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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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  • $\begingroup$ Hint: $\hat f$ is integrable. (And I get much less than 1000 in the RHS.) $\endgroup$ – Did Aug 26 '11 at 20:57
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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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  • $\begingroup$ How would you prove that |f(400,000)-f(1)|<1000? $\endgroup$ – Did Aug 27 '11 at 0:21
  • $\begingroup$ @Didier: $\|f\|_{L^\infty}\le\frac{1}{2\pi}\|\hat{f}\|_{L^1}$ and $\|\hat{f}\|_{L^1}\le\frac{8}{3}$. Therefore, $|f(400000)-f(1)|\le\frac{8}{3\pi}$. $\endgroup$ – robjohn Aug 27 '11 at 0:47
  • $\begingroup$ @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 $|f(x)-f(y)|\le\frac{8}{3\pi}$ $\endgroup$ – robjohn Aug 27 '11 at 0:58

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