The Hat of the Hat

Question

From a textbook on Harmonic Analysis:

(The Hat of the Hat). Let $f$ be the hat function, defined by $f(x) = 1 - |x|$ for $|x| ≤ 1$ and $f(x) = 0$ otherwise. Show that $f$ is a continuous function of moderate decrease but that its Fourier transform $\widehat{f}$ is not.

Attempt. It is clear that $f$ is a continuous function of moderate decrease since $f \in \mathscr{S}(\mathbb{R})$ (i.e., the Schwartz class) implies that $f$ is of moderate decrease.

But how to show that $\widehat{f}$ fails to be continuous of moderate decrease? Again, it must be that $\widehat{f}$ fails to be continuous, since also $f \in \mathscr{S}(\mathbb{R})$ implies that $\widehat{f} \in \mathscr{S}(\mathbb{R})$ so that $\widehat{f}$ is of moderate decrease. So how to show that $\widehat{f}$ fails to be continuous?

EDIT: My book's definition of a function of moderate decrease is as follows:

Answer 1

The question is asking you to prove an incorrect result. It asked you to show $\hat f$ is not "continuous and of moderate decrease". To do this you would have to show either $\hat f$ is not continuous, or $\hat f$ is not of moderate decrease - one of the two. But in fact both properties hold, as I'll show below.

First, $\hat f$ must be continuous: We can give a homemade argument for this:

$$\tag 1\hat f(y) = \int_{-1}^1(1-|x|)e^{-i2\pi yx}\,dx = 2\int_{0}^1(1-x)\cos (2\pi yx)\,dx.$$

The evenness of $f$ was used to obtain the last expression. Now $|\cos b- \cos a|\le |b-a|$ by the mean value theorem. Thus

$$\tag 2|\hat f(y+h) - \hat f(y) | = |2\int_{0}^1(1-x)[\cos (2\pi(y+h)x)-\cos (2\pi yx)]\,dx| $$ $$\le 2\int_{0}^1(1-x)|\cos (2\pi (y+h)x)-\cos (2\pi yx)|\,dx \le 2\int_{0}^1(1-x)|2\pi hx|\,dx.$$

The last expression equals $4\pi |h|\int_{0}^1(1-x)x\,dx.$ This shows the left side of $(2)\to 0$ as $h\to 0,$ giving the desired continuity of $\hat f.$

Second, $\hat f$ is of moderate decrease. To see this, we just evaluate the integral on the right side of $(1).$ I like to change variables first, with $x= t/(2\pi y).$ This gives, for $y\ne 0,$

$$\frac{1}{2\pi y}\int_0^{2\pi y} (1-t/(2\pi y))\cos t\, dt= \frac{1}{2\pi y}\int_0^{2\pi y} \cos t\, dt - \frac{1}{4\pi^2 y^2}\int_0^{2\pi y} t\cos t\, dt.$$

The last two integrals are simple (use integration by parts on the second one). I'll leave this part to you. Remembering to multipy by 2, I'm getting

$$ \hat f (y)=\frac{1-\cos (2\pi y)}{4\pi^2 y^2}.$$

Thus $\hat f$ is of moderate decrease as desired.

Answer 2

First, to simplify notation a bit and allow myself to keep using $f$ as a generic function, I am going to let $H$ denote the hat function, i.e. $$H(x) := \begin{cases} 1-|x| & \text{if $-1\le x \le 1$, and} \\ 0 & \text{otherwise.} \\ \end{cases} $$ I have not seen the phrase "a function of moderate decrease" before, but a quick Google indicates that the definition is something like the following:

Definition: A function of moderate decrease is a function $f$ satisfying the property that there is some constant $C > 0$ such that $$ |f(x)| \le \frac{C}{1+|x|^{2}}$$ for all $x\in \mathbb{R}$.

If this is the correct definition, it is not too difficult to see that $H$ satisfies this property with $C=1$. Hence $H$ is a continuous function of moderate decrease. More generally, any compactly supported continuous function will be of moderate decrease—we just have to choose $C$ large enough so that ${C}(1+|x|^2)^{-1}$ is larger than $\max |f(x)|$ for all $x$ in the support of $f$.

Do note, however, that your argument is fallacious, in that $H$ is not a Schwartz function, as it fails to be smooth (indeed, it is not even continuously differentiable). If it were Schwartz, you would be immediately done (and the exercise would be very clearly misstated), as Schwartz functions are of rapid decay (even better than moderate decrease) and the Fourier transform of a Schwartz function is also a Schwartz function.

Next, it is a basic result that $\mathscr{F} : L^1(\mathbb{R}) \to C_0(\mathbb{R})$. That is, the Fourier transform maps integrable functions to continuous functions that go to zero in the limit. This implies that $\hat{H}$ is continuous. Indeed, it appears that this exercise comes from the book Harmonic Analysis: From Fourier to wavelets by Pereyra and Ward. This exercise is on page 215, just below a table where several major result are summarized, including $\mathscr{F}:L^1 \to C_0$ (this is the Riemann-Lebesgue lemma—the uniform norm of $\hat{f}$ is bounded by the $L^1$ norm of $f$).

It remains to determine whether or not $\hat{H}$ is a function of moderate decrease. Assuming that I have done my arithmetic correctly, we get $$ \hat{H}(\xi) = \int_{\mathbb{R}} H(x) \mathrm{e}^{2\pi ix\xi} \,\mathrm{d}x = \left( \frac{\sin(\pi\xi)}{\pi\xi} \right)^2. $$ Exercise: Make sure this is correct.

Alternatively, as noted by FTC, if we are clever we might notice that $H = \chi_{[0,1]}\ast\chi_{[0,1]}$, i.e. that the hat function is the two-fold convolution of the characteristic function of the unit interval. But then $$ \hat{H}(\xi) = \widehat{\chi_{[0,1]} \ast\chi_{[0,1]}}(\xi) = \hat{\chi}_{[0,1]} \cdot \hat{\chi}_{[0,1]} (\xi) = \operatorname{sinc}(\pi\xi)^2, $$ where $\operatorname{sinc}$ is defined by $$ \operatorname{sinc}(\xi) := \begin{cases} \frac{\sin(\pi\xi)}{\pi \xi} & \text{if $\xi \ne 0$, and} \\ 1 & \text{$\xi = 0$.} \end{cases} $$ Note that this is continuous. Indeed, if we were worried about hitting the previous part of the problem about the head and shoulders with the big hammer of Riemann-Lebesgue lemma, this can instead be used to justify the statement that $\hat{H}$ is continuous.

But then we can bound $|\hat{H}(\xi)|$ by a function of the form $C(1+|\xi|^2)^{-1}$ for an appropriately chosen $C$: for $\xi$ large enough (say $|\xi|>1$), we have the rather sloppy estimate $$ |\hat{H}(\xi)| \le \frac{1}{\pi^2} \frac{1}{|\xi|^2} = \frac{2}{\pi^2} \frac{1}{|\xi|^2 + |\xi|^2} \le \frac{C_1}{1+|\xi|^2}, $$ where $C_1 := \frac{2}{\pi^2}$. On the other hand, if $|\xi| \le 1$, then $$ |\hat{H}(\xi)| \le 1 \le \frac{C_2}{1+|\xi|^2}, $$ where we just need to choose $C_2$ large enough for the inequality to hold ($C_2 = 2$, for example, gets the job done). Taking $C = \max\{C_1, C_2\}$, we have $$ |\hat{H}(\xi)| \le \frac{C}{1+|\xi|^2} $$ for all $\xi\in\mathbb{R}$, which seems to show that $\hat{H}$ is a function of moderate decrease.