Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

3 Answers 3

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 $g$.

share|improve this answer

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?

share|improve this answer
    
+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, $|f(3)-f(1)|\le\frac{2}{\pi}$.

share|improve this answer
    
How would you prove that |f(400,000)-f(1)|<1000? –  Did Aug 27 '11 at 0:21
    
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.