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.
 A: 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 $g$.
A: 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?
A: 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, $|f(3)-f(1)|\le\frac{2}{\pi}$.
