What is the idea to show this property of a function based on its Fourier transform?

Question

Let $f\in \mathcal{M}(\mathbb{R})$ be a function of moderate decrease such that the Fourier transform $\mathfrak{F}(f)$ is continuous and satisfies:

$$\mathfrak{F}(f)(\xi) = O\left(\dfrac{1}{|\xi|^{1+\alpha}}\right), \quad \text{as $|\xi|\to\infty$}$$

where $\alpha \in (0,1)$. I want to show that $f$ satisfies the Hölder condition of order $\alpha$:

$$|f(x+h)-f(x)|\leq M|h|^\alpha, \quad \text{with $M> 0$ and for all $x,h\in \mathbb{R}$}.$$

Now the condition on $\mathfrak{F}(f)$ says that it is continuous and furthermore, there's $K\in \mathbb{R}$ such that if $|\xi|> K$ we have

$$\mathfrak{F}(f)(\xi)\leq \dfrac{A}{|\xi|^{1+\alpha}}, \quad \text{for all $|\xi|> K$ and for some $A\in \mathbb{R}$}.$$

Now, I really have no idea on what to do here. The obvious initial step is to write:

$$|f(x+h)-f(x)| = \left|\int_{-\infty}^{\infty} \mathfrak{F}(f)(\xi)e^{2\pi i (x+h)\xi}d\xi-\int_{-\infty}^{\infty}\mathfrak{F}(f)(\xi)e^{2\pi i x\xi}d\xi\right|$$

So that we have

$$|f(x+h)-f(x)| \leq \int_{-\infty}^{\infty} |\mathfrak{F}(f)(\xi)| | e^{2\pi i h\xi}-1|d\xi.$$

Now one suggestion is to break this integral as follows:

$$|f(x+h)-f(x)|\leq \int_{|\xi|\leq 1/|h|} |\mathfrak{F}(f)(\xi)||e^{2\pi ih\xi -1}|d\xi + \int_{|\xi|\geq 1/|h|} |\mathfrak{F}(f)(\xi)||e^{2\pi ih\xi -1}|d\xi. $$

And now I'm completely stuck. First of all, I really have no idea or intuition whatsoever on what has to be done.

My only guess would be to use continuity on the compact $[-1/|h|,1/|h|]$ to bound $\mathfrak{F}(f)$ there, but this would not produce the desired $|h|^\alpha$. On the other integral I can't also use the other condition over $\mathfrak{F}(f)$ because the integral is for $|\xi| \geq 1/|h|$ not $|\xi| > K$.

Anyway, I'm quite lost here. So how can I show this? But much more importantly than how to prove this, how can I actually arrive at the proof? How can I have the idea on what has to be done? What is the right way to think about this problem, in order to solve it?

I'm much more interested here in how should I reason about this, and how can I have the idea of how to prove it.

Answer 1

First of all, note that because of the assumption on the decay of $\mathcal{F}(f)$ at infinity, there exists a constant $c>0$ such that

$$\mathcal{F}(f)(\xi) \leq \frac{c}{|\xi|^{1+\alpha}} \qquad \text{for all $|\xi| \geq 1$.} \tag{1}$$

(For $|\xi| \gg 1$, say $|\xi| \geq K$, this is a consequence of the decay and for $|\xi| \in [1,K]$ we can use that $|\xi|^{1+\alpha} \mathcal{F}(f)(\xi)$ is a continuous function, hence bounded on compacts.) Moreover, note that it suffices to show

$$|f(x+h)-f(x)| \leq M |h|^{\alpha} \qquad \text{for all $x \in \mathbb{R}$, $h \in (0,1)$} \tag{2}$$

(for $h \geq 1$ we can simply choose the constant $M$ sufficiently large, e.g. $M> 2 \|f\|_{\infty}$).

