that Schwartz functions are invariant under the Fourier transform $\mathcal{F}$. Essentially, we can write $f(x)$ as: $ f(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{ixt}\hat{f}(t)dt$

I am having some difficulties dealing with the estimates after that.

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    Hint: $\hat f$ is integrable. (And I get much less than 1000 in the RHS.)2011-08-26

3 Answers 3

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Hint: $f(3) - f(1) = \int_{\mathbb R} f(x) g(x) \ dx$ for a certain tempered distribution $g$. Express that in terms of the Fourier transforms of $f$ and $g$.

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Hint: Split the interval of integration, and then use the above bounds. We can actually bound $f(3)$ and $f(1)$ individually so well that the triangle inequality gives us the result. Since $f(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{ixt}\hat{f}(t)dt$ we see that for any $x$ $ |f(x)|\leq \frac{1}{2\pi}\int_{\mathbb{R}}|\hat{f}(t)|dt\leq \frac{1}{2\pi}\int_{|t|\geq 1}|t|^{-4}dt+\frac{1}{2\pi}\int_{|t|\leq 1}1dt.$

(Notice I am using the bound $|\hat{f}(t)|\leq t^{-4}$ for one part and $|\hat{f}(t)|\leq 1$ for the other)

Can you finish it from?

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Using the normalization of the Fourier Transform shown above, we get f'(x)=\frac{1}{2\pi}\int_{\mathbb{R}}it\;e^{ixt}\hat{f}(t)dt Therefore, \|f'\|_{L^\infty}\le\frac{1}{2\pi}\|\omega\hat{f}\|_{L^1}\tag{1} Using the estimates on $\hat{f}$ above, we get $ \begin{align} \|\omega\hat{f}\|_{L^1}&\le\int_{-\infty}^{-1}|\omega|^{-3}\mathrm{d}\omega+\int_{-1}^{1}\;|\omega|\;\mathrm{d}\omega+\int_{1}^{\infty}|\omega|^{-3}\mathrm{d}\omega\\ &=\frac{1}{2}+1+\frac{1}{2}\\ &=2\tag{2} \end{align} $ Putting together $(1)$ and $(2)$, we get \|f'\|_{L^\infty}\le\frac{1}{\pi}\tag{3} Thus, by the Mean Value Theorem, $|f(3)-f(1)|\le\frac{2}{\pi}$.

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    @Didier: Eric Naslund had already showed how to bound $\|f\|_{L^\infty}$, so I thought I would show how to bound $\|f'\|_{L^\infty}$ and use that. In different estimates, it might be useful to know that $|f(x)-f(y)|\le|x-y|/\pi$ as well as $|f(x)-f(y)|\le\frac{8}{3\pi}$2011-08-27