Schwartz function problem

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.

-
Hint: $\hat f$ is integrable. (And I get much less than 1000 in the RHS.) –  Did Aug 26 '11 at 20:57

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 $g$.

-

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?

-
+1 for the hint(s) rather than a complete solution. –  Did Aug 26 '11 at 21:07
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, $|f(3)-f(1)|\le\frac{2}{\pi}$.
@Didier: $\|f\|_{L^\infty}\le\frac{1}{2\pi}\|\hat{f}\|_{L^1}$ and $\|\hat{f}\|_{L^1}\le\frac{8}{3}$. Therefore, $|f(400000)-f(1)|\le\frac{8}{3\pi}$. –  robjohn Aug 27 '11 at 0:47
@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 $|f(x)-f(y)|\le\frac{8}{3\pi}$ –  robjohn Aug 27 '11 at 0:58