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As you already pointed out, we have

$$|f(x+h)-f(x)| \leq I_1+I_2$$

where

$$\begin{align*} I_1 &:= \int_{|\xi| \leq |h|^{-1}} |\mathcal{F}f(\xi)| \cdot |e^{2\pi i h \xi}-1| \, d\xi \\ I_2 &:= \int_{|\xi| > |h|^{-1}} |\mathcal{F}f(\xi)| \cdot |e^{2\pi i h \xi}-1| \, d\xi. \end{align*}$$

We estimate both terms separately. We are going to show that there exist constants $C_1,C_2>0$ such that

$$I_1 \leq C_1 (h+h^{\alpha}) \qquad \text{and} \qquad I_2 \leq C_2 h^{\alpha} \tag{$\star$}$$

for all $h \in (0,1)$. First, we estimate $I_2$. Recall that

$$|e^{ix}-1| \leq 2 \min\{|x|,1\} = \begin{cases} 2|x|, & |x| \leq 1, \\ 2, & |x|>1 \end{cases} \tag{3}$$

Now, since $|2\pi h \xi|>1$ for any $|\xi|>h^{-1}$, we use the estimate $|e^{2\pi i \xi h}-1| \leq 2$ for $I_2$. Consequently, we get

$$I_2 \leq 2 \int_{|\xi|>h^{-1}} |\mathcal{F}f(\xi)| \, d\xi.$$

For any $h \in (0,1)$ it now follows from $(1)$ that

$$I_2 \leq 2c \int_{|\xi|>h^{-1}} \frac{1}{|\xi|^{1+\alpha}} \, d\xi = \frac{2c}{\alpha} h^{\alpha}. \tag{4}$$

It remains to estimate $I_1$. We write $$I_1 = I_{11} + I_{12}$$ where

$$\begin{align*} I_{11} &:= \int_{|\xi| \leq 1} |\mathcal{F}f(\xi)| \cdot |e^{2\pi i h \xi}-1| \, d\xi \\ I_{12} &:= \int_{1<|\xi| \leq |h|^{-1}} |\mathcal{F}f(\xi)| \cdot |e^{2\pi i h \xi}-1| \, d\xi. \end{align*}$$

Using $(3)$ and the boundedness of $\mathcal{F}f$, we get

$$I_{11} \leq \|\mathcal{F}f\|_{\infty} \int_{|\xi| \leq 1} |2\pi \xi h| \,d\xi \leq 2\pi \|\mathcal{F}f\|_{\infty} |h|. \tag{5}$$

On the other hand, $(1)$ and $(3)$ give

$$\begin{align*} I_{12} \leq \int_{1 < |\xi| < h^{-1}} |\xi|^{-1-\alpha} |2\pi \xi h| \, d\xi &= 2 \pi h \int_{1<|\xi|<h^{-1}} |\xi|^{-\alpha} \, d\xi \\ &\leq 2\pi h \frac{h^{\alpha-1}}{1-\alpha}. \tag{6} \end{align*}$$

Combining $(4)$, $(5)$ and $(6)$, we get $(\star)$. Consequently, there exists a constant $C=C(f)$ such that

$$|f(x+h)-f(x)| \leq C (h+h^{\alpha})$$

for all $x \in \mathbb{R}$ and $h \in (0,1)$. As $\alpha \in (0,1)$ and $h \in (0,1)$ this implies

$$|f(x+h)-f(x)| \leq 2C h^{\alpha},\qquad x \in \mathbb{R}, h \in (0,1)$$

and this shows $(2)$.

Remark: You asked how to come up with the correct way to split the integrals. Because of $(1)$, it makes sense to consider the domains of integration

$$\{\xi; |\xi| \leq 1\} \qquad \text{and} \qquad \{\xi; |\xi|>1\}$$

separately (for the set on the right-hand side we have our nice decay $(1)$; for the set on left-hand side we can use boundedness of $\mathcal{F}f$ and there is a chance that this gives a nice upper bound because this set is compact.) On the other hand, estimate $(3)$ suggests to consider the domains of integration

$$\{\xi; |\xi h| \leq 1\} \qquad \text{and} \qquad \{\xi; |\xi h| > 1\}$$

separately (for the set on the left-hand side the estimate $|e^{i \xi h}-1| \leq |\xi h|$ is good (because $|\xi h|$ is small) whereas $|e^{ih \xi-1}| \leq 2$ is a useful estimate if $|\xi h|$ is big.)

Answer 2

There is $|\xi_o|$ (which we may assume without loss of generality to be greater than $1$) such that

$$|\mathcal{F}(f)(\xi)| \leq \frac{M}{|\xi|^{1+\alpha}}\leq \frac{2M}{1+|\xi|^{1+\alpha}} \qquad \text{for all $|\xi| \geq |\xi_o|$.} \tag{1}$$

On the other hand, for $0 \leq |\xi| \leq |\xi_o|$, we definetely have

$$|\mathcal{F}(f)(\xi)| \leq \sup_{1 \leq \xi\leq |\xi_o|}\mathcal{F}(f)(\xi)\leq \sup_{1 \leq \xi\leq |\xi_o|}\mathcal{F}(f)(\xi)\frac{1+|\xi_o|^{1+\alpha}}{1+|\xi|^{1+\alpha}} \tag{2}$$

Combininig ${(1)}$ and ${(2)}$ with $M'=2M+(1+|\xi_o|^{1+\alpha})\sup_{1\leq \xi\leq |\xi_o|}\mathcal{F}(f)(\xi)$:

$$|\mathcal{F}(f)(\xi)| \leq \frac{M'}{1+|\xi|^{1+\alpha}} \qquad \text{for all $\xi\in \mathbb{R}$}$$

We note that an explicit calculation yields $|e^{ia}-1|=2|\sin(a/2)|$. This is very useful. Indeed:

$$\left|\int_{-\infty}^\infty \mathcal{F}(f)(y)e^{2\pi i yx}(e^{2\pi i yh}-1)dy \right|\leq \int_{-\infty}^\infty \frac{2|\sin(\pi yh)|M'}{1+|y|^{1+\alpha}}dy=$$ $$4\int_0^\infty \frac{M'\sin(\pi |y h|)}{1+|y|^{1+\alpha}}dy=4M' \left(\int_0^{1/|h|} \frac{\sin(\pi |y h|)}{1+y^{1+\alpha}}dy+\int_{1/|h|}^\infty \frac{\sin(\pi |y h|)}{1+y^{1+\alpha}}dy \right)\leq 4M'\left(\int_0^{1/|h|} \frac{\sin(\pi y|h|)}{1+y^{1+\alpha}}dy+\int_{1/|h|}^\infty \frac{1}{y^{1+\alpha}}dy\right)=$$ $$ 4M'\left(\int_0^{1/|h|} \frac{\sin(\pi y|h|)}{1+y^{1+\alpha}}dy+\frac{|h|^\alpha}{\alpha}\right)\tag{3}$$

Case I: $|h|>1$:

$$\int_0^{1/|h|} \frac{\sin(\pi y|h|)}{1+y^{1+\alpha}}dy\leq \int_0^{1/|h|} \frac{\pi y|h|}{1+y^{1+\alpha}}dy\leq \int_0^{1/|h|} {\pi y|h|}= \frac{\pi}{2|h|}\leq \frac{\pi}{2}|h|^\alpha$$

Case 2: $|h|<1$:

$$\int_0^{1/|h|} \frac{\sin(\pi y|h|)}{1+y^{1+\alpha}}dy\leq \int_0^{1} \sin(\pi y |h|)dy+\int_1^{1/|h|} \frac{\pi y|h|}{1+y^{1+\alpha}}dy\leq$$ $$\frac{\cos(\pi|h|)-\cos(0)}{\pi |h|}+\pi|h|\int_1^{1/|h|}\frac{y}{y^{1+\alpha}}dy=\sin(\pi|h'|)+\frac{|h|^{2-\alpha}}{1-\alpha}-\frac{|h|}{1-\alpha}$$ Because $\alpha<2-\alpha$ and $|h'|<|h|<1$ we have:

$$\int_0^{1/|h|} \frac{\sin(\pi y|h|)}{1+y^{1+\alpha}}dy\leq\pi |h|+\frac{|h|^\alpha}{1-\alpha}\leq|h|^\alpha\left(\pi +\frac{1}{1-\alpha} \right)$$

Combining our bounds we have that in either case:

$$|f(x+h)-f(x)|=\left|\int_{-\infty}^\infty \mathcal{F}(f)(y)e^{2\pi i yx}(e^{2\pi i yh}-1)dy \right|\leq 4M'\left(\pi +\frac{1}{1-\alpha} \right)|h|^\alpha $$